MATH 121 - COLLEGE ALGEBRA -
Equations and properties of conic
sections
Question Bank - Set 9
Liberty University
Question 1
Question
Solve the following system of equations:
{x2+ 4y2= 16
x2−y2= 9
Solution
Step 1: Let’s solve the second equation for y2:
x2−y2= 9
y2=x2−9
Step 2: Substitute y2in terms of xinto the first equation:
x2+ 4(x2−9) = 16
Step 3: Simplify and solve for x:
x2+ 4x2−36 = 16
5x2= 52
x2=52
5
x=±√52
5
Step 4: Now, substitute the values of xback into the equation y2=x2−9
to find y: For x=√52
5:
y2=(√52
5)2
−9
y2=52
5−9
y2=52
5−45
5
y2=7
5
y=±√7
5
For x=−√52
5:
y2=(−√52
5)2
−9
y2=52
5−9
y2=52
5−45
5
y2=7
5
y=±√7
5
Therefore, the solutions to the system of equations are:
(√52
5,√7
5)and (−√52
5,−√7
5)
Question 2
Question
Find the standard form of the equation of the ellipse that satisfies the given
conditions: Foci at (−2,0) and (2,0), and minor axis of length 6.
2
Solution
Step 1: The distance between the foci is equal to the length of the major axis.
Let’s denote the distance between the foci as 2c. Step 2: Since the foci are at
(−2,0) and (2,0), we have 2c= 2 ⇒c= 1. Step 3: The length of the minor
axis is the same as the length of the major axis. Let’s denote the length of the
minor axis as 2b. Step 4: Given that the minor axis has a length of 6, we have
2b= 6 ⇒b= 3. Step 5: The standard form of the equation of an ellipse is
(x−h)2
a2+(y−k)2
b2= 1, where (h, k)is the center of the ellipse, ais the length of
the major axis, and bis the length of the minor axis. Step 6: Since the center
of the ellipse is the midpoint between the foci, the center is at the origin (0,0).
Step 7: We have a2=c2+b2= 12+ 32= 1 + 9 = 10. Step 8: Therefore, the
standard form of the equation of the ellipse is x2
10 +y2
9= 1 .
Question 3
Question
Let f(x) = x3−4x2−5x+ 18. Find all real zeros of the function.
Solution
Step 1: To find the zeros of the function f(x), we set f(x)=0and solve for
x. Step 2: So, we have the equation x3−4x2−5x+ 18 = 0. Step 3: Since
this is a cubic equation, it is difficult to find the zeros directly. Let’s try to use
synthetic division to determine if there are any rational roots. Step 4: By trying
potential factors of the constant term (in this case, 18) divided by the factors
of the leading coefficient (in this case, 1), we can find that x= 2 is a root of
the equation. Step 5: Perform synthetic division with (x−2) as the divisor to
factor out the root. 2 1 −4−5
18
2−4
−18
1−2−9
0
Step 6: The result of the synthetic division is a quadratic equation x2−2x−
9=0. We can solve this equation using the quadratic formula. Step 7: The
quadratic formula states that x=−(−2)±√(−2)2−4∗1∗(−9)
2∗1. Step 8: Simplifying,
we get x=2±√4+36
2, which results in x=2±√40
2. Step 9: This gives us two
potential solutions: x= 1 + √10 and x= 1 −√10. Step 10: Therefore, the real
zeros of the function f(x) = x3−4x2−5x+ 18 are x= 2,1 + √10,1−√10.
3
Question 4
Question
Find the standard form of the equation of the hyperbola with vertices at (−3,2)
and (3,2) and foci at (−5,2) and (5,2).
Solution
Step 1: Identify the center of the hyperbola. The center of the hyperbola is the
midpoint of the segment connecting the vertices, which is given by
(−3+3
2,2+2
2) = (0,2).
Step 2: Find the distance from the center to either of the vertices. Since the
vertices are to the left and right of the center, the distance from the center to a
vertex is the horizontal distance. Thus, the distance is 3units.
Step 3: Find the value of a. The value of ais the distance from the center
to a vertex, which is a= 3.
Step 4: Find the value of c. The value of cis the distance from the center
to a focus, so c= 5.
Step 5: Find the value of b. The relationship between a,b, and cin a
hyperbola is given by the equation c2=a2+b2. Substituting the known values,
we have 52= 32+b2, which simplifies to 25 = 9 + b2. Solving for b, we get
b2= 16, so b= 4.
Step 6: Determine the equation of the hyperbola. The standard form of the
equation of a hyperbola centered at (h, k)with vertices on the transverse axis
is (x−h)2
a2−(y−k)2
b2= 1. Substituting our values, we have
(x−0)2
32−(y−2)2
42= 1.
Simplifying, we get
x2
9−(y−2)2
16 = 1.
Therefore, the standard form of the equation of the hyperbola is
x2
9−(y−2)2
16 = 1 .
Question 5
Question
Find the standard form of the equation of the hyperbola that satisfies the given
conditions: Vertices at (−2,1) and (2,1) and foci at (−4,1) and (4,1).
4
Solution
Step 1: The center of the hyperbola lies at the midpoint of the vertices, which
can be calculated as follows:
(h, k) = (−2+2
2,1+1
2)= (0,1).
Step 2: The distance from the center to a vertex is the distance between two
x-coordinates, which is the value of a. In this case, we have |a|= 2.
Step 3: The distance from the center to a focus is the value of c, which can
be calculated using the distance formula:
c=√(h−f)2+ (k−k)2=√(0 −4)2+ (1 −1)2= 4.
Step 4: The equation of a hyperbola centered at (h, k)with vertices along
the x-axis is given by (x−h)2
a2−(y−k)2
b2= 1, where b2=a2−c2.
Step 5: Substituting the given values of aand cinto the formula for b2:
b2= 22−42= 4 −16 = −12.
Step 6: The equation of the hyperbola can now be written as:
x2
4−y2
−12 = 1.
Step 7: To simplify and rewrite the equation in standard form, multiply
through by −1to eliminate the negative denominator:
−x2
4+y2
12 = 1.
Step 8: Therefore, the standard form of the equation of the hyperbola is
−x2
4+y2
12 = 1 .
Question 6
Question
Find the equation of the hyperbola with vertices at (−3,2) and (3,2), and foci
at (−5,2) and (5,2).
Solution
Step 1: Find the center of the hyperbola. The center of the hyperbola is the
midpoint of the vertices:
Center =(−3+3
2,2+2
2)= (0,2)
5
Step 2: Find the distance between the vertices to get the length of the
transverse axis.
2a= 3 −(−3) = 6 ⇒a= 3
Step 3: Find the distance between the foci to get the constant c.
2c= 5 −(−5) = 10 ⇒c= 5
Step 4: Calculate busing the relationship c2=a2+b2.
b2=c2−a2= 25 −9 = 16 ⇒b= 4
Step 5: Write the equation of the hyperbola in standard form. Since the
hyperbola opens horizontally, the standard form is:
(x−h)2
a2−(y−k)2
b2= 1
where (h, k)is the center of the hyperbola (0,2). So, the equation of the hyper-
bola is: x2
9−(y−2)2
16 = 1
Question 7
Question
Find the standard form equation of the hyperbola that satisfies the given con-
ditions: Center at (−2,3), horizontal transverse axis of length 10, and a vertex
at (−7,3).
Solution
Step 1: Determine the equation for a hyperbola with center at (−2,3) and hori-
zontal transverse axis of length 10. The standard form equation for a hyperbola
with center (h, k), horizontal transverse axis length 2a, and vertical transverse
axis length 2bis:
(x−h)2/a2−(y−k)2/b2= 1
Given that the center is (−2,3) and the horizontal transverse axis length is
10, the equation becomes:
(x+ 2)2/25 −(y−3)2/b2= 1
Step 2: Find the value of busing the information that the vertex is (−7,3).
The distance from the center to the vertex is a= 5, so the distance from the
center to the midpoint between the vertices is c=a= 5. Since the vertex is
located 5 units to the left of the center, we have a2+b2=c2:
52+b2= 52
6
25 + b2= 25
b2= 0
b= 0
Step 3: Substitute the value of binto the equation to get the final equation
of the hyperbola. Substitute b= 0 into the equation:
(x+ 2)2/25 −(y−3)2/0 = 1
This simplifies to:
(x+ 2)2/25 = 1
Therefore, the standard form equation of the hyperbola is (x+ 2)2/25 = 1 .
Question 8
Question
Solve for yin the equation of the parabola x= 2y2−4y+ 3.
Solution
Step 1: Rearrange the equation by completing the square.
x+ 1 = 2(y2−2y)
Step 2: Complete the square inside the parenthesis.
x+ 1 = 2(y2−2y+ 1 −1)
Step 3: Simplify the right side of the equation.
x+ 1 = 2((y−1)2−1)
Step 4: Distribute the 2 on the right side.
x+ 1 = 2(y−1)2−2
Step 5: Add 2 to both sides to isolate the squared term.
x+ 3 = 2(y−1)2
Step 6: Divide both sides by 2.
x+ 3
2= (y−1)2
Step 7: Take the square root of both sides.
±√x+ 3
2=y−1
7
Step 8: Add 1 to both sides to solve for y.
y= 1 ±√x+ 3
2
Therefore, the solutions for yin the equation of the parabola x= 2y2−4y+3
are y= 1 + √x+3
2and y= 1 −√x+3
2.
Question 9
Question
Find the standard form equation of the hyperbola with foci at (-3, 0) and (3,
0) and vertices at (-5, 0) and (5, 0).
Solution
Step 1: Find the center of the hyperbola.
The center of the hyperbola is the midpoint of the segment connecting the
two vertices. The x-coordinate of the center is the average of the x-coordinates
of the vertices, and the y-coordinate is 0 since the hyperbola is centered on the
x-axis.
The center is ((−5 + 5)/2,0) = (0,0).
Step 2: Find the distance between the center and one of the foci.
The distance between the center and one of the foci is the distance from the
center to either of the foci. Since the foci are on the x-axis, the distance is the
absolute value of the difference between the x-coordinate of the center and the
x-coordinate of either focus.
The distance is |0−(−3)|= 3.
Step 3: Find the value of c, the distance from the center to either focus.
In the standard form of the equation of a hyperbola, cis related to the dis-
tances between the center and the foci and between the center and the vertices.
We know that for a hyperbola, c2=a2+b2, where ais half the distance between
the vertices and bis unknown.
Since a= 5, we can calculate b:
c=√a2+b2
3 = √52+b2
3 = √25 + b2
32= 25 + b2
9 = 25 + b2
b2= 9 −25
8
b2=−16
Since bmust be a positive value, we made a mistake in our calculation. This
indicates that there was an error in our initial assumption. Let’s revisit our
steps to determine the correct b value.
Step 4: Identify and correct the error in the previous steps.
When we calculated the distance between the center and one of the foci,
we mistakenly treated it as the distance c. However, we should have used the
formula for the equation of a hyperbola that involves c, a, and b. Let’s correct
this error and go back to Step 3.
Step 5: Correct the calculation of bin Step 3.
We know that for a hyperbola, c2=a2+b2, where ais half the distance
between the vertices and bis unknown. Since c= 3,a= 5, we can calculate b:
c2=a2+b2
32= 52+b2
9 = 25 + b2
b2= 9 −25
b2=−16
Since bmust be a positive value, b= 4.
