MATH 121 - COLLEGE ALGEBRA -
Equations and properties of conic
sections
Question Bank - Set 5
Liberty University
Question 1
Question
Solve the following system of equations:
{x2+y2= 25
3x−4y= 10
Solution
Step 1: Solve the second equation for xin terms of y. Step 2: Substitute this
expression for xinto the first equation to find the values of y. Step 3: Use the
values of yto solve for x. Step 4: Check for extraneous solutions.
Step 1: Solve the second equation for x:
3x−4y= 10
3x= 4y+ 10
x=4
3y+10
3
Step 2: Substitute xinto the first equation:
(4
3y+10
3)2+y2= 25
16
9y2+80
9y+100
9+y2= 25
25
9y2+80
9y+100
9= 25
Step 3: Solve the quadratic equation:
25
9y2+80
9y+100
9−25 = 0
25
9y2+80
9y−125
9= 0
(5y−5)(5y+ 25) = 0
Therefore, y= 1 or y=−5.
For y= 1, substituting into x=4
3y+10
3gives x=4
3·1 + 10
3= 6.
For y=−5, substituting into x=4
3y+10
3gives x=4
3·(−5) + 10
3=−10
3.
Therefore, the solutions are (x, y) = (6,1) and (−10
3,−5).
Step 4: Check the solutions: Substitute (x, y) = (6,1) and (−10
3,−5) into
the original equations to verify if they hold true.
Hence, the solutions are (x, y) = (6,1) and (−10
3,−5).
Question 2
Question
Find the standard form of the equation of the circle with center at (-3, 4) and
passing through the point (1, 2).
Solution
Step 1: Recall that the standard form of the equation of a circle with center
(h, k)and radius ris (x−h)2+ (y−k)2=r2.
Step 2: First, we need to find the radius rof the circle. We can use the
distance formula to find the distance between the center (−3,4) and the point
(1,2), which will be equal to the radius.
r=√(1 −(−3))2+ (2 −4)2
r=√42+ (−2)2
r=√16 + 4
r=√20
r= 2√5
Step 3: Now we can substitute the center (−3,4) and the radius 2√5into
the standard form of the equation of a circle to obtain the equation.
(x+ 3)2+ (y−4)2= (2√5)2
(x+ 3)2+ (y−4)2= 20
Step 4: Therefore, the standard form of the equation of the circle with center
at (-3, 4) and passing through the point (1, 2) is (x+ 3)2+ (y−4)2= 20.
2
Question 3
Question
Find the standard form equation of the ellipse that satisfies the following condi-
tions: the major axis has endpoints (-2,1) and (6,1), the center is at the origin,
and the distance from the center to a focus is √13.
Solution
Step 1: First, let’s determine important information based on the given condi-
tions. Since the major axis lies on the x-axis, we have a=1
2(length of major axis) =
1
2(6 −(−2)) = 4. The distance from the center to the focus is given by
c=√a2−b2, where cis the distance from the center to a focus. We can
solve for busing this information.
Step 2: Using the formula c=√a2−b2, plug in a= 4 and c=√13 to find
b. The equation becomes: √13 = √42−b2.
Step 3: Square both sides to solve for b. This gives us 13 = 16 −b2, which
simplifies to b2= 3.
Step 4: With a= 4 and b2= 3, we can now write the standard form of
the equation of the ellipse. Since the center is at the origin, the equation is
x2
a2+y2
b2= 1.
Step 5: Substitute a= 4 and b2= 3 into the equation to get x2
16 +y2
3= 1.
Therefore, the standard form equation of the ellipse satisfying the given
conditions is x2
16 +y2
3= 1.
Question 4
Question
Find the standard form of the equation of the parabola with a focus at (−2,1)
and a directrix at x= 4.
Solution
Step 1: First, recall that the standard form of the equation of a parabola with
a vertical axis is given by (x−h)2= 4p(y−k), where (h, k)is the vertex and p
is the distance between the vertex and the focus (or directrix).
Step 2: Since the directrix is a vertical line, we can determine that the
parabola opens to the left or right and that it has a vertical axis.
Step 3: The vertex of the parabola is the midpoint between the focus and
the directrix. Thus, the vertex is at (−2+4
2,1) = (1,1).
Step 4: The distance between the focus and the vertex represents p. Given
the focus at (−2,1) and the vertex at (1,1), we find p=|1−(−2)|= 3.
3
Step 5: Since the directrix is at x= 4, we know that the parabola opens to
the left. Therefore, the standard form of the equation is (x−1)2=−12(y−1).
Step 6: Simplifying, we have (x−1)2=−12y+ 12.
Step 7: Thus, the standard form of the equation of the parabola is (x−1)2=−12y+ 12 .
Question 5
Question
Consider the conic section given by the equation 4x2−9y2−16x−18y−36 = 0.
Determine the standard form of the conic section, and then identify the type of
conic section it represents.
Solution
Step 1: To determine the standard form of the conic section, we need to complete
the square for both the xand yterms.
Step 2: Starting with the xterms, we rewrite the equation as (4x2−16x)−
9y2−18y−36 = 0.
Step 3: Completing the square for x, we have 4(x2−4x)−9y2−18y−36 = 0.
To complete the square inside the parentheses, we add and subtract (4/2)2= 4
inside the parentheses: 4[(x−2)2−4] −9y2−18y−36 = 0.
Step 4: Further simplifying, we get 4(x−2)2−16 −9y2−18y−36 = 0.
Step 5: Now, we focus on completing the square for y. We rewrite the
equation as 4(x−2)2−9(y2+ 2y)−52 = 0.
Step 6: Completing the square for y, we have 4(x−2)2−9(y2+ 2y+ 1) −
52 + 9 = 0. This simplifies to 4(x−2)2−9(y+ 1)2−43 = 0.
Step 7: Rearranging the terms, we obtain the standard form of the conic
section as 4(x−2)2−9(y+ 1)2= 43.
Step 8: Finally, we observe that the standard form of the conic section
corresponds to a hyperbola. Therefore, the given conic section represents a
hyperbola.
Question 6
Question
Determine the standard form of the equation of a parabola with a focus at
F(3,−2) and a directrix at y=−6.
Solution
Step 1: Find the vertex of the parabola. Since the vertex lies in the middle
of the focus and the directrix, the x-coordinate of the vertex will be the same
as the x-coordinate of the focus, which is 3. The y-coordinate of the vertex
4
will be the average of the y-coordinate of the focus and the y-coordinate of the
directrix, which is −2+(−6)
2=−4. Therefore, the vertex is at V(3,−4).
Step 2: Determine the distance between the focus and the vertex. Since
the focus is above the vertex, the distance is equal to the difference between
the y-coordinates of the focus and the vertex, which is | − 2−(−4)|= 2. This
distance is also equal to the distance between the vertex and the directrix.
Step 3: Identify the axis of symmetry. Since the parabola opens either
upward or downward, the axis of symmetry is parallel to the y-axis and passes
through the vertex.
Step 4: Determine if the parabola opens upward or downward. Since the
focus lies below the vertex and the directrix is above the vertex, the parabola
opens downward.
Step 5: Write the equation of the parabola in standard form. Given that
the parabola opens downward, the standard form of the equation of a parabola
is (x−h)2= 4p(y−k), where (h, k)is the vertex, and pis the distance between
the vertex and the focus (or the vertex and the directrix).
Plugging in the values, the equation of the parabola is
(x−3)2= 4(−2−(−4))(y+ 4).
Therefore, the standard form of the equation of the parabola is
(x−3)2=−8(y+ 4).
Question 7
Question
Find the standard form of the equation of the hyperbola with vertices at (−3,0)
and (3,0) and asymptotes y=±2
3x.
Solution
Step 1: Find the center of the hyperbola by using the midpoint formula:
Midpoint =(x1+x2
2,y1+y2
2)
Step 2: Use the center to write the general equation of the hyperbola:
(x−h)2
a2−(y−k)2
b2= 1
Step 3: Find aand busing the properties of the hyperbola (distance from the
center to the vertices and the slopes of the asymptotes). Step 4: Substitute the
values of h,k,a, and binto the general equation to obtain the standard form
of the equation of the hyperbola.
5
Question 8
Question
Find the standard form of the equation of the ellipse with foci at (±4,0) and
co-vertices at (0,±2).
Solution
Step 1: Determine the center of the ellipse. Since the foci are at (±4,0), the
center of the ellipse is at the midpoint of the foci. Therefore, the center is (0,0).
Step 2: Determine the lengths of the major and minor axes. The distance
between the foci is 2c= 8, so c= 4. The distance between the co-vertices is
2b= 4, so b= 2. Therefore, the major axis is 2a= 8 so a= 4.
Step 3: Write the standard form of the equation. 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. Substituting the values for h, k, a, and b, we get:
x2
16 +y2
4= 1
Question 9
Question
Solve the following system of equations in xand y:
{x2+y2= 25
x2−y2= 9
Solution
Step 1: Let’s solve the second equation for x2:
x2=y2+ 9
Step 2: Substitute x2into the first equation:
(y2+ 9) + y2= 25
Step 3: Simplify the equation:
2y2+ 9 = 25
Step 4: Subtract 9 from both sides:
2y2= 16
6
Step 5: Divide by 2:
y2= 8
Step 6: Take the square root of both sides:
y=±√8
Step 7: Simplify the square root:
y=±2√2
Step 8: Substitute the values of yback into the equation x2=y2+ 9:
x2= (2√2)2+ 9 and x2= (−2√2)2+ 9
Step 9: Simplify the equations:
x2= 13 and x2= 13
Step 10: Take the square root of both sides:
x=±√13
Step 11: Therefore, the solutions to the system of equations are:
(√13,2√2),(√13,−2√2),(−√13,2√2),(−√13,−2√2)
Question 10
Question
Solve the system of equations:
{x2+y2= 25
x−y= 3
Solution
Let’s solve this system of equations by substitution.