Step 6: Write the equation of the hyperbola in standard form.
The standard form of the equation of a hyperbola centered at the origin with
foci on the x-axis is: x2
a2−y2
b2= 1
Substitute the values of a= 5 and b= 4 into the equation:
x2
52−y2
42= 1
x2
25 −y2
16 = 1
Therefore, the standard form equation of the hyperbola is x2
25 −y2
16 = 1.
Question 10
Question
Find the standard form of the equation of the parabola with a focus at (−2,4)
and a directrix y=−6.
9
Solution
Step 1: First, determine whether the parabola opens vertically or horizontally.
The parabola opens towards the directrix and away from the focus. Since the
focus is above the directrix, the parabola must open upwards.
Step 2: The distance between the vertex and the focus is equal to the dis-
tance between the vertex and the directrix. In this case, the vertex lies halfway
between the focus and the directrix. Therefore, the vertex is at (−2,4+(−6)
2) =
(−2,−1).
Step 3: The equation of a parabola that opens upwards with vertex at (h, k)
is given by (x−h)2= 4p(y−k), where pis the distance between the vertex and
the focus (or the vertex and the directrix in absolute value).
Step 4: Since the distance between the vertex and the focus is 5 units (from
(−2,−1) to (−2,4)), we have |p|= 5.
Step 5: Substituting the vertex (h, k) = (−2,−1) and p= 5 into the equation
gives (x+ 2)2= 4 ·5(y+ 1).
Step 6: Simplifying the equation gives (x+ 2)2= 20(y+ 1).
Therefore, the standard form of the equation of the parabola is (x+ 2)2=
20(y+ 1).
Question 11
Question
Determine the standard form equation of the hyperbola with vertices at (−5,1)
and (3,1) and foci at (−7,1) and (5,1).
Solution
Step 1: Find the center of the hyperbola. The center of the hyperbola is the
midpoint of the line segment connecting the vertices. Use the midpoint formula
(x1+x2
2,y1+y2
2).
Midpoint =(−5+3
2,1+1
2)= (−1,1)
Step 2: Determine the distance between the center and one of the vertices
to find a(the distance from the center to a vertex). Since the vertices are along
the x-axis, the distance is the difference in x-coordinates.
a= 3 −(−1) = 4
Step 3: Determine the distance between the center and one of the foci to
find c(the distance from the center to a focus).
c= 5 −(−1) = 6
10
Step 4: Use the relationship for a hyperbola to find b(the distance from the
center to the vertices along the y-axis).
c2=a2+b2⇒62= 42+b2⇒36 = 16 + b2
b2= 36 −16 = 20 ⇒b=√20 = 2√5
Step 5: Write the standard form equation of a hyperbola.
(x−h)2
a2−(y−k)2
b2= 1
Where the center of the hyperbola is (h, k).
Plugging in the values:
(x+ 1)2
42−(y−1)2
(2√5)2= 1
Therefore, the equation of the hyperbola is:
(x+ 1)2
16 −(y−1)2
20 = 1
Question 12
Question
Determine the standard form of the equation of the hyperbola with vertices at
(−1,3) and (−1,−3) and foci at (−1,5) and (−1,−5).
Solution
Step 1: Find the center of the hyperbola by taking the average of the vertices.
Center =(−1+(−1)
2,3+(−3)
2)= (−1,0)
Step 2: Determine the distance between the center and one of the vertices
to find a(the distance from the center to the vertices).
a= 3 −0 = 3
Step 3: Determine the distance between the center and one of the foci to
find c(the distance from the center to the foci).
c= 5 −0 = 5
Step 4: Use the relationship between a,b, and cfor a hyperbola: c2=a2+b2.
Solve for b.
b=√c2−a2=√25 −9 = √16 = 4
11
Step 5: Write the equation of the hyperbola in standard form ((x−h)2
a2−(y−k)2
b2= 1).
(x+ 1)2
9−y2
16 = 1
Therefore, the standard form of the equation of the hyperbola is (x+1)2
9−y2
16 =
1.
Question 13
Question
Solve the equation 5x2+ 12xy + 7y2−28x−38y−37 = 0 for yin terms of x.
Solution
Step 1: Rearrange the given equation into a standard conic form:
5x2+ 12xy + 7y2−28x−38y−37 = 0
⇒5x2+ 12xy + 7y2= 28x+ 38y+ 37
Step 2: Complete the square for the terms involving xand y:
5x2+ 12xy + 7y2= 28x+ 38y+ 37
⇒5(x2+12
5x·y+?) + 7(y2+12
7x·y+?) = 28x+ 38y+ 37
Step 3: To complete the square for the xterms, we add (12
5)2=144
25 inside
the parentheses:
5(x2+12
5x·y+144
25 ) + 7(y2+12
7x·y+?) = 28x+ 38y+ 37
Step 4: To complete the square for the yterms, we add (12
7)2=144
49 inside
the parentheses:
5(x2+12
5x·y+144
25 ) + 7(y2+12
7x·y+144
49 ) = 28x+ 38y+ 37
Step 5: Write the expressions as squares of binomials and simplify:
5(x+12
5y)2
+ 7 (y+12
7x)2
= 28x+ 38y+ 37
Step 6: Divide through by the constant term on the right side:
(x+12
5y)2
+7
5(y+12
7x)2
=28x+ 38y+ 37
5
Step 7: Factor the expression and simplify the right side further if needed.
12
Question 14
Question
Solve the system of equations for xand y:
{x2+y2= 25
x2−y2= 9
Solution
To solve the system of equations, we will use the method of substitution.
Step 1: Solve the second equation for x2:
x2= 9 + y2
Step 2: Substitute x2= 9 + y2into the first equation:
(9 + y2) + y2= 25
Step 3: Simplify the equation:
9+2y2= 25
Step 4: Solve for y2:
2y2= 16
y2= 8
y=±√8 = ±2√2
Step 5: Substitute y=±2√2back into x2= 9 + y2to find x:
x2= 9 + (2√2)2= 9 + 8 = 17
x=±√17
Therefore, the solutions to the system of equations are:
(x, y) = (√17,2√2),(√17,−2√2),(−√17,2√2),(−√17,−2√2)
Question 15
Question
Solve the system of equations:
{x2+ 4y2= 16
y=x2−4
13
Solution
Step 1: Substitute the second equation into the first equation to eliminate y.
x2+ 4(x2−4)2= 16
Step 2: Expand and simplify the equation.
x2+ 4(x2−4)2= 16 =⇒x2+ 4(x4−8x2+ 16) = 16
Step 3: Distribute and combine like terms.
x2+ 4x4−32x2+ 64 = 16 =⇒4x4−31x2+ 48 = 0
Step 4: Factor the quadratic in terms of x2.
(4x2−3)(x2−16) = 0
Step 5: Solve for x2.
{4x2−3 = 0
x2−16 = 0
Step 6: Solve for x.
{4x2−3 = 0 =⇒x=±√3
2
x2−16 = 0 =⇒x=±4
Step 7: Substitute the values of xback into the second equation to find the
corresponding yvalues.
For x=−4y= (−4)2−4y= 12
For x=−√3
2y=(−√3
2)2−4y=−7
4
For x=√3
2y=(√3
2)2−4y=−7
4
For x= 4 y= 42−4y= 12
Therefore, the solutions to the system of equations are (−4,12),(−√3
2,−7
4),
and (√3
2,−7
4).
Question 16
Question
Solve the system of equations:
{x2−y2= 4
xy = 6
14
Solution
Step 1: We start by solving the second equation, xy = 6, for yin terms of x:
We can rewrite the second equation as y=6
x.
Step 2: Substitute y=6
xinto the first equation x2−y2= 4:x2−(6
x)2= 4
Expanding and simplifying, we get: x2−36
x2= 4
Step 3: Multiply through by x2to clear the fraction: x4−36 = 4x2
Step 4: Rearrange the equation to get a quadratic equation: x4−4x2−36 = 0
Step 5: Let u=x2, then the equation becomes: u2−4u−36 = 0
Step 6: Solve the quadratic equation for uby factoring: (u−9)(u+ 4) = 0
This gives us u= 9 or u=−4.
Step 7: Substitute back u=x2: For u= 9:x2= 9 =⇒x=±3, and since
y=6
x, we get y=±2.
For u=−4, this solution is extraneous since x2cannot be negative.
Therefore, the solution to the system of equations are:
{x= 3, y = 2
x=−3, y =−2
Question 17
Question
Solve the following equation for x:
(2x+ 1)2−3(2x+ 1) −4 = 0
Solution
Step 1: Let’s simplify the left side of the equation by expanding and combining
like terms:
(2x+ 1)2−3(2x+ 1) −4 = 4x2+ 4x+ 1 −6x−3−4
= 4x2−2x−6
Step 2: Substitute the simplified expression back into the equation:
4x2−2x−6 = 0
Step 3: Next, we solve the quadratic equation by factoring. Since the coef-
ficient of x2is not 1, we need to use the AC method to factor:
4x2−2x−6 = 0
4·(−6) = −24
Factors of −24 that add up to −2are −6and 4
4x2−6x+ 4x−6 = 0
4x(x−1) + 2(x−3) = 0
(4x+ 2)(x−3) = 0
15
Step 4: Set each factor to zero and solve for x:
4x+ 2 = 0 or x−3 = 0
4x=−2or x= 3
x=−1
2or x= 3
Therefore, the solutions to the equation are x=−1
2and x= 3.
Question 18
Question
Let x2+ 4y2−4x+ 16y+ 4 = 0 be the equation of a conic section. Determine
the type of conic section and sketch its graph.
Solution
Step 1: Rewrite the equation in standard form by completing the square for
both the xand yterms.
x2−4x+4+4y2+ 16y+ 16 = 0
(x−2)2+ 4(y+ 2)2= 4
Step 2: Divide both sides of the equation by 4 to get the standard form of
the equation.
(x−2)2
4+(y+ 2)2
1= 1
Step 3: Compare the equation to the standard form equations of conic sec-
tions to determine the type. Since the xterm is positive and the yterm is
negative, this is the equation of an ellipse.
Step 4: Determine the major and minor axes of the ellipse. The length of
the major axis is 2a= 4, so a= 2. The length of the minor axis is 2b= 2, so
b= 1.
Step 5: Sketch the ellipse with center at (2,−2), major axis along the x-axis,
and minor axis along the y-axis.
The sketch will show an ellipse with major axis length 4 along the x-axis
and minor axis length 2 along the y-axis.
Question 19
Question
Solve the system of equations:
{x2+y2= 25
4x−3y= 7
16
Solution
Step 1: Rearrange the second equation to solve for xin terms of y.
4x−3y= 7 =⇒4x= 3y+ 7 =⇒x=3y+ 7
4
Step 2: Substitute the expression for xinto the first equation.
(3y+ 7
4)2
+y2= 25
Step 3: Simplify the equation by expanding and solving for y.