Step 1: Solve the second equation for xin terms of y.
x=y+ 3
Step 2: Substitute y+ 3 for xin the first equation:
(y+ 3)2+y2= 25
Step 3: Expand and simplify the equation:
y2+ 6y+9+y2= 25
7
2y2+ 6y−16 = 0
Step 4: Solve the quadratic equation for y:
y=−6±√62−4(2)(−16)
2(2)
y=−6±√36 + 128
4
y=−6±√164
4
y=−6±2√41
4
Step 5: Find the corresponding values of xby using x=y+ 3:
x1=−6+2√41
4=−3 + √41
2
x2=−6−2√41
4=−3−√41
2
Step 6: Therefore, the solutions to the system of equations are:
{x=−3+√41
2, y =−3+√41
2
x=−3−√41
2, y =−3−√41
2
Question 11
Question
Solve the following equation for x:
4x2−9 = 0
Solution
Step 1: Begin by adding 9 to both sides of the equation to isolate the term with
x2.
4x2= 9
Step 2: Divide both sides by 4 to solve for x2.
x2=9
4
Step 3: Take the square root of both sides to solve for x.
x=±√9
4
8
Step 4: Simplify the square root.
x=±3
2
Therefore, the solutions to the equation 4x2−9 = 0 are x=3
2and x=−3
2.
Question 12
Question
Solve the system of equations:
{4x2+y2= 16
x2−4y2= 16
Solution
Step 1: Rewrite the system of equations in standard form.
We rewrite the given system as:
{x2
4+y2
16 = 1
x2
16 −y2
4= 1
Step 2: Identify the conic sections represented by each equation.
The first equation represents an ellipse, while the second equation represents a
hyperbola.
Step 3: Determine the graphical relationship between the two conic sections.
Since the signs of the x2and y2terms differ, the two conic sections are orthogonal
to each other.
Step 4: Solve the system of equations using the properties of conic sections.
We can infer that the center of the ellipse is at the origin since the larger
coefficient is under y2. For the hyperbola, since the larger coefficient is under
x2, we can determine that the center is also at the origin.
Step 5: Summarize the solution.
The system of equations has two different conic sections that do not intersect.
Therefore, the system has no solution.
Question 13
Question
Solve the equation 9x2−16y2= 144 for y.
9
Solution
Step 1: Rewrite the given equation in standard form by dividing both sides by
144.
(9x2−16y2)/144 = 1
x2/16 −y2/9 = 1
Step 2: Identify the conic section represented by the equation. Since the
coefficient of x2is positive, while the coefficient of y2is negative, this equation
represents a hyperbola.
Step 3: To solve for y, isolate y2by switching the terms to the other side
and then dividing by 9.
y2=−9(x2/16 −1)
Step 4: Take the square root of both sides to solve for y.
y=±√−9(x2/16 −1)
Step 5: Simplify the square root expression.
y=±√−9x2/16 + 9
Therefore, the solution to the equation 9x2−16y2= 144 for yis y=
±√−9x2/16 + 9.
Question 14
Question
Find the equation of the hyperbola with vertices at (−5,0) and (5,0) and foci
at (−3,0) and (3,0).
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: The distance between the center and each vertex is the absolute
value of the difference between the x-coordinates. Since the vertices lie on the
x-axis, the distance between the center and each vertex is equal to 5−0 = 5.
Step 3: The distance between the center and each focus is given by the
absolute value of the difference between the x-coordinates of the foci. This
distance is equal to 3−0 = 3.
Step 4: Define aas the distance between the center and one of the vertices,
which is a= 5. Define cas the distance between the center and one of the foci,
which is c= 3.
10
Step 5: Use the relationship c2=a2+b2for a hyperbola, where bis the
distance from the center to the transverse axis. Since the transverse axis is
along the x-axis, bis along the y-axis. However, since the vertices also lie along
the x-axis, the hyperbola is vertical (bis the y-coordinate).
Step 6: Plug in the known values to solve for b:
32= 52+b2
9 = 25 + b2
b2= 9 −25
b2=−16
b=√−16 = 4i
Step 7: The equation of the hyperbola with a vertical transverse axis is:
(x−h)2
a2−(y−k)2
b2= 1
Plugging in the values:
x2
25 −y2
−16 = 1
x2
25 +y2
16 = 1
Question 15
Question
Let xand ybe real numbers such that 3x2−8xy + 3y2= 4. Determine the
center and nature of the conic section represented by this equation.
Solution
To determine the center and nature of the conic section represented by the given
equation, we’ll first complete the square to rewrite the equation in standard
form.
Step 1: Rewrite the given equation by rearranging the terms:
3x2−8xy + 3y2= 4
Step 2: Now, complete the square separately for xand yterms. For the x
terms, we have:
3x2−8xy = 4 −3y2
3(x2−8
3xy) = 4 −3y2
11
Step 3: Complete the square for xby adding and subtracting (8
6y)2inside
the parentheses:
3(x2−8
3xy + (4
3y)2)= 4 −3y2
Step 4: Simplify the right-hand side:
3(x−4
3y)2
= 4 −3y2
Step 5: Divide through by 3 to obtain the standard form of the equation:
(x−4
3y)2=4−3y2
3
Step 6: The equation is now in the standard form for a conic section cen-
tered at (4
3,0). Since the coefficient of x2and y2are the same and have opposite
signs, the conic section is a hyperbola with center (4
3,0).
Question 16
Question
Solve the following system of equations:
{3x2+ 8y2= 24
x−2y= 3
Solution
Step 1: To eliminate a variable, let’s solve the second equation for x:
x−2y= 3
x= 2y+ 3
Step 2: Now, substitute xin terms of yinto the first equation:
3(2y+ 3)2+ 8y2= 24
Step 3: Simplify the equation:
3(4y2+ 12y+9)+8y2= 24
12y2+ 36y+27+8y2= 24
20y2+ 36y+ 27 = 24
Step 4: Rearrange terms to form a quadratic equation:
20y2+ 36y+ 3 = 0
12
Step 5: Solve for yusing the quadratic formula:
y=−b±√b2−4ac
2a
where a= 20,b= 36, and c= 3.
Step 6: Substitute the values of a,b, and cinto the quadratic formula:
y=−36 ±√362−4·20 ·3
2·20
y=−36 ±√1296 −240
40
y=−36 ±√1056
40
y=−36 ±32√3
40
Step 7: Simplify the solutions for y:
y1=−36 + 32√3
40 =−9+8√3
10
y2=−36 −32√3
40 =−9−8√3
10
Step 8: Finally, substitute the values of yback into the equation x= 2y+ 3
to find the corresponding values of x. Thus, the solutions are:
(x1, y1) = (6 + 16√3
10 ,−9+8√3
10 )
(x2, y2) = (6−16√3
10 ,−9−8√3
10 )
Question 17
Question
Find the equation of the hyperbola with vertices at (3,2) and (−1,2), and
passing through the point (−2,5).
13
Solution
Step 1: Find the center of the hyperbola.
• The center of the hyperbola is the midpoint of the line segment connecting
the vertices.
• The coordinates of the center are:
(3+(−1)
2,2+2
2)= (1,2)
Step 2: Find the distance between the center and one of the vertices to
determine the value of a.
• The distance between the center (1,2) and the vertex (3,2) is √(3 −1)2+ (2 −2)2=
2.
• So, a= 2.
Step 3: Find the equation of the hyperbola.
• The general equation of a hyperbola centered at (h, k)is (x−h)2
a2−(y−k)2
b2=
1.
• Since the hyperbola passes through (−2,5), we can substitute these coor-
dinates into the equation and solve for b:
(−2−1)2
22−(5 −2)2
b2= 1
9
4−9
b2= 1
−9
b2=−5
4
b2=36
5
b=±6
√5
• The equation of the hyperbola is:
(x−1)2
4−(y−2)2
36
5
= 1
(x−1)2
4−5(y−2)2
36 = 1
14
Question 18
Question
Consider the conic section given by the equation 2x2−3xy+2y2+4x−4y+5 = 0.
Determine whether this conic section is an ellipse, a hyperbola, a parabola, or
a degenerate case. If it is an ellipse or a hyperbola, find its standard form.
Solution
Step 1: To determine the type of conic section, we need to calculate the dis-
criminant ∆ = B2−4AC, where A,B, and Care the coefficients of x2,xy, and
y2respectively. Calculating A= 2,B=−3, and C= 2, we have:
∆ = (−3)2−4(2)(2) = 9 −16 = −7
Step 2: Since ∆<0, the conic section is neither an ellipse nor a hyperbola.
To determine if it is a parabola or a degenerate case, we analyze its conic form.
The general form of a conic section is given by Ax2+Bxy+Cy2+Dx+Ey+F=
0.