9y2+ 42y+ 49
16 +y2= 25
9y2+ 42y+ 49 + 16y2= 400
25y2+ 42y+ 49 = 400
25y2+ 42y−351 = 0
Step 4: Solve the quadratic equation for yby using the quadratic formula.
y=−42 ±√422−4(25)(−351)
2(25)
y=−42 ±√1764 + 35100
50
y=−42 ±√36864
50
y=−42 ±192
50
Step 5: Find the two possible values for y.
y1=−42 + 192
50 =150
50 = 3
y2=−42 −192
50 =−234
50 =−4.68
Step 6: Substitute the values of yback into the equation 4x= 3y+ 7 to find
the corresponding values of x.
x1=3(3) + 7
4=9+7
4=16
4= 4
x2=3(−4.68) + 7
4=−14.04 + 7
4=−7.04
4=−1.76
Therefore, the solutions to the system of equations are (4,3) and (−1.76,−4.68).
17
Question 20
Question
Find the standard form of the equation of the hyperbola with vertices at (−5,0)
and (5,0), and foci at (−7,0) and (7,0).
Solution
Step 1: Find the center of the hyperbola by averaging the coordinates of the
vertices:
Center =(−5+5
2,0+0
2)= (0,0)
Step 2: Find the distance between the center and either vertex to determine
a(distance to vertices):
a= 5 −0 = 5
Step 3: Find the distance between the center and either focus to determine
c(distance to foci):
c= 7 −0 = 7
Step 4: Use the relationship c2=a2+b2to solve for b:
b2=c2−a2= 72−52= 49 −25 = 24
b=√24 = 2√6
Step 5: The standard form of the equation of a hyperbola centered at (h, k)
is: (x−h)2
a2−(y−k)2
b2= 1
Step 6: Plug in the values of h= 0,k= 0,a= 5, and b= 2√6:
x2
52−y2
(2√6)2= 1
x2
25 −y2
24 = 1
Therefore, the standard form of the equation of the hyperbola is x2
25 −y2
24 = 1.
Question 21
Question
Determine the standard form equation of the ellipse with foci (−3,0) and (3,0),
passing through the point (6,−4).
18
Solution
Step 1: Determine the center of the ellipse. The center is the midpoint of the
foci, which is ((−3 + 3)/2,(0 + 0)/2) = (0,0).
Step 2: Determine the distance between the foci. The distance between the
foci is 2a=|6−(−6)|= 12, where 2ais the length of the major axis.
Step 3: Determine the distance between the center and either focus, which
is a= 6.
Step 4: Determine the value of c, the distance between the center and a
focus: c= 3.
Step 5: Recall the relationship between a,b, and c:c2=a2−b2. Substituting
the known values, we get 32= 62−b2, which simplifies to 9 = 36 −b2.
Step 6: Solve for b2:b2= 36 −9 = 27.
Step 7: The standard form equation of an ellipse with center (h, k), major
axis along the x-axis, and minor axis along the y-axis is (x−h)2
a2+(y−k)2
b2= 1.
Substituting the known values, we have x2
36 +y2
27 = 1. Therefore, the standard
form equation of the ellipse is x2
36 +y2
27 = 1 .
Question 22
Question
Solve the following system of equations:
{x2+y2−10x+ 4y+ 9 = 0
x2−2x−y2+ 2y−15 = 0
Solution
Step 1: We can start solving the system of equations by completing the square
for both equations.
For the first equation, we have:
x2+y2−10x+ 4y+ 9 = 0
x2−10x+y2+ 4y=−9
x2−10x+ 25 + y2+ 4y+ 4 = −9+25+4
(x−5)2+ (y+ 2)2= 20
For the second equation, we have:
19
x2−2x−y2+ 2y−15 = 0
x2−2x+y2+ 2y= 15
x2−2x+1+y2+ 2y+ 1 = 15 + 1 + 1
(x−1)2+ (y+ 1)2= 17
Step 2: Now we have the system of equations in standard form:
{(x−5)2+ (y+ 2)2= 20
(x−1)2+ (y+ 1)2= 17
Step 3: The first equation represents a circle with center at (5,−2) and
radius √20. The second equation represents a circle with center at (1,−1) and
radius √17.
Since circles can intersect at most two points, we need to find the points of
intersection between these two circles.
Step 4: To solve for the points of intersection, we set the equations equal to
each other:
(x−5)2+ (y+ 2)2= (x−1)2+ (y+ 1)2
Expanding and simplifying, we get:
x2−10x+ 25 + y2+ 4y+ 4 = x2−2x+1+y2+ 2y+ 1
Step 5: Simplifying further, we get:
−10x+25+4y+ 4 = −2x+1+2y+ 1
Step 6: Rearranging terms, we further simplify:
−8x+ 4y+ 28 = −2x+ 2y+ 2
−6x+ 2y+ 26 = 0
3x−y−13 = 0
Therefore, the system of equations has been reduced to the equation 3x−
y−13 = 0.
Question 23
Question
Consider the equation of a hyperbola in standard form: (x−3)2
4−(y+1)2
9= 1.
Determine the center, vertices, foci, and asymptotes of the hyperbola.
20
Solution
Step 1: First, identify the center of the hyperbola. The center is given by the
values inside the parentheses in the standard form. Therefore, the center is
(3,−1).
Step 2: Next, we can find the vertices of the hyperbola. Since the hyperbola
opens horizontally, the vertices are located ±the square root of the denominator
under xunits from the center. So, the vertices are at (3 ±2,−1) = (5,−1) and
(1,−1).
Step 3: To find the foci of the hyperbola, we need to use the relationship
between the distances a,b, and cin a hyperbola. The distance formula is given
by c2=a2+b2, where aand bare the denominators under x2and y2respectively.
Thus, c2= 4 + 9 = 13, which implies c=√13. Therefore, the foci are located
at (3 ±√13,−1) ≈(6.61,−1) and (−0.61,−1).
Step 4: To determine the equations of the asymptotes, we use the formula:
y=±(b
a)(x−h) + k, where (h, k)is the center of the hyperbola. Substituting
the values, we get y=±(3
2)(x−3)−1. Therefore, the asymptotes are y=3
2x−4
and y=−3
2x+ 2.
Question 24
Question
Solve the equation 2x2+ 3xy + 2y2−4x−5y+ 4 = 0 for xin terms of y.
Solution
To solve this equation for xin terms of y, we will treat the equation as a
quadratic in xand use the quadratic formula.
Step 1: Rewrite the equation in the standard form of a quadratic equation
in x:
2x2+ (3y−4)x+ 2y2−5y+ 4 = 0
Step 2: Use the quadratic formula, x=−b±√b2−4ac
2a, where a= 2,b= 3y−4,
and c= 2y2−5y+ 4.
Step 3: Substitute a= 2,b= 3y−4, and c= 2y2−5y+4 into the quadratic
formula:
x=−(3y−4) ±√(3y−4)2−4(2)(2y2−5y+ 4)
2(2)
Step 4: Simplify under the square root:
(3y−4)2−4(2)(2y2−5y+ 4) = 9y2−24y+ 16 −16(2y2−5y+ 4)
= 9y2−24y+ 16 −32y2+ 80y−64 = −23y2+ 56y−48
21
Step 5: Substitute the simplified expression back into the quadratic formula:
x=−(3y−4) ±√−23y2+ 56y−48
4
Therefore, the solution to the equation 2x2+ 3xy + 2y2−4x−5y+ 4 = 0
for xin terms of yis x=−(3y−4)±√−23y2+56y−48
4.
Question 25
Question
Find the equation of the hyperbola with vertices at (−3,1) and (1,1) and foci
at (−5,1) and (3,1).
Solution
Step 1: Find the center of the hyperbola. Since the hyperbola has vertices at
(−3,1) and (1,1), the center will be the midpoint of these points.
Center =(−3+1
2,1+1
2)= (−1,1)
Step 2: Calculate the distance between the center and one of the vertices to
find a, the distance from the center to one of the vertices.
Distance between center and vertex =√(−1 + 1)2+ (1 −1)2= 4
So, a= 4.
Step 3: Since the foci are located at (−5,1) and (3,1), the value of cis the
distance between the center and one of the foci.
Distance between center and focus =√(−1 + 5)2+ (1 −1)2= 4
So, c= 4.
Step 4: Use the relationship c2=a2+b2for a hyperbola to find b.
b2=c2−a2= 42−42= 0
So, b= 0.
Step 5: Write the equation of the hyperbola in standard form (x−h)2/a2−
(y−k)2/b2= 1 where (h, k)is the center, ais the distance from the center to
the vertex, and bis the distance from the center to the co-vertex.
(x+ 1)2
16 −(y−1)2
0= 1
22
Solution
Step 1: The distance between the foci is equal to the length of the major axis.
Let’s denote the distance between the foci as 2c. Step 2: Since the foci are at
(−2,0) and (2,0), we have 2c= 2 ⇒c= 1. Step 3: The length of the minor
axis is the same as the length of the major axis. Let’s denote the length of the
minor axis as 2b. Step 4: Given that the minor axis has a length of 6, we have
2b= 6 ⇒b= 3. Step 5: The standard form of the equation of an ellipse is
(x−h)2
a2+(y−k)2
b2= 1, where (h, k)is the center of the ellipse, ais the length of
the major axis, and bis the length of the minor axis. Step 6: Since the center
of the ellipse is the midpoint between the foci, the center is at the origin (0,0).
Step 7: We have a2=c2+b2= 12+ 32= 1 + 9 = 10. Step 8: Therefore, the
standard form of the equation of the ellipse is x2
10 +y2
9= 1 .
Question 3
Question
Let f(x) = x3−4x2−5x+ 18. Find all real zeros of the function.
Solution
Step 1: To find the zeros of the function f(x), we set f(x)=0and solve for
x. Step 2: So, we have the equation x3−4x2−5x+ 18 = 0. Step 3: Since
this is a cubic equation, it is difficult to find the zeros directly. Let’s try to use
synthetic division to determine if there are any rational roots. Step 4: By trying
potential factors of the constant term (in this case, 18) divided by the factors
of the leading coefficient (in this case, 1), we can find that x= 2 is a root of
the equation. Step 5: Perform synthetic division with (x−2) as the divisor to
factor out the root. 2 1 −4−5
18
2−4
−18
1−2−9
0
Step 6: The result of the synthetic division is a quadratic equation x2−2x−
9=0. We can solve this equation using the quadratic formula. Step 7: The
quadratic formula states that x=−(−2)±√(−2)2−4∗1∗(−9)
2∗1. Step 8: Simplifying,
we get x=2±√4+36
2, which results in x=2±√40
2. Step 9: This gives us two
potential solutions: x= 1 + √10 and x= 1 −√10. Step 10: Therefore, the real
zeros of the function f(x) = x3−4x2−5x+ 18 are x= 2,1 + √10,1−√10.
3
Question 4
Question
Find the standard form of the equation of the hyperbola with vertices at (−3,2)
and (3,2) and foci at (−5,2) and (5,2).
Solution
Step 1: Identify the center of the hyperbola. The center of the hyperbola is the
midpoint of the segment connecting the vertices, which is given by
(−3+3
2,2+2
2) = (0,2).
Step 2: Find the distance from the center to either of the vertices. Since the
vertices are to the left and right of the center, the distance from the center to a
vertex is the horizontal distance. Thus, the distance is 3units.
Step 3: Find the value of a. The value of ais the distance from the center
to a vertex, which is a= 3.
Step 4: Find the value of c. The value of cis the distance from the center
to a focus, so c= 5.