Step 3: We complete the square to simplify the equation. Group the xterms
and yterms together and complete the square for both variables:
2(x2+ 2x)−3xy + 2(y2−2y) = −5
2(x+ 1)2−1−2(y−1)2−1 = −5
2(x+ 1)2−2(y−1)2= 3
Step 4: Comparing our completed square form to the standard equation
for conic sections, we see that this conic section is a degenerate case of two
intersecting lines since there are no xy terms and the coefficients of x2and y2
have opposite signs.
Question 19
Question
Find the standard form equation of the hyperbola that satisfies the given con-
ditions: vertices at (5,2) and (9,2), and passes through the point (2,6).
Solution
Step 1: Determine the center of the hyperbola by finding the midpoint of the
line segment connecting the vertices.
Midpoint =(5+9
2,2+2
2)= (7,2)
15
Step 2: Calculate the distance between the center and one of the vertices to
find the value of a.
a=distance between the vertices
2=9−5
2= 2
Step 3: Use the distance formula to find the value of c, which is the distance
between the center and one of the foci.
c=√a2+b2
Step 4: Write the general equation of a hyperbola in standard form.
(x−h)2
a2−(y−k)2
b2= 1
Step 5: Find the value of busing the relationship between a,b, and cfor a
hyperbola: c2=a2+b2.
Step 6: Substitute the known values into the standard form equation to find
the specific equation of the hyperbola.
Step 7: Use the fact that the hyperbola passes through the point (2,6) to
solve for the remaining unknowns and finalize the equation of the hyperbola.
Question 20
Question
Solve the following system of equations for xand y:
{x2−y2= 4
x2+y2= 10
Solution
Step 1: First, rewrite the system of equations:
{x2−y2= 4
x2+y2= 10
Step 2: Add the two equations together to eliminate y2:
2x2= 14
Step 3: Divide by 2 to find the value of x2:
x2= 7
Step 4: Substitute x2= 7 into one of the original equations (e.g. x2−y2= 4)
to solve for y2:
7−y2= 4 =⇒y2= 3
16
Step 5: Take the square root of both sides to find the possible values for y:
y=±√3
Step 6: Therefore, the solutions to the system of equations are:
x=±√7and y=±√3
Question 21
Question
Find the standard form equation of the hyperbola that satisfies the following
conditions: the distance between its foci F1and F2is 10, the distance between
the vertices V1and V2is 14, and one of its asymptotes has the equation y= 2x.
Solution
Step 1: Recall the standard form of the equation for a hyperbola centered at
the origin with transverse axis along the x-axis is
(x
a)2−(y
b)2= 1
where ais half the distance between the vertices and bis half the distance
between the foci.
Step 2: We are given that the distance between the vertices, V1and V2, is
14. This means 2a= 14, so a= 7.
Step 3: We are given that the distance between the foci, F1and F2, is 10.
This means 2ae = 10, where eis the eccentricity of the hyperbola.
Step 4: Since one of the asymptotes has the equation y= 2x, we know that
the slopes of the asymptotes are ±b
a, so b
a= 2.
Step 5: From step 4, we have b
7= 2, so b= 14.
Step 6: Substitute a= 7 and b= 14 into the standard form equation of the
hyperbola to get
(x
7)2−(y
14)2= 1
Step 7: The standard equation for a hyperbola centered at the origin but
with the transverse axis along the y-axis is
(y
a)2−(x
b)2= 1
so the equation for the given hyperbola can be rewritten as
(y
14)2−(x
7)2= 1
Therefore, the standard form equation of the hyperbola that satisfies the
given conditions is
(y
14)2−(x
7)2= 1
17
Question 22
Question
Solve the system of equations:
{x2+y2= 25
4x+ 3y= 10
Solution
Step 1: The first equation x2+y2= 25 represents a circle centered at the origin
with radius 5.
Step 2: We can solve the second equation 4x+3y= 10 for xto get x=10−3y
4.
Step 3: Substitute x=10−3y
4into x2+y2= 25:
(10 −3y
4)2
+y2= 25
Step 4: Simplify the equation:
(10 −3y)2
16 +y2= 25
Step 5: Expand and simplify:
100 −60y+ 9y2
16 +y2= 25
Step 6: Multiply both sides by 16 to eliminate the denominator:
100 −60y+ 9y2+ 16y2= 400
Step 7: Combine like terms:
25y2−60y−300 = 0
Step 8: Divide by 25 to simplify the equation:
y2−60
25y−300
25 = 0
Step 9: Simplify further:
y2−2.4y−12 = 0
Step 10: Use the quadratic formula to solve for y:
y=−(−2.4) ±√(−2.4)2−4(1)(−12)
2(1)
Step 11: Calculate the solutions for yand then substitute back to find the
corresponding xvalues.
18
Question 23
Question
Consider the ellipse with the equation x2
4+y2
9= 1. Find the foci of the ellipse.
Solution
Step 1: Recall that the standard form of an ellipse centered at the origin is
x2
a2+y2
b2= 1, where ais the semi-major axis and bis the semi-minor axis.
Step 2: By comparing the given equation x2
4+y2
9= 1 with the standard
form, we can see that a2= 4 and b2= 9.
Step 3: The distance from the center to the foci is given by c=√a2−b2.
Step 4: Plugging in the values of a2= 4 and b2= 9 into the formula for c,
we get c=√4−9 = √−5.
Step 5: Since the square root of a negative number is imaginary, the foci of
the ellipse are at points (0,±i√5).
Question 24
Question
Solve the following system of equations:
{x2+y2= 25
2x−3y= 5
Solution
Step 1: Start by solving the second equation for xin terms of y:
2x−3y= 5
2x= 3y+ 5
x=3y+ 5
2
Step 2: Substitute this expression for xinto the first equation:
(3y+ 5
2)2
+y2= 25
9y2+ 30y+ 25
4+y2= 25
Step 3: Multiply by 4 to get rid of the fractions:
9y2+ 30y+25+4y2= 100
19
13y2+ 30y+ 25 = 100
Step 4: Rearrange the equation into standard form:
13y2+ 30y+ 25 −100 = 0
13y2+ 30y−75 = 0
Step 5: Now, we can solve this quadratic equation for y. The solutions will
be:
y=−30 ±√302−4(13)(−75)
2(13)
y=−30 ±√900 + 3900
26
y=−30 ±√4800
26
y=−30 ±20√3
26
y=−15 ±10√3
13
Step 6: Substitute the values of yback into the equation we found for xin
terms of yto find the corresponding values of x.
Therefore, the solutions to the system of equations are:
(−15 + 10√3
13 ,2(−15 + 10√3) + 5
3)and (−15 −10√3
13 ,2(−15 −10√3) + 5
3)
Question 25
Question
Find the equation of the circle that passes through the points A(3,2) and
B(−1,−4).
Solution
Step 1: Find the midpoint of the line segment connecting points Aand B. Let
this midpoint be point M.
Step 2: Find the slope of the line passing through points Aand B. This will
be the negative reciprocal of the slope of the line passing through the center of
the circle, M, and the midpoint of the chord formed by connecting Aand B
through the center of the circle.
Step 3: Using the midpoint, M, and the slope of the line passing through
points Aand B, find the equation of the circle passing through points Aand B.
Step 4: Write down the equation of the circle in standard form.
20
Let’s complete these steps in detail.
Step 1: Find the midpoint of the line segment connecting points Aand B.
Let this midpoint be point M. The midpoint formula is given by:
M(x1+x2
2,y1+y2
2)
Substitute the coordinates of points A(3,2) and B(−1,−4) into the formula:
M(3+(−1)
2,2+(−4)
2)=M(2
2,−2
2)=M(1,−1)
So, the midpoint Mis M(1,−1).
Step 2: Find the slope of the line passing through points Aand B. The
slope formula is given by:
m=y2−y1
x2−x1
Substitute the coordinates of points A(3,2) and B(−1,−4) into the formula:
m=−4−2
−1−3=−6
−4=3
2
Thus, the slope of the line passing through points Aand Bis 3
2.
Step 3: Using the midpoint, M(1,−1), and the slope m=3
2, find the
equation of the circle passing through points Aand B. The equation of the
circle can be written in the form (x−h)2+ (y−k)2=r2, where (h, k)is the
center of the circle and ris the radius of the circle.
Since the circle passes through point M(1,−1), the coordinates of the center
are (h, k) = (1,−1).
The negative reciprocal of the slope m=3
2is −2
3, which represents the slope
of the perpendicular bisector of the chord AB passing through M(1,−1).
Since the equation of a line passing through point (1,−1) with slope −2
3that
is perpendicular to the line passing through Aand B(m=3
2) has the equation:
y−(−1) = −2
3(x−1)
On simplifying, we get:
y+ 1 = −2
3x+2
3
y=−2
3x−5
3
Let this line intersect the circle at points Dand E. As this line is the
perpendicular bisector of AB, we can find the distance from Ato Mand set it
equal to the distance from Ato the circle to find r.
r=√(1 −3)2+ (−1−2)2=√(−2)2+ (−3)2=√13
Thus, the equation of the circle is (x−1)2+ (y+ 1)2= 13.
21
25
9y2+80
9y+100
9= 25
Step 3: Solve the quadratic equation:
25
9y2+80
9y+100
9−25 = 0
25
9y2+80
9y−125
9= 0
(5y−5)(5y+ 25) = 0
Therefore, y= 1 or y=−5.
For y= 1, substituting into x=4
3y+10
3gives x=4
3·1 + 10
3= 6.
For y=−5, substituting into x=4
3y+10
3gives x=4
3·(−5) + 10
3=−10
3.
Therefore, the solutions are (x, y) = (6,1) and (−10
3,−5).
Step 4: Check the solutions: Substitute (x, y) = (6,1) and (−10
3,−5) into
the original equations to verify if they hold true.
Hence, the solutions are (x, y) = (6,1) and (−10
3,−5).
Question 2
Question
Find the standard form of the equation of the circle with center at (-3, 4) and
passing through the point (1, 2).
Solution
Step 1: Recall that the standard form of the equation of a circle with center
(h, k)and radius ris (x−h)2+ (y−k)2=r2.
Step 2: First, we need to find the radius rof the circle. We can use the
distance formula to find the distance between the center (−3,4) and the point
(1,2), which will be equal to the radius.
r=√(1 −(−3))2+ (2 −4)2
r=√42+ (−2)2
r=√16 + 4
r=√20
r= 2√5
Step 3: Now we can substitute the center (−3,4) and the radius 2√5into
the standard form of the equation of a circle to obtain the equation.
(x+ 3)2+ (y−4)2= (2√5)2
(x+ 3)2+ (y−4)2= 20
Step 4: Therefore, the standard form of the equation of the circle with center
at (-3, 4) and passing through the point (1, 2) is (x+ 3)2+ (y−4)2= 20.
2
Question 3
Question
Find the standard form equation of the ellipse that satisfies the following condi-
tions: the major axis has endpoints (-2,1) and (6,1), the center is at the origin,
and the distance from the center to a focus is √13.
Solution
Step 1: First, let’s determine important information based on the given condi-
tions. Since the major axis lies on the x-axis, we have a=1
2(length of major axis) =
1
2(6 −(−2)) = 4. The distance from the center to the focus is given by
c=√a2−b2, where cis the distance from the center to a focus. We can
solve for busing this information.
Step 2: Using the formula c=√a2−b2, plug in a= 4 and c=√13 to find
b. The equation becomes: √13 = √42−b2.
Step 3: Square both sides to solve for b. This gives us 13 = 16 −b2, which
simplifies to b2= 3.
Step 4: With a= 4 and b2= 3, we can now write the standard form of
the equation of the ellipse. Since the center is at the origin, the equation is
x2
a2+y2
b2= 1.
Step 5: Substitute a= 4 and b2= 3 into the equation to get x2
16 +y2
3= 1.
Therefore, the standard form equation of the ellipse satisfying the given
conditions is x2
16 +y2
3= 1.
Question 4
Question
Find the standard form of the equation of the parabola with a focus at (−2,1)
and a directrix at x= 4.
Solution
Step 1: First, recall that the standard form of the equation of a parabola with
a vertical axis is given by (x−h)2= 4p(y−k), where (h, k)is the vertex and p
is the distance between the vertex and the focus (or directrix).
Step 2: Since the directrix is a vertical line, we can determine that the
parabola opens to the left or right and that it has a vertical axis.
Step 3: The vertex of the parabola is the midpoint between the focus and
the directrix. Thus, the vertex is at (−2+4
2,1) = (1,1).
Step 4: The distance between the focus and the vertex represents p. Given
the focus at (−2,1) and the vertex at (1,1), we find p=|1−(−2)|= 3.
3
Step 5: Since the directrix is at x= 4, we know that the parabola opens to
the left. Therefore, the standard form of the equation is (x−1)2=−12(y−1).
Step 6: Simplifying, we have (x−1)2=−12y+ 12.
Step 7: Thus, the standard form of the equation of the parabola is (x−1)2=−12y+ 12 .
Question 5
Question
Consider the conic section given by the equation 4x2−9y2−16x−18y−36 = 0.
Determine the standard form of the conic section, and then identify the type of
conic section it represents.
Solution
Step 1: To determine the standard form of the conic section, we need to complete
the square for both the xand yterms.
Step 2: Starting with the xterms, we rewrite the equation as (4x2−16x)−
9y2−18y−36 = 0.
Step 3: Completing the square for x, we have 4(x2−4x)−9y2−18y−36 = 0.
To complete the square inside the parentheses, we add and subtract (4/2)2= 4
inside the parentheses: 4[(x−2)2−4] −9y2−18y−36 = 0.
Step 4: Further simplifying, we get 4(x−2)2−16 −9y2−18y−36 = 0.
Step 5: Now, we focus on completing the square for y. We rewrite the
equation as 4(x−2)2−9(y2+ 2y)−52 = 0.
Step 6: Completing the square for y, we have 4(x−2)2−9(y2+ 2y+ 1) −
52 + 9 = 0. This simplifies to 4(x−2)2−9(y+ 1)2−43 = 0.
Step 7: Rearranging the terms, we obtain the standard form of the conic
section as 4(x−2)2−9(y+ 1)2= 43.
Step 8: Finally, we observe that the standard form of the conic section
corresponds to a hyperbola. Therefore, the given conic section represents a
hyperbola.
Question 6
Question
Determine the standard form of the equation of a parabola with a focus at
F(3,−2) and a directrix at y=−6.
Solution
Step 1: Find the vertex of the parabola. Since the vertex lies in the middle
of the focus and the directrix, the x-coordinate of the vertex will be the same
as the x-coordinate of the focus, which is 3. The y-coordinate of the vertex
4
will be the average of the y-coordinate of the focus and the y-coordinate of the
directrix, which is −2+(−6)
2=−4. Therefore, the vertex is at V(3,−4).
Step 2: Determine the distance between the focus and the vertex. Since
the focus is above the vertex, the distance is equal to the difference between
the y-coordinates of the focus and the vertex, which is | − 2−(−4)|= 2. This
distance is also equal to the distance between the vertex and the directrix.
Step 3: Identify the axis of symmetry. Since the parabola opens either
upward or downward, the axis of symmetry is parallel to the y-axis and passes
through the vertex.
Step 4: Determine if the parabola opens upward or downward. Since the
focus lies below the vertex and the directrix is above the vertex, the parabola
opens downward.
Step 5: Write the equation of the parabola in standard form. Given that
the parabola opens downward, the standard form of the equation of a parabola
is (x−h)2= 4p(y−k), where (h, k)is the vertex, and pis the distance between
the vertex and the focus (or the vertex and the directrix).
Plugging in the values, the equation of the parabola is
(x−3)2= 4(−2−(−4))(y+ 4).
Therefore, the standard form of the equation of the parabola is
(x−3)2=−8(y+ 4).
Question 7
Question
Find the standard form of the equation of the hyperbola with vertices at (−3,0)
and (3,0) and asymptotes y=±2
3x.
Solution
Step 1: Find the center of the hyperbola by using the midpoint formula:
Midpoint =(x1+x2
2,y1+y2
2)
Step 2: Use the center to write the general equation of the hyperbola:
(x−h)2
a2−(y−k)2
b2= 1
Step 3: Find aand busing the properties of the hyperbola (distance from the
center to the vertices and the slopes of the asymptotes). Step 4: Substitute the
values of h,k,a, and binto the general equation to obtain the standard form
of the equation of the hyperbola.
5
Question 8
Question
Find the standard form of the equation of the ellipse with foci at (±4,0) and
co-vertices at (0,±2).
Solution
Step 1: Determine the center of the ellipse. Since the foci are at (±4,0), the
center of the ellipse is at the midpoint of the foci. Therefore, the center is (0,0).
Step 2: Determine the lengths of the major and minor axes. The distance
between the foci is 2c= 8, so c= 4. The distance between the co-vertices is
2b= 4, so b= 2. Therefore, the major axis is 2a= 8 so a= 4.
Step 3: Write the standard form of the equation. 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. Substituting the values for h, k, a, and b, we get:
x2
16 +y2
4= 1
Question 9
Question
Solve the following system of equations in xand y:
{x2+y2= 25
x2−y2= 9
Solution
Step 1: Let’s solve the second equation for x2:
x2=y2+ 9
Step 2: Substitute x2into the first equation:
(y2+ 9) + y2= 25
Step 3: Simplify the equation:
2y2+ 9 = 25
Step 4: Subtract 9 from both sides:
2y2= 16
6
Step 5: Divide by 2:
y2= 8
Step 6: Take the square root of both sides:
y=±√8
Step 7: Simplify the square root:
y=±2√2
Step 8: Substitute the values of yback into the equation x2=y2+ 9:
x2= (2√2)2+ 9 and x2= (−2√2)2+ 9
Step 9: Simplify the equations:
x2= 13 and x2= 13
Step 10: Take the square root of both sides:
x=±√13
Step 11: Therefore, the solutions to the system of equations are:
(√13,2√2),(√13,−2√2),(−√13,2√2),(−√13,−2√2)
Question 10
Question
Solve the system of equations:
{x2+y2= 25
x−y= 3
Solution
Let’s solve this system of equations by substitution.
Step 1: Solve the second equation for xin terms of y.
x=y+ 3
Step 2: Substitute y+ 3 for xin the first equation:
(y+ 3)2+y2= 25
Step 3: Expand and simplify the equation:
y2+ 6y+9+y2= 25
7
2y2+ 6y−16 = 0
Step 4: Solve the quadratic equation for y:
y=−6±√62−4(2)(−16)
2(2)
y=−6±√36 + 128
4
y=−6±√164
4
y=−6±2√41
4
Step 5: Find the corresponding values of xby using x=y+ 3:
x1=−6+2√41
4=−3 + √41
2
x2=−6−2√41
4=−3−√41
2
Step 6: Therefore, the solutions to the system of equations are:
{x=−3+√41
2, y =−3+√41
2
x=−3−√41
2, y =−3−√41
2
Question 11
Question
Solve the following equation for x:
4x2−9 = 0
Solution
Step 1: Begin by adding 9 to both sides of the equation to isolate the term with
x2.