Step 5: Find the value of b. The relationship between a,b, and cin a
hyperbola is given by the equation c2=a2+b2. Substituting the known values,
we have 52= 32+b2, which simplifies to 25 = 9 + b2. Solving for b, we get
b2= 16, so b= 4.
Step 6: Determine the equation of the hyperbola. The standard form of the
equation of a hyperbola centered at (h, k)with vertices on the transverse axis
is (x−h)2
a2−(y−k)2
b2= 1. Substituting our values, we have
(x−0)2
32−(y−2)2
42= 1.
Simplifying, we get
x2
9−(y−2)2
16 = 1.
Therefore, the standard form of the equation of the hyperbola is
x2
9−(y−2)2
16 = 1 .
Question 5
Question
Find the standard form of the equation of the hyperbola that satisfies the given
conditions: Vertices at (−2,1) and (2,1) and foci at (−4,1) and (4,1).
4
Solution
Step 1: The center of the hyperbola lies at the midpoint of the vertices, which
can be calculated as follows:
(h, k) = (−2+2
2,1+1
2)= (0,1).
Step 2: The distance from the center to a vertex is the distance between two
x-coordinates, which is the value of a. In this case, we have |a|= 2.
Step 3: The distance from the center to a focus is the value of c, which can
be calculated using the distance formula:
c=√(h−f)2+ (k−k)2=√(0 −4)2+ (1 −1)2= 4.
Step 4: The equation of a hyperbola centered at (h, k)with vertices along
the x-axis is given by (x−h)2
a2−(y−k)2
b2= 1, where b2=a2−c2.
Step 5: Substituting the given values of aand cinto the formula for b2:
b2= 22−42= 4 −16 = −12.
Step 6: The equation of the hyperbola can now be written as:
x2
4−y2
−12 = 1.
Step 7: To simplify and rewrite the equation in standard form, multiply
through by −1to eliminate the negative denominator:
−x2
4+y2
12 = 1.
Step 8: Therefore, the standard form of the equation of the hyperbola is
−x2
4+y2
12 = 1 .
Question 6
Question
Find the equation of the hyperbola with vertices at (−3,2) and (3,2), and foci
at (−5,2) and (5,2).
Solution
Step 1: Find the center of the hyperbola. The center of the hyperbola is the
midpoint of the vertices:
Center =(−3+3
2,2+2
2)= (0,2)
5
Step 2: Find the distance between the vertices to get the length of the
transverse axis.
2a= 3 −(−3) = 6 ⇒a= 3
Step 3: Find the distance between the foci to get the constant c.
2c= 5 −(−5) = 10 ⇒c= 5
Step 4: Calculate busing the relationship c2=a2+b2.
b2=c2−a2= 25 −9 = 16 ⇒b= 4
Step 5: Write the equation of the hyperbola in standard form. Since the
hyperbola opens horizontally, the standard form is:
(x−h)2
a2−(y−k)2
b2= 1
where (h, k)is the center of the hyperbola (0,2). So, the equation of the hyper-
bola is: x2
9−(y−2)2
16 = 1
Question 7
Question
Find the standard form equation of the hyperbola that satisfies the given con-
ditions: Center at (−2,3), horizontal transverse axis of length 10, and a vertex
at (−7,3).
Solution
Step 1: Determine the equation for a hyperbola with center at (−2,3) and hori-
zontal transverse axis of length 10. The standard form equation for a hyperbola
with center (h, k), horizontal transverse axis length 2a, and vertical transverse
axis length 2bis:
(x−h)2/a2−(y−k)2/b2= 1
Given that the center is (−2,3) and the horizontal transverse axis length is
10, the equation becomes:
(x+ 2)2/25 −(y−3)2/b2= 1
Step 2: Find the value of busing the information that the vertex is (−7,3).
The distance from the center to the vertex is a= 5, so the distance from the
center to the midpoint between the vertices is c=a= 5. Since the vertex is
located 5 units to the left of the center, we have a2+b2=c2:
52+b2= 52
6
25 + b2= 25
b2= 0
b= 0
Step 3: Substitute the value of binto the equation to get the final equation
of the hyperbola. Substitute b= 0 into the equation:
(x+ 2)2/25 −(y−3)2/0 = 1
This simplifies to:
(x+ 2)2/25 = 1
Therefore, the standard form equation of the hyperbola is (x+ 2)2/25 = 1 .
Question 8
Question
Solve for yin the equation of the parabola x= 2y2−4y+ 3.
Solution
Step 1: Rearrange the equation by completing the square.
x+ 1 = 2(y2−2y)
Step 2: Complete the square inside the parenthesis.
x+ 1 = 2(y2−2y+ 1 −1)
Step 3: Simplify the right side of the equation.
x+ 1 = 2((y−1)2−1)
Step 4: Distribute the 2 on the right side.
x+ 1 = 2(y−1)2−2
Step 5: Add 2 to both sides to isolate the squared term.
x+ 3 = 2(y−1)2
Step 6: Divide both sides by 2.
x+ 3
2= (y−1)2
Step 7: Take the square root of both sides.
±√x+ 3
2=y−1
7
Step 8: Add 1 to both sides to solve for y.
y= 1 ±√x+ 3
2
Therefore, the solutions for yin the equation of the parabola x= 2y2−4y+3
are y= 1 + √x+3
2and y= 1 −√x+3
2.
Question 9
Question
Find the standard form equation of the hyperbola with foci at (-3, 0) and (3,
0) and vertices at (-5, 0) and (5, 0).
Solution
Step 1: Find the center of the hyperbola.
The center of the hyperbola is the midpoint of the segment connecting the
two vertices. The x-coordinate of the center is the average of the x-coordinates
of the vertices, and the y-coordinate is 0 since the hyperbola is centered on the
x-axis.
The center is ((−5 + 5)/2,0) = (0,0).
Step 2: Find the distance between the center and one of the foci.
The distance between the center and one of the foci is the distance from the
center to either of the foci. Since the foci are on the x-axis, the distance is the
absolute value of the difference between the x-coordinate of the center and the
x-coordinate of either focus.
The distance is |0−(−3)|= 3.
Step 3: Find the value of c, the distance from the center to either focus.
In the standard form of the equation of a hyperbola, cis related to the dis-
tances between the center and the foci and between the center and the vertices.
We know that for a hyperbola, c2=a2+b2, where ais half the distance between
the vertices and bis unknown.
Since a= 5, we can calculate b:
c=√a2+b2
3 = √52+b2
3 = √25 + b2
32= 25 + b2
9 = 25 + b2
b2= 9 −25
8
b2=−16
Since bmust be a positive value, we made a mistake in our calculation. This
indicates that there was an error in our initial assumption. Let’s revisit our
steps to determine the correct b value.
Step 4: Identify and correct the error in the previous steps.
When we calculated the distance between the center and one of the foci,
we mistakenly treated it as the distance c. However, we should have used the
formula for the equation of a hyperbola that involves c, a, and b. Let’s correct
this error and go back to Step 3.
Step 5: Correct the calculation of bin Step 3.
We know that for a hyperbola, c2=a2+b2, where ais half the distance
between the vertices and bis unknown. Since c= 3,a= 5, we can calculate b:
c2=a2+b2
32= 52+b2
9 = 25 + b2
b2= 9 −25
b2=−16
Since bmust be a positive value, b= 4.
Step 6: Write the equation of the hyperbola in standard form.
The standard form of the equation of a hyperbola centered at the origin with
foci on the x-axis is: x2
a2−y2
b2= 1
Substitute the values of a= 5 and b= 4 into the equation:
x2
52−y2
42= 1
x2
25 −y2
16 = 1
Therefore, the standard form equation of the hyperbola is x2
25 −y2
16 = 1.
Question 10
Question
Find the standard form of the equation of the parabola with a focus at (−2,4)
and a directrix y=−6.
9
Solution
Step 1: First, determine whether the parabola opens vertically or horizontally.
The parabola opens towards the directrix and away from the focus. Since the
focus is above the directrix, the parabola must open upwards.
Step 2: The distance between the vertex and the focus is equal to the dis-
tance between the vertex and the directrix. In this case, the vertex lies halfway
between the focus and the directrix. Therefore, the vertex is at (−2,4+(−6)
2) =
(−2,−1).
Step 3: The equation of a parabola that opens upwards with vertex at (h, k)
is given by (x−h)2= 4p(y−k), where pis the distance between the vertex and
the focus (or the vertex and the directrix in absolute value).
Step 4: Since the distance between the vertex and the focus is 5 units (from
(−2,−1) to (−2,4)), we have |p|= 5.
Step 5: Substituting the vertex (h, k) = (−2,−1) and p= 5 into the equation
gives (x+ 2)2= 4 ·5(y+ 1).
Step 6: Simplifying the equation gives (x+ 2)2= 20(y+ 1).
Therefore, the standard form of the equation of the parabola is (x+ 2)2=
20(y+ 1).
Question 11
Question
Determine the standard form equation of the hyperbola with vertices at (−5,1)
and (3,1) and foci at (−7,1) and (5,1).
Solution
Step 1: Find the center of the hyperbola. The center of the hyperbola is the
midpoint of the line segment connecting the vertices. Use the midpoint formula
(x1+x2
2,y1+y2
2).
Midpoint =(−5+3
2,1+1
2)= (−1,1)
Step 2: Determine the distance between the center and one of the vertices
to find a(the distance from the center to a vertex). Since the vertices are along
the x-axis, the distance is the difference in x-coordinates.
a= 3 −(−1) = 4
Step 3: Determine the distance between the center and one of the foci to
find c(the distance from the center to a focus).
c= 5 −(−1) = 6
10
Step 4: Use the relationship for a hyperbola to find b(the distance from the
center to the vertices along the y-axis).
c2=a2+b2⇒62= 42+b2⇒36 = 16 + b2
b2= 36 −16 = 20 ⇒b=√20 = 2√5
Step 5: Write the standard form equation of a hyperbola.
(x−h)2
a2−(y−k)2
b2= 1
Where the center of the hyperbola is (h, k).
Plugging in the values:
(x+ 1)2
42−(y−1)2
(2√5)2= 1
Therefore, the equation of the hyperbola is:
(x+ 1)2
16 −(y−1)2
20 = 1
Question 12
Question
Determine the standard form of the equation of the hyperbola with vertices at
(−1,3) and (−1,−3) and foci at (−1,5) and (−1,−5).
Solution
Step 1: Find the center of the hyperbola by taking the average of the vertices.
Center =(−1+(−1)
2,3+(−3)
2)= (−1,0)
Step 2: Determine the distance between the center and one of the vertices
to find a(the distance from the center to the vertices).
a= 3 −0 = 3
Step 3: Determine the distance between the center and one of the foci to
find c(the distance from the center to the foci).
c= 5 −0 = 5
Step 4: Use the relationship between a,b, and cfor a hyperbola: c2=a2+b2.
Solve for b.
b=√c2−a2=√25 −9 = √16 = 4
11
Step 5: Write the equation of the hyperbola in standard form ((x−h)2
a2−(y−k)2
b2= 1).
(x+ 1)2
9−y2
16 = 1
Therefore, the standard form of the equation of the hyperbola is (x+1)2
9−y2
16 =
1.
Question 13
Question
Solve the equation 5x2+ 12xy + 7y2−28x−38y−37 = 0 for yin terms of x.
Solution
Step 1: Rearrange the given equation into a standard conic form:
5x2+ 12xy + 7y2−28x−38y−37 = 0
⇒5x2+ 12xy + 7y2= 28x+ 38y+ 37
Step 2: Complete the square for the terms involving xand y:
5x2+ 12xy + 7y2= 28x+ 38y+ 37
⇒5(x2+12
5x·y+?) + 7(y2+12
7x·y+?) = 28x+ 38y+ 37
Step 3: To complete the square for the xterms, we add (12
5)2=144
25 inside
the parentheses:
5(x2+12
5x·y+144
25 ) + 7(y2+12
7x·y+?) = 28x+ 38y+ 37
Step 4: To complete the square for the yterms, we add (12
7)2=144
49 inside
the parentheses:
5(x2+12
5x·y+144
25 ) + 7(y2+12
7x·y+144
49 ) = 28x+ 38y+ 37
Step 5: Write the expressions as squares of binomials and simplify:
5(x+12
5y)2
+ 7 (y+12
7x)2
= 28x+ 38y+ 37
Step 6: Divide through by the constant term on the right side:
(x+12
5y)2
+7
5(y+12
7x)2
=28x+ 38y+ 37
5
Step 7: Factor the expression and simplify the right side further if needed.
12
Question 14
Question
Solve the system of equations for xand y:
{x2+y2= 25
x2−y2= 9
Solution
To solve the system of equations, we will use the method of substitution.
Step 1: Solve the second equation for x2:
x2= 9 + y2
Step 2: Substitute x2= 9 + y2into the first equation:
(9 + y2) + y2= 25
Step 3: Simplify the equation:
9+2y2= 25
Step 4: Solve for y2:
2y2= 16
y2= 8
y=±√8 = ±2√2
Step 5: Substitute y=±2√2back into x2= 9 + y2to find x:
x2= 9 + (2√2)2= 9 + 8 = 17
x=±√17
Therefore, the solutions to the system of equations are:
(x, y) = (√17,2√2),(√17,−2√2),(−√17,2√2),(−√17,−2√2)
Question 15
Question
Solve the system of equations:
{x2+ 4y2= 16
y=x2−4
13
Solution
Step 1: Substitute the second equation into the first equation to eliminate y.
x2+ 4(x2−4)2= 16
Step 2: Expand and simplify the equation.
x2+ 4(x2−4)2= 16 =⇒x2+ 4(x4−8x2+ 16) = 16
Step 3: Distribute and combine like terms.
x2+ 4x4−32x2+ 64 = 16 =⇒4x4−31x2+ 48 = 0
Step 4: Factor the quadratic in terms of x2.
(4x2−3)(x2−16) = 0
Step 5: Solve for x2.
{4x2−3 = 0
x2−16 = 0
Step 6: Solve for x.
{4x2−3 = 0 =⇒x=±√3
2
x2−16 = 0 =⇒x=±4
Step 7: Substitute the values of xback into the second equation to find the
corresponding yvalues.
For x=−4y= (−4)2−4y= 12
For x=−√3
2y=(−√3
2)2−4y=−7
4
For x=√3
2y=(√3
2)2−4y=−7
4
For x= 4 y= 42−4y= 12
Therefore, the solutions to the system of equations are (−4,12),(−√3
2,−7
4),
and (√3
2,−7
4).
Question 16
Question
Solve the system of equations:
{x2−y2= 4
xy = 6
14
Solution
Step 1: We start by solving the second equation, xy = 6, for yin terms of x:
We can rewrite the second equation as y=6
x.
Step 2: Substitute y=6
xinto the first equation x2−y2= 4:x2−(6
x)2= 4
Expanding and simplifying, we get: x2−36
x2= 4
Step 3: Multiply through by x2to clear the fraction: x4−36 = 4x2
Step 4: Rearrange the equation to get a quadratic equation: x4−4x2−36 = 0
Step 5: Let u=x2, then the equation becomes: u2−4u−36 = 0
Step 6: Solve the quadratic equation for uby factoring: (u−9)(u+ 4) = 0
This gives us u= 9 or u=−4.
Step 7: Substitute back u=x2: For u= 9:x2= 9 =⇒x=±3, and since
y=6
x, we get y=±2.
For u=−4, this solution is extraneous since x2cannot be negative.
Therefore, the solution to the system of equations are:
{x= 3, y = 2
x=−3, y =−2
Question 17
Question
Solve the following equation for x:
(2x+ 1)2−3(2x+ 1) −4 = 0
Solution
Step 1: Let’s simplify the left side of the equation by expanding and combining
like terms:
(2x+ 1)2−3(2x+ 1) −4 = 4x2+ 4x+ 1 −6x−3−4
= 4x2−2x−6
Step 2: Substitute the simplified expression back into the equation:
4x2−2x−6 = 0
Step 3: Next, we solve the quadratic equation by factoring. Since the coef-
ficient of x2is not 1, we need to use the AC method to factor:
4x2−2x−6 = 0
4·(−6) = −24
Factors of −24 that add up to −2are −6and 4
4x2−6x+ 4x−6 = 0
4x(x−1) + 2(x−3) = 0
(4x+ 2)(x−3) = 0
15
Step 4: Set each factor to zero and solve for x:
4x+ 2 = 0 or x−3 = 0
4x=−2or x= 3
x=−1
2or x= 3
Therefore, the solutions to the equation are x=−1
2and x= 3.
Question 18
Question
Let x2+ 4y2−4x+ 16y+ 4 = 0 be the equation of a conic section. Determine
the type of conic section and sketch its graph.
Solution
Step 1: Rewrite the equation in standard form by completing the square for
both the xand yterms.
x2−4x+4+4y2+ 16y+ 16 = 0
(x−2)2+ 4(y+ 2)2= 4
Step 2: Divide both sides of the equation by 4 to get the standard form of
the equation.
(x−2)2
4+(y+ 2)2
1= 1
Step 3: Compare the equation to the standard form equations of conic sec-
tions to determine the type. Since the xterm is positive and the yterm is
negative, this is the equation of an ellipse.
Step 4: Determine the major and minor axes of the ellipse. The length of
the major axis is 2a= 4, so a= 2. The length of the minor axis is 2b= 2, so
b= 1.
Step 5: Sketch the ellipse with center at (2,−2), major axis along the x-axis,
and minor axis along the y-axis.
The sketch will show an ellipse with major axis length 4 along the x-axis
and minor axis length 2 along the y-axis.
Question 19
Question
Solve the system of equations:
{x2+y2= 25
4x−3y= 7
16
Solution
Step 1: Rearrange the second equation to solve for xin terms of y.
4x−3y= 7 =⇒4x= 3y+ 7 =⇒x=3y+ 7
4
Step 2: Substitute the expression for xinto the first equation.
(3y+ 7
4)2
+y2= 25
Step 3: Simplify the equation by expanding and solving for y.
9y2+ 42y+ 49
16 +y2= 25
9y2+ 42y+ 49 + 16y2= 400
25y2+ 42y+ 49 = 400
25y2+ 42y−351 = 0
Step 4: Solve the quadratic equation for yby using the quadratic formula.
y=−42 ±√422−4(25)(−351)
2(25)
y=−42 ±√1764 + 35100
50
y=−42 ±√36864
50
y=−42 ±192
50
Step 5: Find the two possible values for y.
y1=−42 + 192
50 =150
50 = 3
y2=−42 −192
50 =−234
50 =−4.68
Step 6: Substitute the values of yback into the equation 4x= 3y+ 7 to find
the corresponding values of x.
x1=3(3) + 7
4=9+7
4=16
4= 4
x2=3(−4.68) + 7
4=−14.04 + 7
4=−7.04
4=−1.76
Therefore, the solutions to the system of equations are (4,3) and (−1.76,−4.68).
17
Question 20
Question
Find the standard form of the equation of the hyperbola with vertices at (−5,0)
and (5,0), and foci at (−7,0) and (7,0).
Solution
Step 1: Find the center of the hyperbola by averaging the coordinates of the
vertices:
Center =(−5+5
2,0+0
2)= (0,0)
Step 2: Find the distance between the center and either vertex to determine
a(distance to vertices):
a= 5 −0 = 5
Step 3: Find the distance between the center and either focus to determine
c(distance to foci):
c= 7 −0 = 7
Step 4: Use the relationship c2=a2+b2to solve for b:
b2=c2−a2= 72−52= 49 −25 = 24
b=√24 = 2√6
Step 5: The standard form of the equation of a hyperbola centered at (h, k)
is: (x−h)2
a2−(y−k)2
b2= 1
Step 6: Plug in the values of h= 0,k= 0,a= 5, and b= 2√6:
x2
52−y2
(2√6)2= 1
x2
25 −y2
24 = 1
Therefore, the standard form of the equation of the hyperbola is x2
25 −y2
24 = 1.
Question 21
Question
Determine the standard form equation of the ellipse with foci (−3,0) and (3,0),
passing through the point (6,−4).
18
Solution
Step 1: Determine the center of the ellipse. The center is the midpoint of the
foci, which is ((−3 + 3)/2,(0 + 0)/2) = (0,0).
Step 2: Determine the distance between the foci. The distance between the
foci is 2a=|6−(−6)|= 12, where 2ais the length of the major axis.
Step 3: Determine the distance between the center and either focus, which
is a= 6.
Step 4: Determine the value of c, the distance between the center and a
focus: c= 3.
Step 5: Recall the relationship between a,b, and c:c2=a2−b2. Substituting
the known values, we get 32= 62−b2, which simplifies to 9 = 36 −b2.
Step 6: Solve for b2:b2= 36 −9 = 27.
Step 7: The standard form equation of an ellipse with center (h, k), major
axis along the x-axis, and minor axis along the y-axis is (x−h)2
a2+(y−k)2
b2= 1.
Substituting the known values, we have x2
36 +y2
27 = 1. Therefore, the standard
form equation of the ellipse is x2
36 +y2
27 = 1 .
Question 22
Question
Solve the following system of equations:
{x2+y2−10x+ 4y+ 9 = 0
x2−2x−y2+ 2y−15 = 0
Solution
Step 1: We can start solving the system of equations by completing the square
for both equations.
For the first equation, we have:
x2+y2−10x+ 4y+ 9 = 0
x2−10x+y2+ 4y=−9
x2−10x+ 25 + y2+ 4y+ 4 = −9+25+4
(x−5)2+ (y+ 2)2= 20
For the second equation, we have:
19
x2−2x−y2+ 2y−15 = 0
x2−2x+y2+ 2y= 15
x2−2x+1+y2+ 2y+ 1 = 15 + 1 + 1
(x−1)2+ (y+ 1)2= 17
Step 2: Now we have the system of equations in standard form:
{(x−5)2+ (y+ 2)2= 20
(x−1)2+ (y+ 1)2= 17
Step 3: The first equation represents a circle with center at (5,−2) and
radius √20. The second equation represents a circle with center at (1,−1) and
radius √17.
Since circles can intersect at most two points, we need to find the points of
intersection between these two circles.