4x2= 9
Step 2: Divide both sides by 4 to solve for x2.
x2=9
4
Step 3: Take the square root of both sides to solve for x.
x=±√9
4
8
Step 4: Simplify the square root.
x=±3
2
Therefore, the solutions to the equation 4x2−9 = 0 are x=3
2and x=−3
2.
Question 12
Question
Solve the system of equations:
{4x2+y2= 16
x2−4y2= 16
Solution
Step 1: Rewrite the system of equations in standard form.
We rewrite the given system as:
{x2
4+y2
16 = 1
x2
16 −y2
4= 1
Step 2: Identify the conic sections represented by each equation.
The first equation represents an ellipse, while the second equation represents a
hyperbola.
Step 3: Determine the graphical relationship between the two conic sections.
Since the signs of the x2and y2terms differ, the two conic sections are orthogonal
to each other.
Step 4: Solve the system of equations using the properties of conic sections.
We can infer that the center of the ellipse is at the origin since the larger
coefficient is under y2. For the hyperbola, since the larger coefficient is under
x2, we can determine that the center is also at the origin.
Step 5: Summarize the solution.
The system of equations has two different conic sections that do not intersect.
Therefore, the system has no solution.
Question 13
Question
Solve the equation 9x2−16y2= 144 for y.
9
Solution
Step 1: Rewrite the given equation in standard form by dividing both sides by
144.
(9x2−16y2)/144 = 1
x2/16 −y2/9 = 1
Step 2: Identify the conic section represented by the equation. Since the
coefficient of x2is positive, while the coefficient of y2is negative, this equation
represents a hyperbola.
Step 3: To solve for y, isolate y2by switching the terms to the other side
and then dividing by 9.
y2=−9(x2/16 −1)
Step 4: Take the square root of both sides to solve for y.
y=±√−9(x2/16 −1)
Step 5: Simplify the square root expression.
y=±√−9x2/16 + 9
Therefore, the solution to the equation 9x2−16y2= 144 for yis y=
±√−9x2/16 + 9.
Question 14
Question
Find the equation of the hyperbola with vertices at (−5,0) and (5,0) and foci
at (−3,0) and (3,0).
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: The distance between the center and each vertex is the absolute
value of the difference between the x-coordinates. Since the vertices lie on the
x-axis, the distance between the center and each vertex is equal to 5−0 = 5.
Step 3: The distance between the center and each focus is given by the
absolute value of the difference between the x-coordinates of the foci. This
distance is equal to 3−0 = 3.
Step 4: Define aas the distance between the center and one of the vertices,
which is a= 5. Define cas the distance between the center and one of the foci,
which is c= 3.
10
Step 5: Use the relationship c2=a2+b2for a hyperbola, where bis the
distance from the center to the transverse axis. Since the transverse axis is
along the x-axis, bis along the y-axis. However, since the vertices also lie along
the x-axis, the hyperbola is vertical (bis the y-coordinate).
Step 6: Plug in the known values to solve for b:
32= 52+b2
9 = 25 + b2
b2= 9 −25
b2=−16
b=√−16 = 4i
Step 7: The equation of the hyperbola with a vertical transverse axis is:
(x−h)2
a2−(y−k)2
b2= 1
Plugging in the values:
x2
25 −y2
−16 = 1
x2
25 +y2
16 = 1
Question 15
Question
Let xand ybe real numbers such that 3x2−8xy + 3y2= 4. Determine the
center and nature of the conic section represented by this equation.
Solution
To determine the center and nature of the conic section represented by the given
equation, we’ll first complete the square to rewrite the equation in standard
form.
Step 1: Rewrite the given equation by rearranging the terms:
3x2−8xy + 3y2= 4
Step 2: Now, complete the square separately for xand yterms. For the x
terms, we have:
3x2−8xy = 4 −3y2
3(x2−8
3xy) = 4 −3y2
11
Step 3: Complete the square for xby adding and subtracting (8
6y)2inside
the parentheses:
3(x2−8
3xy + (4
3y)2)= 4 −3y2
Step 4: Simplify the right-hand side:
3(x−4
3y)2
= 4 −3y2
Step 5: Divide through by 3 to obtain the standard form of the equation:
(x−4
3y)2=4−3y2
3
Step 6: The equation is now in the standard form for a conic section cen-
tered at (4
3,0). Since the coefficient of x2and y2are the same and have opposite
signs, the conic section is a hyperbola with center (4
3,0).
Question 16
Question
Solve the following system of equations:
{3x2+ 8y2= 24
x−2y= 3
Solution
Step 1: To eliminate a variable, let’s solve the second equation for x:
x−2y= 3
x= 2y+ 3
Step 2: Now, substitute xin terms of yinto the first equation:
3(2y+ 3)2+ 8y2= 24
Step 3: Simplify the equation:
3(4y2+ 12y+9)+8y2= 24
12y2+ 36y+27+8y2= 24
20y2+ 36y+ 27 = 24
Step 4: Rearrange terms to form a quadratic equation:
20y2+ 36y+ 3 = 0
12
Step 5: Solve for yusing the quadratic formula:
y=−b±√b2−4ac
2a
where a= 20,b= 36, and c= 3.
Step 6: Substitute the values of a,b, and cinto the quadratic formula:
y=−36 ±√362−4·20 ·3
2·20
y=−36 ±√1296 −240
40
y=−36 ±√1056
40
y=−36 ±32√3
40
Step 7: Simplify the solutions for y:
y1=−36 + 32√3
40 =−9+8√3
10
y2=−36 −32√3
40 =−9−8√3
10
Step 8: Finally, substitute the values of yback into the equation x= 2y+ 3
to find the corresponding values of x. Thus, the solutions are:
(x1, y1) = (6 + 16√3
10 ,−9+8√3
10 )
(x2, y2) = (6−16√3
10 ,−9−8√3
10 )
Question 17
Question
Find the equation of the hyperbola with vertices at (3,2) and (−1,2), and
passing through the point (−2,5).
13
Solution
Step 1: Find the center of the hyperbola.
• The center of the hyperbola is the midpoint of the line segment connecting
the vertices.
• The coordinates of the center are:
(3+(−1)
2,2+2
2)= (1,2)
Step 2: Find the distance between the center and one of the vertices to
determine the value of a.
• The distance between the center (1,2) and the vertex (3,2) is √(3 −1)2+ (2 −2)2=
2.
• So, a= 2.
Step 3: Find the equation of the hyperbola.
• The general equation of a hyperbola centered at (h, k)is (x−h)2
a2−(y−k)2
b2=
1.
• Since the hyperbola passes through (−2,5), we can substitute these coor-
dinates into the equation and solve for b:
(−2−1)2
22−(5 −2)2
b2= 1
9
4−9
b2= 1
−9
b2=−5
4
b2=36
5
b=±6
√5
• The equation of the hyperbola is:
(x−1)2
4−(y−2)2
36
5
= 1
(x−1)2
4−5(y−2)2
36 = 1
14
Question 18
Question
Consider the conic section given by the equation 2x2−3xy+2y2+4x−4y+5 = 0.
Determine whether this conic section is an ellipse, a hyperbola, a parabola, or
a degenerate case. If it is an ellipse or a hyperbola, find its standard form.
Solution
Step 1: To determine the type of conic section, we need to calculate the dis-
criminant ∆ = B2−4AC, where A,B, and Care the coefficients of x2,xy, and
y2respectively. Calculating A= 2,B=−3, and C= 2, we have:
∆ = (−3)2−4(2)(2) = 9 −16 = −7
Step 2: Since ∆<0, the conic section is neither an ellipse nor a hyperbola.
To determine if it is a parabola or a degenerate case, we analyze its conic form.
The general form of a conic section is given by Ax2+Bxy+Cy2+Dx+Ey+F=
0.
Step 3: We complete the square to simplify the equation. Group the xterms
and yterms together and complete the square for both variables:
2(x2+ 2x)−3xy + 2(y2−2y) = −5
2(x+ 1)2−1−2(y−1)2−1 = −5
2(x+ 1)2−2(y−1)2= 3
Step 4: Comparing our completed square form to the standard equation
for conic sections, we see that this conic section is a degenerate case of two
intersecting lines since there are no xy terms and the coefficients of x2and y2
have opposite signs.
Question 19
Question
Find the standard form equation of the hyperbola that satisfies the given con-
ditions: vertices at (5,2) and (9,2), and passes through the point (2,6).
Solution
Step 1: Determine the center of the hyperbola by finding the midpoint of the
line segment connecting the vertices.
Midpoint =(5+9
2,2+2
2)= (7,2)
15
Step 2: Calculate the distance between the center and one of the vertices to
find the value of a.
a=distance between the vertices
2=9−5
2= 2
Step 3: Use the distance formula to find the value of c, which is the distance
between the center and one of the foci.
c=√a2+b2
Step 4: Write the general equation of a hyperbola in standard form.
(x−h)2
a2−(y−k)2
b2= 1
Step 5: Find the value of busing the relationship between a,b, and cfor a
hyperbola: c2=a2+b2.
Step 6: Substitute the known values into the standard form equation to find
the specific equation of the hyperbola.