Step 4: To solve for the points of intersection, we set the equations equal to
each other:
(x−5)2+ (y+ 2)2= (x−1)2+ (y+ 1)2
Expanding and simplifying, we get:
x2−10x+ 25 + y2+ 4y+ 4 = x2−2x+1+y2+ 2y+ 1
Step 5: Simplifying further, we get:
−10x+25+4y+ 4 = −2x+1+2y+ 1
Step 6: Rearranging terms, we further simplify:
−8x+ 4y+ 28 = −2x+ 2y+ 2
−6x+ 2y+ 26 = 0
3x−y−13 = 0
Therefore, the system of equations has been reduced to the equation 3x−
y−13 = 0.
Question 23
Question
Consider the equation of a hyperbola in standard form: (x−3)2
4−(y+1)2
9= 1.
Determine the center, vertices, foci, and asymptotes of the hyperbola.
20
Solution
Step 1: First, identify the center of the hyperbola. The center is given by the
values inside the parentheses in the standard form. Therefore, the center is
(3,−1).
Step 2: Next, we can find the vertices of the hyperbola. Since the hyperbola
opens horizontally, the vertices are located ±the square root of the denominator
under xunits from the center. So, the vertices are at (3 ±2,−1) = (5,−1) and
(1,−1).
Step 3: To find the foci of the hyperbola, we need to use the relationship
between the distances a,b, and cin a hyperbola. The distance formula is given
by c2=a2+b2, where aand bare the denominators under x2and y2respectively.
Thus, c2= 4 + 9 = 13, which implies c=√13. Therefore, the foci are located
at (3 ±√13,−1) ≈(6.61,−1) and (−0.61,−1).
Step 4: To determine the equations of the asymptotes, we use the formula:
y=±(b
a)(x−h) + k, where (h, k)is the center of the hyperbola. Substituting
the values, we get y=±(3
2)(x−3)−1. Therefore, the asymptotes are y=3
2x−4
and y=−3
2x+ 2.
Question 24
Question
Solve the equation 2x2+ 3xy + 2y2−4x−5y+ 4 = 0 for xin terms of y.
Solution
To solve this equation for xin terms of y, we will treat the equation as a
quadratic in xand use the quadratic formula.
Step 1: Rewrite the equation in the standard form of a quadratic equation
in x:
2x2+ (3y−4)x+ 2y2−5y+ 4 = 0
Step 2: Use the quadratic formula, x=−b±√b2−4ac
2a, where a= 2,b= 3y−4,
and c= 2y2−5y+ 4.
Step 3: Substitute a= 2,b= 3y−4, and c= 2y2−5y+4 into the quadratic
formula:
x=−(3y−4) ±√(3y−4)2−4(2)(2y2−5y+ 4)
2(2)
Step 4: Simplify under the square root:
(3y−4)2−4(2)(2y2−5y+ 4) = 9y2−24y+ 16 −16(2y2−5y+ 4)
= 9y2−24y+ 16 −32y2+ 80y−64 = −23y2+ 56y−48
21
Step 5: Substitute the simplified expression back into the quadratic formula:
x=−(3y−4) ±√−23y2+ 56y−48
4
Therefore, the solution to the equation 2x2+ 3xy + 2y2−4x−5y+ 4 = 0
for xin terms of yis x=−(3y−4)±√−23y2+56y−48
4.
Question 25
Question
Find the equation of the hyperbola with vertices at (−3,1) and (1,1) and foci
at (−5,1) and (3,1).
Solution
Step 1: Find the center of the hyperbola. Since the hyperbola has vertices at
(−3,1) and (1,1), the center will be the midpoint of these points.
Center =(−3+1
2,1+1
2)= (−1,1)
Step 2: Calculate the distance between the center and one of the vertices to
find a, the distance from the center to one of the vertices.
Distance between center and vertex =√(−1 + 1)2+ (1 −1)2= 4
So, a= 4.
Step 3: Since the foci are located at (−5,1) and (3,1), the value of cis the
distance between the center and one of the foci.
Distance between center and focus =√(−1 + 5)2+ (1 −1)2= 4
So, c= 4.
Step 4: Use the relationship c2=a2+b2for a hyperbola to find b.
b2=c2−a2= 42−42= 0
So, b= 0.
Step 5: Write the equation of the hyperbola in standard form (x−h)2/a2−
(y−k)2/b2= 1 where (h, k)is the center, ais the distance from the center to
the vertex, and bis the distance from the center to the co-vertex.
(x+ 1)2
16 −(y−1)2
0= 1
22
Solution
Step 1: The distance between the foci is equal to the length of the major axis.
Let’s denote the distance between the foci as 2c. Step 2: Since the foci are at
(−2,0) and (2,0), we have 2c= 2 ⇒c= 1. Step 3: The length of the minor
axis is the same as the length of the major axis. Let’s denote the length of the
minor axis as 2b. Step 4: Given that the minor axis has a length of 6, we have
2b= 6 ⇒b= 3. Step 5: The standard form of the equation of an ellipse is
(x−h)2
a2+(y−k)2
b2= 1, where (h, k)is the center of the ellipse, ais the length of
the major axis, and bis the length of the minor axis. Step 6: Since the center
of the ellipse is the midpoint between the foci, the center is at the origin (0,0).
Step 7: We have a2=c2+b2= 12+ 32= 1 + 9 = 10. Step 8: Therefore, the
standard form of the equation of the ellipse is x2
10 +y2
9= 1 .
Question 3
Question
Let f(x) = x3−4x2−5x+ 18. Find all real zeros of the function.
Solution
Step 1: To find the zeros of the function f(x), we set f(x)=0and solve for
x. Step 2: So, we have the equation x3−4x2−5x+ 18 = 0. Step 3: Since
this is a cubic equation, it is difficult to find the zeros directly. Let’s try to use
synthetic division to determine if there are any rational roots. Step 4: By trying
potential factors of the constant term (in this case, 18) divided by the factors
of the leading coefficient (in this case, 1), we can find that x= 2 is a root of
the equation. Step 5: Perform synthetic division with (x−2) as the divisor to
factor out the root. 2 1 −4−5
18
2−4
−18
1−2−9
0
Step 6: The result of the synthetic division is a quadratic equation x2−2x−
9=0. We can solve this equation using the quadratic formula. Step 7: The
quadratic formula states that x=−(−2)±√(−2)2−4∗1∗(−9)
2∗1. Step 8: Simplifying,
we get x=2±√4+36
2, which results in x=2±√40
2. Step 9: This gives us two
potential solutions: x= 1 + √10 and x= 1 −√10. Step 10: Therefore, the real
zeros of the function f(x) = x3−4x2−5x+ 18 are x= 2,1 + √10,1−√10.
3
Question 4
Question
Find the standard form of the equation of the hyperbola with vertices at (−3,2)
and (3,2) and foci at (−5,2) and (5,2).
Solution
Step 1: Identify the center of the hyperbola. The center of the hyperbola is the
midpoint of the segment connecting the vertices, which is given by
(−3+3
2,2+2
2) = (0,2).
Step 2: Find the distance from the center to either of the vertices. Since the
vertices are to the left and right of the center, the distance from the center to a
vertex is the horizontal distance. Thus, the distance is 3units.
Step 3: Find the value of a. The value of ais the distance from the center
to a vertex, which is a= 3.
Step 4: Find the value of c. The value of cis the distance from the center
to a focus, so c= 5.
Step 5: Find the value of b. The relationship between a,b, and cin a
hyperbola is given by the equation c2=a2+b2. Substituting the known values,
we have 52= 32+b2, which simplifies to 25 = 9 + b2. Solving for b, we get
b2= 16, so b= 4.
Step 6: Determine the equation of the hyperbola. The standard form of the
equation of a hyperbola centered at (h, k)with vertices on the transverse axis
is (x−h)2
a2−(y−k)2
b2= 1. Substituting our values, we have
(x−0)2
32−(y−2)2
42= 1.
Simplifying, we get
x2
9−(y−2)2
16 = 1.
Therefore, the standard form of the equation of the hyperbola is
x2
9−(y−2)2
16 = 1 .
Question 5
Question
Find the standard form of the equation of the hyperbola that satisfies the given
conditions: Vertices at (−2,1) and (2,1) and foci at (−4,1) and (4,1).
4
Solution
Step 1: The center of the hyperbola lies at the midpoint of the vertices, which
can be calculated as follows:
(h, k) = (−2+2
2,1+1
2)= (0,1).
Step 2: The distance from the center to a vertex is the distance between two
x-coordinates, which is the value of a. In this case, we have |a|= 2.
Step 3: The distance from the center to a focus is the value of c, which can
be calculated using the distance formula:
c=√(h−f)2+ (k−k)2=√(0 −4)2+ (1 −1)2= 4.
Step 4: The equation of a hyperbola centered at (h, k)with vertices along
the x-axis is given by (x−h)2
a2−(y−k)2
b2= 1, where b2=a2−c2.
Step 5: Substituting the given values of aand cinto the formula for b2:
b2= 22−42= 4 −16 = −12.
Step 6: The equation of the hyperbola can now be written as:
x2
4−y2
−12 = 1.
Step 7: To simplify and rewrite the equation in standard form, multiply
through by −1to eliminate the negative denominator:
−x2
4+y2
12 = 1.
Step 8: Therefore, the standard form of the equation of the hyperbola is
−x2
4+y2
12 = 1 .
Question 6
Question
Find the equation of the hyperbola with vertices at (−3,2) and (3,2), and foci
at (−5,2) and (5,2).
Solution
Step 1: Find the center of the hyperbola. The center of the hyperbola is the
midpoint of the vertices:
Center =(−3+3
2,2+2
2)= (0,2)
5
Step 2: Find the distance between the vertices to get the length of the
transverse axis.
2a= 3 −(−3) = 6 ⇒a= 3
Step 3: Find the distance between the foci to get the constant c.
2c= 5 −(−5) = 10 ⇒c= 5
Step 4: Calculate busing the relationship c2=a2+b2.
b2=c2−a2= 25 −9 = 16 ⇒b= 4
Step 5: Write the equation of the hyperbola in standard form. Since the
hyperbola opens horizontally, the standard form is:
(x−h)2
a2−(y−k)2
b2= 1
where (h, k)is the center of the hyperbola (0,2). So, the equation of the hyper-
bola is: x2
9−(y−2)2
16 = 1
Question 7
Question
Find the standard form equation of the hyperbola that satisfies the given con-
ditions: Center at (−2,3), horizontal transverse axis of length 10, and a vertex
at (−7,3).
Solution
Step 1: Determine the equation for a hyperbola with center at (−2,3) and hori-
zontal transverse axis of length 10. The standard form equation for a hyperbola
with center (h, k), horizontal transverse axis length 2a, and vertical transverse
axis length 2bis:
(x−h)2/a2−(y−k)2/b2= 1
Given that the center is (−2,3) and the horizontal transverse axis length is
10, the equation becomes:
(x+ 2)2/25 −(y−3)2/b2= 1
Step 2: Find the value of busing the information that the vertex is (−7,3).
The distance from the center to the vertex is a= 5, so the distance from the
center to the midpoint between the vertices is c=a= 5. Since the vertex is
located 5 units to the left of the center, we have a2+b2=c2:
52+b2= 52
6
25 + b2= 25
b2= 0
b= 0
Step 3: Substitute the value of binto the equation to get the final equation
of the hyperbola. Substitute b= 0 into the equation:
(x+ 2)2/25 −(y−3)2/0 = 1
This simplifies to:
(x+ 2)2/25 = 1
Therefore, the standard form equation of the hyperbola is (x+ 2)2/25 = 1 .
Question 8
Question
Solve for yin the equation of the parabola x= 2y2−4y+ 3.
Solution
Step 1: Rearrange the equation by completing the square.
x+ 1 = 2(y2−2y)
Step 2: Complete the square inside the parenthesis.
x+ 1 = 2(y2−2y+ 1 −1)
Step 3: Simplify the right side of the equation.
x+ 1 = 2((y−1)2−1)
Step 4: Distribute the 2 on the right side.
x+ 1 = 2(y−1)2−2
Step 5: Add 2 to both sides to isolate the squared term.
x+ 3 = 2(y−1)2
Step 6: Divide both sides by 2.
x+ 3
2= (y−1)2
Step 7: Take the square root of both sides.
±√x+ 3
2=y−1
7
Step 8: Add 1 to both sides to solve for y.
y= 1 ±√x+ 3
2
Therefore, the solutions for yin the equation of the parabola x= 2y2−4y+3
are y= 1 + √x+3
2and y= 1 −√x+3
2.
Question 9
Question
Find the standard form equation of the hyperbola with foci at (-3, 0) and (3,
0) and vertices at (-5, 0) and (5, 0).
Solution
Step 1: Find the center of the hyperbola.
The center of the hyperbola is the midpoint of the segment connecting the
two vertices. The x-coordinate of the center is the average of the x-coordinates
of the vertices, and the y-coordinate is 0 since the hyperbola is centered on the
x-axis.
The center is ((−5 + 5)/2,0) = (0,0).
Step 2: Find the distance between the center and one of the foci.
The distance between the center and one of the foci is the distance from the
center to either of the foci. Since the foci are on the x-axis, the distance is the
absolute value of the difference between the x-coordinate of the center and the
x-coordinate of either focus.
The distance is |0−(−3)|= 3.
Step 3: Find the value of c, the distance from the center to either focus.
In the standard form of the equation of a hyperbola, cis related to the dis-
tances between the center and the foci and between the center and the vertices.
We know that for a hyperbola, c2=a2+b2, where ais half the distance between
the vertices and bis unknown.
Since a= 5, we can calculate b:
c=√a2+b2
3 = √52+b2
3 = √25 + b2
32= 25 + b2
9 = 25 + b2
b2= 9 −25
8
b2=−16
Since bmust be a positive value, we made a mistake in our calculation. This
indicates that there was an error in our initial assumption. Let’s revisit our
steps to determine the correct b value.
Step 4: Identify and correct the error in the previous steps.
When we calculated the distance between the center and one of the foci,
we mistakenly treated it as the distance c. However, we should have used the
formula for the equation of a hyperbola that involves c, a, and b. Let’s correct
this error and go back to Step 3.
Step 5: Correct the calculation of bin Step 3.
We know that for a hyperbola, c2=a2+b2, where ais half the distance
between the vertices and bis unknown. Since c= 3,a= 5, we can calculate b:
c2=a2+b2
32= 52+b2
9 = 25 + b2
b2= 9 −25
b2=−16
Since bmust be a positive value, b= 4.
Step 6: Write the equation of the hyperbola in standard form.
The standard form of the equation of a hyperbola centered at the origin with
foci on the x-axis is: x2
a2−y2
b2= 1
Substitute the values of a= 5 and b= 4 into the equation:
x2
52−y2
42= 1
x2
25 −y2
16 = 1
Therefore, the standard form equation of the hyperbola is x2
25 −y2
16 = 1.
Question 10
Question
Find the standard form of the equation of the parabola with a focus at (−2,4)
and a directrix y=−6.
9
Solution
Step 1: First, determine whether the parabola opens vertically or horizontally.
The parabola opens towards the directrix and away from the focus. Since the
focus is above the directrix, the parabola must open upwards.
Step 2: The distance between the vertex and the focus is equal to the dis-
tance between the vertex and the directrix. In this case, the vertex lies halfway
between the focus and the directrix. Therefore, the vertex is at (−2,4+(−6)
2) =
(−2,−1).
Step 3: The equation of a parabola that opens upwards with vertex at (h, k)
is given by (x−h)2= 4p(y−k), where pis the distance between the vertex and
the focus (or the vertex and the directrix in absolute value).
Step 4: Since the distance between the vertex and the focus is 5 units (from
(−2,−1) to (−2,4)), we have |p|= 5.
Step 5: Substituting the vertex (h, k) = (−2,−1) and p= 5 into the equation
gives (x+ 2)2= 4 ·5(y+ 1).
Step 6: Simplifying the equation gives (x+ 2)2= 20(y+ 1).
Therefore, the standard form of the equation of the parabola is (x+ 2)2=
20(y+ 1).
Question 11
Question
Determine the standard form equation of the hyperbola with vertices at (−5,1)
and (3,1) and foci at (−7,1) and (5,1).
Solution
Step 1: Find the center of the hyperbola. The center of the hyperbola is the
midpoint of the line segment connecting the vertices. Use the midpoint formula
(x1+x2
2,y1+y2
2).
Midpoint =(−5+3
2,1+1
2)= (−1,1)
Step 2: Determine the distance between the center and one of the vertices
to find a(the distance from the center to a vertex). Since the vertices are along
the x-axis, the distance is the difference in x-coordinates.
a= 3 −(−1) = 4
Step 3: Determine the distance between the center and one of the foci to
find c(the distance from the center to a focus).
c= 5 −(−1) = 6
10
Step 4: Use the relationship for a hyperbola to find b(the distance from the
center to the vertices along the y-axis).
c2=a2+b2⇒62= 42+b2⇒36 = 16 + b2
b2= 36 −16 = 20 ⇒b=√20 = 2√5
Step 5: Write the standard form equation of a hyperbola.
(x−h)2
a2−(y−k)2
b2= 1
Where the center of the hyperbola is (h, k).
Plugging in the values:
(x+ 1)2
42−(y−1)2
(2√5)2= 1
Therefore, the equation of the hyperbola is:
(x+ 1)2
16 −(y−1)2
20 = 1
Question 12
Question
Determine the standard form of the equation of the hyperbola with vertices at
(−1,3) and (−1,−3) and foci at (−1,5) and (−1,−5).
Solution
Step 1: Find the center of the hyperbola by taking the average of the vertices.
Center =(−1+(−1)
2,3+(−3)
2)= (−1,0)
Step 2: Determine the distance between the center and one of the vertices
to find a(the distance from the center to the vertices).
a= 3 −0 = 3
Step 3: Determine the distance between the center and one of the foci to
find c(the distance from the center to the foci).
c= 5 −0 = 5
Step 4: Use the relationship between a,b, and cfor a hyperbola: c2=a2+b2.
Solve for b.
b=√c2−a2=√25 −9 = √16 = 4
11
Step 5: Write the equation of the hyperbola in standard form ((x−h)2
a2−(y−k)2
b2= 1).
(x+ 1)2
9−y2
16 = 1
Therefore, the standard form of the equation of the hyperbola is (x+1)2
9−y2
16 =
1.
Question 13
Question
Solve the equation 5x2+ 12xy + 7y2−28x−38y−37 = 0 for yin terms of x.
Solution
Step 1: Rearrange the given equation into a standard conic form:
5x2+ 12xy + 7y2−28x−38y−37 = 0
⇒5x2+ 12xy + 7y2= 28x+ 38y+ 37
Step 2: Complete the square for the terms involving xand y:
5x2+ 12xy + 7y2= 28x+ 38y+ 37
⇒5(x2+12
5x·y+?) + 7(y2+12
7x·y+?) = 28x+ 38y+ 37
Step 3: To complete the square for the xterms, we add (12
5)2=144
25 inside
the parentheses:
5(x2+12
5x·y+144
25 ) + 7(y2+12
7x·y+?) = 28x+ 38y+ 37
Step 4: To complete the square for the yterms, we add (12
7)2=144
49 inside
the parentheses:
5(x2+12
5x·y+144
25 ) + 7(y2+12
7x·y+144
49 ) = 28x+ 38y+ 37
Step 5: Write the expressions as squares of binomials and simplify:
5(x+12
5y)2
+ 7 (y+12
7x)2
= 28x+ 38y+ 37
Step 6: Divide through by the constant term on the right side:
(x+12
5y)2
+7
5(y+12
7x)2
=28x+ 38y+ 37
5
Step 7: Factor the expression and simplify the right side further if needed.
12
Question 14
Question
Solve the system of equations for xand y:
{x2+y2= 25
x2−y2= 9
Solution
To solve the system of equations, we will use the method of substitution.
Step 1: Solve the second equation for x2:
x2= 9 + y2
Step 2: Substitute x2= 9 + y2into the first equation:
(9 + y2) + y2= 25
Step 3: Simplify the equation:
9+2y2= 25
Step 4: Solve for y2:
2y2= 16
y2= 8
y=±√8 = ±2√2
Step 5: Substitute y=±2√2back into x2= 9 + y2to find x:
x2= 9 + (2√2)2= 9 + 8 = 17
x=±√17
Therefore, the solutions to the system of equations are:
(x, y) = (√17,2√2),(√17,−2√2),(−√17,2√2),(−√17,−2√2)
Question 15
Question
Solve the system of equations:
{x2+ 4y2= 16
y=x2−4
13
Solution
Step 1: Substitute the second equation into the first equation to eliminate y.
x2+ 4(x2−4)2= 16
Step 2: Expand and simplify the equation.
x2+ 4(x2−4)2= 16 =⇒x2+ 4(x4−8x2+ 16) = 16
Step 3: Distribute and combine like terms.
x2+ 4x4−32x2+ 64 = 16 =⇒4x4−31x2+ 48 = 0
Step 4: Factor the quadratic in terms of x2.
(4x2−3)(x2−16) = 0
Step 5: Solve for x2.
{4x2−3 = 0
x2−16 = 0
Step 6: Solve for x.
{4x2−3 = 0 =⇒x=±√3
2
x2−16 = 0 =⇒x=±4
Step 7: Substitute the values of xback into the second equation to find the
corresponding yvalues.
For x=−4y= (−4)2−4y= 12
For x=−√3
2y=(−√3
2)2−4y=−7
4
For x=√3
2y=(√3
2)2−4y=−7
4
For x= 4 y= 42−4y= 12
Therefore, the solutions to the system of equations are (−4,12),(−√3
2,−7
4),
and (√3
2,−7
4).
Question 16
Question
Solve the system of equations:
{x2−y2= 4
xy = 6
14
Solution
Step 1: We start by solving the second equation, xy = 6, for yin terms of x:
We can rewrite the second equation as y=6
x.
Step 2: Substitute y=6
xinto the first equation x2−y2= 4:x2−(6
x)2= 4
Expanding and simplifying, we get: x2−36
x2= 4
Step 3: Multiply through by x2to clear the fraction: x4−36 = 4x2
Step 4: Rearrange the equation to get a quadratic equation: x4−4x2−36 = 0
Step 5: Let u=x2, then the equation becomes: u2−4u−36 = 0
Step 6: Solve the quadratic equation for uby factoring: (u−9)(u+ 4) = 0
This gives us u= 9 or u=−4.
Step 7: Substitute back u=x2: For u= 9:x2= 9 =⇒x=±3, and since
y=6
x, we get y=±2.