Step 7: Use the fact that the hyperbola passes through the point (2,6) to
solve for the remaining unknowns and finalize the equation of the hyperbola.
Question 20
Question
Solve the following system of equations for xand y:
{x2−y2= 4
x2+y2= 10
Solution
Step 1: First, rewrite the system of equations:
{x2−y2= 4
x2+y2= 10
Step 2: Add the two equations together to eliminate y2:
2x2= 14
Step 3: Divide by 2 to find the value of x2:
x2= 7
Step 4: Substitute x2= 7 into one of the original equations (e.g. x2−y2= 4)
to solve for y2:
7−y2= 4 =⇒y2= 3
16
Step 5: Take the square root of both sides to find the possible values for y:
y=±√3
Step 6: Therefore, the solutions to the system of equations are:
x=±√7and y=±√3
Question 21
Question
Find the standard form equation of the hyperbola that satisfies the following
conditions: the distance between its foci F1and F2is 10, the distance between
the vertices V1and V2is 14, and one of its asymptotes has the equation y= 2x.
Solution
Step 1: Recall the standard form of the equation for a hyperbola centered at
the origin with transverse axis along the x-axis is
(x
a)2−(y
b)2= 1
where ais half the distance between the vertices and bis half the distance
between the foci.
Step 2: We are given that the distance between the vertices, V1and V2, is
14. This means 2a= 14, so a= 7.
Step 3: We are given that the distance between the foci, F1and F2, is 10.
This means 2ae = 10, where eis the eccentricity of the hyperbola.
Step 4: Since one of the asymptotes has the equation y= 2x, we know that
the slopes of the asymptotes are ±b
a, so b
a= 2.
Step 5: From step 4, we have b
7= 2, so b= 14.
Step 6: Substitute a= 7 and b= 14 into the standard form equation of the
hyperbola to get
(x
7)2−(y
14)2= 1
Step 7: The standard equation for a hyperbola centered at the origin but
with the transverse axis along the y-axis is
(y
a)2−(x
b)2= 1
so the equation for the given hyperbola can be rewritten as
(y
14)2−(x
7)2= 1
Therefore, the standard form equation of the hyperbola that satisfies the
given conditions is
(y
14)2−(x
7)2= 1
17
Question 22
Question
Solve the system of equations:
{x2+y2= 25
4x+ 3y= 10
Solution
Step 1: The first equation x2+y2= 25 represents a circle centered at the origin
with radius 5.
Step 2: We can solve the second equation 4x+3y= 10 for xto get x=10−3y
4.
Step 3: Substitute x=10−3y
4into x2+y2= 25:
(10 −3y
4)2
+y2= 25
Step 4: Simplify the equation:
(10 −3y)2
16 +y2= 25
Step 5: Expand and simplify:
100 −60y+ 9y2
16 +y2= 25
Step 6: Multiply both sides by 16 to eliminate the denominator:
100 −60y+ 9y2+ 16y2= 400
Step 7: Combine like terms:
25y2−60y−300 = 0
Step 8: Divide by 25 to simplify the equation:
y2−60
25y−300
25 = 0
Step 9: Simplify further:
y2−2.4y−12 = 0
Step 10: Use the quadratic formula to solve for y:
y=−(−2.4) ±√(−2.4)2−4(1)(−12)
2(1)
Step 11: Calculate the solutions for yand then substitute back to find the
corresponding xvalues.
18
Question 23
Question
Consider the ellipse with the equation x2
4+y2
9= 1. Find the foci of the ellipse.
Solution
Step 1: Recall that the standard form of an ellipse centered at the origin is
x2
a2+y2
b2= 1, where ais the semi-major axis and bis the semi-minor axis.
Step 2: By comparing the given equation x2
4+y2
9= 1 with the standard
form, we can see that a2= 4 and b2= 9.
Step 3: The distance from the center to the foci is given by c=√a2−b2.
Step 4: Plugging in the values of a2= 4 and b2= 9 into the formula for c,
we get c=√4−9 = √−5.
Step 5: Since the square root of a negative number is imaginary, the foci of
the ellipse are at points (0,±i√5).
Question 24
Question
Solve the following system of equations:
{x2+y2= 25
2x−3y= 5
Solution
Step 1: Start by solving the second equation for xin terms of y:
2x−3y= 5
2x= 3y+ 5
x=3y+ 5
2
Step 2: Substitute this expression for xinto the first equation:
(3y+ 5
2)2
+y2= 25
9y2+ 30y+ 25
4+y2= 25
Step 3: Multiply by 4 to get rid of the fractions:
9y2+ 30y+25+4y2= 100
19
13y2+ 30y+ 25 = 100
Step 4: Rearrange the equation into standard form:
13y2+ 30y+ 25 −100 = 0
13y2+ 30y−75 = 0
Step 5: Now, we can solve this quadratic equation for y. The solutions will
be:
y=−30 ±√302−4(13)(−75)
2(13)
y=−30 ±√900 + 3900
26
y=−30 ±√4800
26
y=−30 ±20√3
26
y=−15 ±10√3
13
Step 6: Substitute the values of yback into the equation we found for xin
terms of yto find the corresponding values of x.
Therefore, the solutions to the system of equations are:
(−15 + 10√3
13 ,2(−15 + 10√3) + 5
3)and (−15 −10√3
13 ,2(−15 −10√3) + 5
3)
Question 25
Question
Find the equation of the circle that passes through the points A(3,2) and
B(−1,−4).
Solution
Step 1: Find the midpoint of the line segment connecting points Aand B. Let
this midpoint be point M.
Step 2: Find the slope of the line passing through points Aand B. This will
be the negative reciprocal of the slope of the line passing through the center of
the circle, M, and the midpoint of the chord formed by connecting Aand B
through the center of the circle.
Step 3: Using the midpoint, M, and the slope of the line passing through
points Aand B, find the equation of the circle passing through points Aand B.
Step 4: Write down the equation of the circle in standard form.
20
Let’s complete these steps in detail.
Step 1: Find the midpoint of the line segment connecting points Aand B.
Let this midpoint be point M. The midpoint formula is given by:
M(x1+x2
2,y1+y2
2)
Substitute the coordinates of points A(3,2) and B(−1,−4) into the formula:
M(3+(−1)
2,2+(−4)
2)=M(2
2,−2
2)=M(1,−1)
So, the midpoint Mis M(1,−1).
Step 2: Find the slope of the line passing through points Aand B. The
slope formula is given by:
m=y2−y1
x2−x1
Substitute the coordinates of points A(3,2) and B(−1,−4) into the formula:
m=−4−2
−1−3=−6
−4=3
2
Thus, the slope of the line passing through points Aand Bis 3
2.
Step 3: Using the midpoint, M(1,−1), and the slope m=3
2, find the
equation of the circle passing through points Aand B. The equation of the
circle can be written in the form (x−h)2+ (y−k)2=r2, where (h, k)is the
center of the circle and ris the radius of the circle.
Since the circle passes through point M(1,−1), the coordinates of the center
are (h, k) = (1,−1).
The negative reciprocal of the slope m=3
2is −2
3, which represents the slope
of the perpendicular bisector of the chord AB passing through M(1,−1).
Since the equation of a line passing through point (1,−1) with slope −2
3that
is perpendicular to the line passing through Aand B(m=3
2) has the equation:
y−(−1) = −2
3(x−1)
On simplifying, we get:
y+ 1 = −2
3x+2
3
y=−2
3x−5
3
Let this line intersect the circle at points Dand E. As this line is the
perpendicular bisector of AB, we can find the distance from Ato Mand set it
equal to the distance from Ato the circle to find r.
r=√(1 −3)2+ (−1−2)2=√(−2)2+ (−3)2=√13
Thus, the equation of the circle is (x−1)2+ (y+ 1)2= 13.
21
25
9y2+80
9y+100
9= 25
Step 3: Solve the quadratic equation:
25
9y2+80
9y+100
9−25 = 0
25
9y2+80
9y−125
9= 0
(5y−5)(5y+ 25) = 0
Therefore, y= 1 or y=−5.
For y= 1, substituting into x=4
3y+10
3gives x=4
3·1 + 10
3= 6.
For y=−5, substituting into x=4
3y+10
3gives x=4
3·(−5) + 10
3=−10
3.
Therefore, the solutions are (x, y) = (6,1) and (−10
3,−5).
Step 4: Check the solutions: Substitute (x, y) = (6,1) and (−10
3,−5) into
the original equations to verify if they hold true.
Hence, the solutions are (x, y) = (6,1) and (−10
3,−5).
Question 2
Question
Find the standard form of the equation of the circle with center at (-3, 4) and
passing through the point (1, 2).
Solution
Step 1: Recall that the standard form of the equation of a circle with center
(h, k)and radius ris (x−h)2+ (y−k)2=r2.
Step 2: First, we need to find the radius rof the circle. We can use the
distance formula to find the distance between the center (−3,4) and the point
(1,2), which will be equal to the radius.
r=√(1 −(−3))2+ (2 −4)2
r=√42+ (−2)2
r=√16 + 4
r=√20
r= 2√5
Step 3: Now we can substitute the center (−3,4) and the radius 2√5into
the standard form of the equation of a circle to obtain the equation.
(x+ 3)2+ (y−4)2= (2√5)2
(x+ 3)2+ (y−4)2= 20
Step 4: Therefore, the standard form of the equation of the circle with center
at (-3, 4) and passing through the point (1, 2) is (x+ 3)2+ (y−4)2= 20.