For u=−4, this solution is extraneous since x2cannot be negative.
Therefore, the solution to the system of equations are:
{x= 3, y = 2
x=−3, y =−2
Question 17
Question
Solve the following equation for x:
(2x+ 1)2−3(2x+ 1) −4 = 0
Solution
Step 1: Let’s simplify the left side of the equation by expanding and combining
like terms:
(2x+ 1)2−3(2x+ 1) −4 = 4x2+ 4x+ 1 −6x−3−4
= 4x2−2x−6
Step 2: Substitute the simplified expression back into the equation:
4x2−2x−6 = 0
Step 3: Next, we solve the quadratic equation by factoring. Since the coef-
ficient of x2is not 1, we need to use the AC method to factor:
4x2−2x−6 = 0
4·(−6) = −24
Factors of −24 that add up to −2are −6and 4
4x2−6x+ 4x−6 = 0
4x(x−1) + 2(x−3) = 0
(4x+ 2)(x−3) = 0
15
Step 4: Set each factor to zero and solve for x:
4x+ 2 = 0 or x−3 = 0
4x=−2or x= 3
x=−1
2or x= 3
Therefore, the solutions to the equation are x=−1
2and x= 3.
Question 18
Question
Let x2+ 4y2−4x+ 16y+ 4 = 0 be the equation of a conic section. Determine
the type of conic section and sketch its graph.
Solution
Step 1: Rewrite the equation in standard form by completing the square for
both the xand yterms.
x2−4x+4+4y2+ 16y+ 16 = 0
(x−2)2+ 4(y+ 2)2= 4
Step 2: Divide both sides of the equation by 4 to get the standard form of
the equation.
(x−2)2
4+(y+ 2)2
1= 1
Step 3: Compare the equation to the standard form equations of conic sec-
tions to determine the type. Since the xterm is positive and the yterm is
negative, this is the equation of an ellipse.
Step 4: Determine the major and minor axes of the ellipse. The length of
the major axis is 2a= 4, so a= 2. The length of the minor axis is 2b= 2, so
b= 1.
Step 5: Sketch the ellipse with center at (2,−2), major axis along the x-axis,
and minor axis along the y-axis.
The sketch will show an ellipse with major axis length 4 along the x-axis
and minor axis length 2 along the y-axis.
Question 19
Question
Solve the system of equations:
{x2+y2= 25
4x−3y= 7
16
Solution
Step 1: Rearrange the second equation to solve for xin terms of y.
4x−3y= 7 =⇒4x= 3y+ 7 =⇒x=3y+ 7
4
Step 2: Substitute the expression for xinto the first equation.
(3y+ 7
4)2
+y2= 25
Step 3: Simplify the equation by expanding and solving for y.
9y2+ 42y+ 49
16 +y2= 25
9y2+ 42y+ 49 + 16y2= 400
25y2+ 42y+ 49 = 400
25y2+ 42y−351 = 0
Step 4: Solve the quadratic equation for yby using the quadratic formula.
y=−42 ±√422−4(25)(−351)
2(25)
y=−42 ±√1764 + 35100
50
y=−42 ±√36864
50
y=−42 ±192
50
Step 5: Find the two possible values for y.
y1=−42 + 192
50 =150
50 = 3
y2=−42 −192
50 =−234
50 =−4.68
Step 6: Substitute the values of yback into the equation 4x= 3y+ 7 to find
the corresponding values of x.
x1=3(3) + 7
4=9+7
4=16
4= 4
x2=3(−4.68) + 7
4=−14.04 + 7
4=−7.04
4=−1.76
Therefore, the solutions to the system of equations are (4,3) and (−1.76,−4.68).
17
Question 20
Question
Find the standard form of the equation of the hyperbola with vertices at (−5,0)
and (5,0), and foci at (−7,0) and (7,0).
Solution
Step 1: Find the center of the hyperbola by averaging the coordinates of the
vertices:
Center =(−5+5
2,0+0
2)= (0,0)
Step 2: Find the distance between the center and either vertex to determine
a(distance to vertices):
a= 5 −0 = 5
Step 3: Find the distance between the center and either focus to determine
c(distance to foci):
c= 7 −0 = 7
Step 4: Use the relationship c2=a2+b2to solve for b:
b2=c2−a2= 72−52= 49 −25 = 24
b=√24 = 2√6
Step 5: The standard form of the equation of a hyperbola centered at (h, k)
is: (x−h)2
a2−(y−k)2
b2= 1
Step 6: Plug in the values of h= 0,k= 0,a= 5, and b= 2√6:
x2
52−y2
(2√6)2= 1
x2
25 −y2
24 = 1
Therefore, the standard form of the equation of the hyperbola is x2
25 −y2
24 = 1.
Question 21
Question
Determine the standard form equation of the ellipse with foci (−3,0) and (3,0),
passing through the point (6,−4).
18
Solution
Step 1: Determine the center of the ellipse. The center is the midpoint of the
foci, which is ((−3 + 3)/2,(0 + 0)/2) = (0,0).
Step 2: Determine the distance between the foci. The distance between the
foci is 2a=|6−(−6)|= 12, where 2ais the length of the major axis.
Step 3: Determine the distance between the center and either focus, which
is a= 6.
Step 4: Determine the value of c, the distance between the center and a
focus: c= 3.
Step 5: Recall the relationship between a,b, and c:c2=a2−b2. Substituting
the known values, we get 32= 62−b2, which simplifies to 9 = 36 −b2.
Step 6: Solve for b2:b2= 36 −9 = 27.
Step 7: The standard form equation of an ellipse with center (h, k), major
axis along the x-axis, and minor axis along the y-axis is (x−h)2
a2+(y−k)2
b2= 1.
Substituting the known values, we have x2
36 +y2
27 = 1. Therefore, the standard
form equation of the ellipse is x2
36 +y2
27 = 1 .
Question 22
Question
Solve the following system of equations:
{x2+y2−10x+ 4y+ 9 = 0
x2−2x−y2+ 2y−15 = 0
Solution
Step 1: We can start solving the system of equations by completing the square
for both equations.
For the first equation, we have:
x2+y2−10x+ 4y+ 9 = 0
x2−10x+y2+ 4y=−9
x2−10x+ 25 + y2+ 4y+ 4 = −9+25+4
(x−5)2+ (y+ 2)2= 20
For the second equation, we have:
19
x2−2x−y2+ 2y−15 = 0
x2−2x+y2+ 2y= 15
x2−2x+1+y2+ 2y+ 1 = 15 + 1 + 1
(x−1)2+ (y+ 1)2= 17
Step 2: Now we have the system of equations in standard form:
{(x−5)2+ (y+ 2)2= 20
(x−1)2+ (y+ 1)2= 17
Step 3: The first equation represents a circle with center at (5,−2) and
radius √20. The second equation represents a circle with center at (1,−1) and
radius √17.
Since circles can intersect at most two points, we need to find the points of
intersection between these two circles.
Step 4: To solve for the points of intersection, we set the equations equal to
each other:
(x−5)2+ (y+ 2)2= (x−1)2+ (y+ 1)2
Expanding and simplifying, we get:
x2−10x+ 25 + y2+ 4y+ 4 = x2−2x+1+y2+ 2y+ 1
Step 5: Simplifying further, we get:
−10x+25+4y+ 4 = −2x+1+2y+ 1
Step 6: Rearranging terms, we further simplify:
−8x+ 4y+ 28 = −2x+ 2y+ 2
−6x+ 2y+ 26 = 0
3x−y−13 = 0
Therefore, the system of equations has been reduced to the equation 3x−
y−13 = 0.
Question 23
Question
Consider the equation of a hyperbola in standard form: (x−3)2
4−(y+1)2
9= 1.
Determine the center, vertices, foci, and asymptotes of the hyperbola.
20
Solution
Step 1: First, identify the center of the hyperbola. The center is given by the
values inside the parentheses in the standard form. Therefore, the center is
(3,−1).
Step 2: Next, we can find the vertices of the hyperbola. Since the hyperbola
opens horizontally, the vertices are located ±the square root of the denominator
under xunits from the center. So, the vertices are at (3 ±2,−1) = (5,−1) and
(1,−1).
Step 3: To find the foci of the hyperbola, we need to use the relationship
between the distances a,b, and cin a hyperbola. The distance formula is given
by c2=a2+b2, where aand bare the denominators under x2and y2respectively.
Thus, c2= 4 + 9 = 13, which implies c=√13. Therefore, the foci are located
at (3 ±√13,−1) ≈(6.61,−1) and (−0.61,−1).
Step 4: To determine the equations of the asymptotes, we use the formula:
y=±(b
a)(x−h) + k, where (h, k)is the center of the hyperbola. Substituting
the values, we get y=±(3
2)(x−3)−1. Therefore, the asymptotes are y=3
2x−4
and y=−3
2x+ 2.
Question 24
Question
Solve the equation 2x2+ 3xy + 2y2−4x−5y+ 4 = 0 for xin terms of y.
Solution
To solve this equation for xin terms of y, we will treat the equation as a
quadratic in xand use the quadratic formula.
Step 1: Rewrite the equation in the standard form of a quadratic equation
in x:
2x2+ (3y−4)x+ 2y2−5y+ 4 = 0
Step 2: Use the quadratic formula, x=−b±√b2−4ac
2a, where a= 2,b= 3y−4,
and c= 2y2−5y+ 4.
Step 3: Substitute a= 2,b= 3y−4, and c= 2y2−5y+4 into the quadratic
formula:
x=−(3y−4) ±√(3y−4)2−4(2)(2y2−5y+ 4)
2(2)
Step 4: Simplify under the square root:
(3y−4)2−4(2)(2y2−5y+ 4) = 9y2−24y+ 16 −16(2y2−5y+ 4)
= 9y2−24y+ 16 −32y2+ 80y−64 = −23y2+ 56y−48
21
Step 5: Substitute the simplified expression back into the quadratic formula:
x=−(3y−4) ±√−23y2+ 56y−48
4
Therefore, the solution to the equation 2x2+ 3xy + 2y2−4x−5y+ 4 = 0
for xin terms of yis x=−(3y−4)±√−23y2+56y−48
4.
Question 25
Question
Find the equation of the hyperbola with vertices at (−3,1) and (1,1) and foci
at (−5,1) and (3,1).
Solution
Step 1: Find the center of the hyperbola. Since the hyperbola has vertices at
(−3,1) and (1,1), the center will be the midpoint of these points.
Center =(−3+1
2,1+1
2)= (−1,1)
Step 2: Calculate the distance between the center and one of the vertices to
find a, the distance from the center to one of the vertices.
Distance between center and vertex =√(−1 + 1)2+ (1 −1)2= 4
So, a= 4.
Step 3: Since the foci are located at (−5,1) and (3,1), the value of cis the
distance between the center and one of the foci.
Distance between center and focus =√(−1 + 5)2+ (1 −1)2= 4
So, c= 4.
Step 4: Use the relationship c2=a2+b2for a hyperbola to find b.
b2=c2−a2= 42−42= 0
So, b= 0.
Step 5: Write the equation of the hyperbola in standard form (x−h)2/a2−
(y−k)2/b2= 1 where (h, k)is the center, ais the distance from the center to
the vertex, and bis the distance from the center to the co-vertex.
(x+ 1)2
16 −(y−1)2
0= 1
22