2
Question 3
Question
Find the standard form equation of the ellipse that satisfies the following condi-
tions: the major axis has endpoints (-2,1) and (6,1), the center is at the origin,
and the distance from the center to a focus is √13.
Solution
Step 1: First, let’s determine important information based on the given condi-
tions. Since the major axis lies on the x-axis, we have a=1
2(length of major axis) =
1
2(6 −(−2)) = 4. The distance from the center to the focus is given by
c=√a2−b2, where cis the distance from the center to a focus. We can
solve for busing this information.
Step 2: Using the formula c=√a2−b2, plug in a= 4 and c=√13 to find
b. The equation becomes: √13 = √42−b2.
Step 3: Square both sides to solve for b. This gives us 13 = 16 −b2, which
simplifies to b2= 3.
Step 4: With a= 4 and b2= 3, we can now write the standard form of
the equation of the ellipse. Since the center is at the origin, the equation is
x2
a2+y2
b2= 1.
Step 5: Substitute a= 4 and b2= 3 into the equation to get x2
16 +y2
3= 1.
Therefore, the standard form equation of the ellipse satisfying the given
conditions is x2
16 +y2
3= 1.
Question 4
Question
Find the standard form of the equation of the parabola with a focus at (−2,1)
and a directrix at x= 4.
Solution
Step 1: First, recall that the standard form of the equation of a parabola with
a vertical axis is given by (x−h)2= 4p(y−k), where (h, k)is the vertex and p
is the distance between the vertex and the focus (or directrix).
Step 2: Since the directrix is a vertical line, we can determine that the
parabola opens to the left or right and that it has a vertical axis.
Step 3: The vertex of the parabola is the midpoint between the focus and
the directrix. Thus, the vertex is at (−2+4
2,1) = (1,1).
Step 4: The distance between the focus and the vertex represents p. Given
the focus at (−2,1) and the vertex at (1,1), we find p=|1−(−2)|= 3.
3
Step 5: Since the directrix is at x= 4, we know that the parabola opens to
the left. Therefore, the standard form of the equation is (x−1)2=−12(y−1).
Step 6: Simplifying, we have (x−1)2=−12y+ 12.
Step 7: Thus, the standard form of the equation of the parabola is (x−1)2=−12y+ 12 .
Question 5
Question
Consider the conic section given by the equation 4x2−9y2−16x−18y−36 = 0.
Determine the standard form of the conic section, and then identify the type of
conic section it represents.
Solution
Step 1: To determine the standard form of the conic section, we need to complete
the square for both the xand yterms.
Step 2: Starting with the xterms, we rewrite the equation as (4x2−16x)−
9y2−18y−36 = 0.
Step 3: Completing the square for x, we have 4(x2−4x)−9y2−18y−36 = 0.
To complete the square inside the parentheses, we add and subtract (4/2)2= 4
inside the parentheses: 4[(x−2)2−4] −9y2−18y−36 = 0.
Step 4: Further simplifying, we get 4(x−2)2−16 −9y2−18y−36 = 0.
Step 5: Now, we focus on completing the square for y. We rewrite the
equation as 4(x−2)2−9(y2+ 2y)−52 = 0.
Step 6: Completing the square for y, we have 4(x−2)2−9(y2+ 2y+ 1) −
52 + 9 = 0. This simplifies to 4(x−2)2−9(y+ 1)2−43 = 0.
Step 7: Rearranging the terms, we obtain the standard form of the conic
section as 4(x−2)2−9(y+ 1)2= 43.
Step 8: Finally, we observe that the standard form of the conic section
corresponds to a hyperbola. Therefore, the given conic section represents a
hyperbola.
Question 6
Question
Determine the standard form of the equation of a parabola with a focus at
F(3,−2) and a directrix at y=−6.
Solution
Step 1: Find the vertex of the parabola. Since the vertex lies in the middle
of the focus and the directrix, the x-coordinate of the vertex will be the same
as the x-coordinate of the focus, which is 3. The y-coordinate of the vertex
4
will be the average of the y-coordinate of the focus and the y-coordinate of the
directrix, which is −2+(−6)
2=−4. Therefore, the vertex is at V(3,−4).
Step 2: Determine the distance between the focus and the vertex. Since
the focus is above the vertex, the distance is equal to the difference between
the y-coordinates of the focus and the vertex, which is | − 2−(−4)|= 2. This
distance is also equal to the distance between the vertex and the directrix.
Step 3: Identify the axis of symmetry. Since the parabola opens either
upward or downward, the axis of symmetry is parallel to the y-axis and passes
through the vertex.
Step 4: Determine if the parabola opens upward or downward. Since the
focus lies below the vertex and the directrix is above the vertex, the parabola
opens downward.
Step 5: Write the equation of the parabola in standard form. Given that
the parabola opens downward, the standard form of the equation of a parabola
is (x−h)2= 4p(y−k), where (h, k)is the vertex, and pis the distance between
the vertex and the focus (or the vertex and the directrix).
Plugging in the values, the equation of the parabola is
(x−3)2= 4(−2−(−4))(y+ 4).
Therefore, the standard form of the equation of the parabola is
(x−3)2=−8(y+ 4).
Question 7
Question
Find the standard form of the equation of the hyperbola with vertices at (−3,0)
and (3,0) and asymptotes y=±2
3x.
Solution
Step 1: Find the center of the hyperbola by using the midpoint formula:
Midpoint =(x1+x2
2,y1+y2
2)
Step 2: Use the center to write the general equation of the hyperbola:
(x−h)2
a2−(y−k)2
b2= 1
Step 3: Find aand busing the properties of the hyperbola (distance from the
center to the vertices and the slopes of the asymptotes). Step 4: Substitute the
values of h,k,a, and binto the general equation to obtain the standard form
of the equation of the hyperbola.
5
Question 8
Question
Find the standard form of the equation of the ellipse with foci at (±4,0) and
co-vertices at (0,±2).
Solution
Step 1: Determine the center of the ellipse. Since the foci are at (±4,0), the
center of the ellipse is at the midpoint of the foci. Therefore, the center is (0,0).
Step 2: Determine the lengths of the major and minor axes. The distance
between the foci is 2c= 8, so c= 4. The distance between the co-vertices is
2b= 4, so b= 2. Therefore, the major axis is 2a= 8 so a= 4.
Step 3: Write the standard form of the equation. 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. Substituting the values for h, k, a, and b, we get:
x2
16 +y2
4= 1
Question 9
Question
Solve the following system of equations in xand y:
{x2+y2= 25
x2−y2= 9
Solution
Step 1: Let’s solve the second equation for x2:
x2=y2+ 9
Step 2: Substitute x2into the first equation:
(y2+ 9) + y2= 25
Step 3: Simplify the equation:
2y2+ 9 = 25
Step 4: Subtract 9 from both sides:
2y2= 16
6
Step 5: Divide by 2:
y2= 8
Step 6: Take the square root of both sides:
y=±√8
Step 7: Simplify the square root:
y=±2√2
Step 8: Substitute the values of yback into the equation x2=y2+ 9:
x2= (2√2)2+ 9 and x2= (−2√2)2+ 9
Step 9: Simplify the equations:
x2= 13 and x2= 13
Step 10: Take the square root of both sides:
x=±√13
Step 11: Therefore, the solutions to the system of equations are:
(√13,2√2),(√13,−2√2),(−√13,2√2),(−√13,−2√2)
Question 10
Question
Solve the system of equations:
{x2+y2= 25
x−y= 3
Solution
Let’s solve this system of equations by substitution.
Step 1: Solve the second equation for xin terms of y.
x=y+ 3
Step 2: Substitute y+ 3 for xin the first equation:
(y+ 3)2+y2= 25
Step 3: Expand and simplify the equation:
y2+ 6y+9+y2= 25
7
2y2+ 6y−16 = 0
Step 4: Solve the quadratic equation for y:
y=−6±√62−4(2)(−16)
2(2)
y=−6±√36 + 128
4
y=−6±√164
4
y=−6±2√41
4
Step 5: Find the corresponding values of xby using x=y+ 3:
x1=−6+2√41
4=−3 + √41
2
x2=−6−2√41
4=−3−√41
2
Step 6: Therefore, the solutions to the system of equations are:
{x=−3+√41
2, y =−3+√41
2
x=−3−√41
2, y =−3−√41
2
Question 11
Question
Solve the following equation for x:
4x2−9 = 0
Solution
Step 1: Begin by adding 9 to both sides of the equation to isolate the term with
x2.
4x2= 9
Step 2: Divide both sides by 4 to solve for x2.
x2=9
4
Step 3: Take the square root of both sides to solve for x.
x=±√9
4
8
Step 4: Simplify the square root.
x=±3
2
Therefore, the solutions to the equation 4x2−9 = 0 are x=3
2and x=−3
2.
Question 12
Question
Solve the system of equations:
{4x2+y2= 16
x2−4y2= 16
Solution
Step 1: Rewrite the system of equations in standard form.
We rewrite the given system as:
{x2
4+y2
16 = 1
x2
16 −y2
4= 1
Step 2: Identify the conic sections represented by each equation.
The first equation represents an ellipse, while the second equation represents a
hyperbola.
Step 3: Determine the graphical relationship between the two conic sections.
Since the signs of the x2and y2terms differ, the two conic sections are orthogonal
to each other.
Step 4: Solve the system of equations using the properties of conic sections.
We can infer that the center of the ellipse is at the origin since the larger
coefficient is under y2. For the hyperbola, since the larger coefficient is under
x2, we can determine that the center is also at the origin.
Step 5: Summarize the solution.
The system of equations has two different conic sections that do not intersect.
Therefore, the system has no solution.
Question 13
Question
Solve the equation 9x2−16y2= 144 for y.
9
Solution
Step 1: Rewrite the given equation in standard form by dividing both sides by
144.
(9x2−16y2)/144 = 1
x2/16 −y2/9 = 1
Step 2: Identify the conic section represented by the equation. Since the
coefficient of x2is positive, while the coefficient of y2is negative, this equation
represents a hyperbola.
Step 3: To solve for y, isolate y2by switching the terms to the other side
and then dividing by 9.
y2=−9(x2/16 −1)
Step 4: Take the square root of both sides to solve for y.
y=±√−9(x2/16 −1)
Step 5: Simplify the square root expression.
y=±√−9x2/16 + 9
Therefore, the solution to the equation 9x2−16y2= 144 for yis y=
±√−9x2/16 + 9.
Question 14
Question
Find the equation of the hyperbola with vertices at (−5,0) and (5,0) and foci
at (−3,0) and (3,0).
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: The distance between the center and each vertex is the absolute
value of the difference between the x-coordinates. Since the vertices lie on the
x-axis, the distance between the center and each vertex is equal to 5−0 = 5.
Step 3: The distance between the center and each focus is given by the
absolute value of the difference between the x-coordinates of the foci. This
distance is equal to 3−0 = 3.
Step 4: Define aas the distance between the center and one of the vertices,
which is a= 5. Define cas the distance between the center and one of the foci,
which is c= 3.
10
Step 5: Use the relationship c2=a2+b2for a hyperbola, where bis the
distance from the center to the transverse axis. Since the transverse axis is
along the x-axis, bis along the y-axis. However, since the vertices also lie along
the x-axis, the hyperbola is vertical (bis the y-coordinate).
Step 6: Plug in the known values to solve for b:
32= 52+b2
9 = 25 + b2
b2= 9 −25
b2=−16
b=√−16 = 4i
Step 7: The equation of the hyperbola with a vertical transverse axis is:
(x−h)2
a2−(y−k)2
b2= 1
Plugging in the values:
x2
25 −y2
−16 = 1
x2
25 +y2
16 = 1
Question 15
Question
Let xand ybe real numbers such that 3x2−8xy + 3y2= 4. Determine the
center and nature of the conic section represented by this equation.
Solution
To determine the center and nature of the conic section represented by the given
equation, we’ll first complete the square to rewrite the equation in standard
form.
Step 1: Rewrite the given equation by rearranging the terms:
3x2−8xy + 3y2= 4
Step 2: Now, complete the square separately for xand yterms. For the x
terms, we have:
3x2−8xy = 4 −3y2
3(x2−8
3xy) = 4 −3y2
11
Step 3: Complete the square for xby adding and subtracting (8
6y)2inside
the parentheses:
3(x2−8
3xy + (4
3y)2)= 4 −3y2
Step 4: Simplify the right-hand side:
3(x−4
3y)2
= 4 −3y2
Step 5: Divide through by 3 to obtain the standard form of the equation:
(x−4
3y)2=4−3y2
3
Step 6: The equation is now in the standard form for a conic section cen-
tered at (4
3,0). Since the coefficient of x2and y2are the same and have opposite
signs, the conic section is a hyperbola with center (4
3,0).
Question 16
Question
Solve the following system of equations:
{3x2+ 8y2= 24
x−2y= 3
Solution
Step 1: To eliminate a variable, let’s solve the second equation for x:
x−2y= 3
x= 2y+ 3
Step 2: Now, substitute xin terms of yinto the first equation:
3(2y+ 3)2+ 8y2= 24
Step 3: Simplify the equation:
3(4y2+ 12y+9)+8y2= 24
12y2+ 36y+27+8y2= 24
20y2+ 36y+ 27 = 24
Step 4: Rearrange terms to form a quadratic equation:
20y2+ 36y+ 3 = 0
12
Step 5: Solve for yusing the quadratic formula:
y=−b±√b2−4ac
2a
where a= 20,b= 36, and c= 3.
Step 6: Substitute the values of a,b, and cinto the quadratic formula:
y=−36 ±√362−4·20 ·3
2·20
y=−36 ±√1296 −240
40
y=−36 ±√1056
40
y=−36 ±32√3
40
Step 7: Simplify the solutions for y:
y1=−36 + 32√3
40 =−9+8√3
10
y2=−36 −32√3
40 =−9−8√3
10
Step 8: Finally, substitute the values of yback into the equation x= 2y+ 3
to find the corresponding values of x. Thus, the solutions are:
(x1, y1) = (6 + 16√3
10 ,−9+8√3
10 )
(x2, y2) = (6−16√3
10 ,−9−8√3
10 )
Question 17
Question
Find the equation of the hyperbola with vertices at (3,2) and (−1,2), and
passing through the point (−2,5).
13
Solution
Step 1: Find the center of the hyperbola.
• The center of the hyperbola is the midpoint of the line segment connecting
the vertices.
• The coordinates of the center are:
(3+(−1)
2,2+2
2)= (1,2)
Step 2: Find the distance between the center and one of the vertices to
determine the value of a.
• The distance between the center (1,2) and the vertex (3,2) is √(3 −1)2+ (2 −2)2=
2.
• So, a= 2.
Step 3: Find the equation of the hyperbola.
• The general equation of a hyperbola centered at (h, k)is (x−h)2
a2−(y−k)2
b2=
1.
• Since the hyperbola passes through (−2,5), we can substitute these coor-
dinates into the equation and solve for b:
(−2−1)2
22−(5 −2)2
b2= 1
9
4−9
b2= 1
−9
b2=−5
4
b2=36
5
b=±6
√5
• The equation of the hyperbola is:
(x−1)2
4−(y−2)2
36
5
= 1
(x−1)2
4−5(y−2)2
36 = 1
14
Question 18
Question
Consider the conic section given by the equation 2x2−3xy+2y2+4x−4y+5 = 0.
Determine whether this conic section is an ellipse, a hyperbola, a parabola, or
a degenerate case. If it is an ellipse or a hyperbola, find its standard form.
Solution
Step 1: To determine the type of conic section, we need to calculate the dis-
criminant ∆ = B2−4AC, where A,B, and Care the coefficients of x2,xy, and
y2respectively. Calculating A= 2,B=−3, and C= 2, we have:
∆ = (−3)2−4(2)(2) = 9 −16 = −7
Step 2: Since ∆<0, the conic section is neither an ellipse nor a hyperbola.
To determine if it is a parabola or a degenerate case, we analyze its conic form.
The general form of a conic section is given by Ax2+Bxy+Cy2+Dx+Ey+F=
0.
Step 3: We complete the square to simplify the equation. Group the xterms
and yterms together and complete the square for both variables:
2(x2+ 2x)−3xy + 2(y2−2y) = −5
2(x+ 1)2−1−2(y−1)2−1 = −5
2(x+ 1)2−2(y−1)2= 3
Step 4: Comparing our completed square form to the standard equation
for conic sections, we see that this conic section is a degenerate case of two
intersecting lines since there are no xy terms and the coefficients of x2and y2
have opposite signs.
Question 19
Question
Find the standard form equation of the hyperbola that satisfies the given con-
ditions: vertices at (5,2) and (9,2), and passes through the point (2,6).
Solution
Step 1: Determine the center of the hyperbola by finding the midpoint of the
line segment connecting the vertices.
Midpoint =(5+9
2,2+2
2)= (7,2)
15
Step 2: Calculate the distance between the center and one of the vertices to
find the value of a.
a=distance between the vertices
2=9−5
2= 2
Step 3: Use the distance formula to find the value of c, which is the distance
between the center and one of the foci.
c=√a2+b2
Step 4: Write the general equation of a hyperbola in standard form.
(x−h)2
a2−(y−k)2
b2= 1
Step 5: Find the value of busing the relationship between a,b, and cfor a
hyperbola: c2=a2+b2.
Step 6: Substitute the known values into the standard form equation to find
the specific equation of the hyperbola.
Step 7: Use the fact that the hyperbola passes through the point (2,6) to
solve for the remaining unknowns and finalize the equation of the hyperbola.
Question 20
Question
Solve the following system of equations for xand y:
{x2−y2= 4
x2+y2= 10
Solution
Step 1: First, rewrite the system of equations:
{x2−y2= 4
x2+y2= 10
Step 2: Add the two equations together to eliminate y2:
2x2= 14
Step 3: Divide by 2 to find the value of x2:
x2= 7
Step 4: Substitute x2= 7 into one of the original equations (e.g. x2−y2= 4)
to solve for y2:
7−y2= 4 =⇒y2= 3
16
Step 5: Take the square root of both sides to find the possible values for y:
y=±√3
Step 6: Therefore, the solutions to the system of equations are:
x=±√7and y=±√3
Question 21
Question
Find the standard form equation of the hyperbola that satisfies the following
conditions: the distance between its foci F1and F2is 10, the distance between
the vertices V1and V2is 14, and one of its asymptotes has the equation y= 2x.
Solution
Step 1: Recall the standard form of the equation for a hyperbola centered at
the origin with transverse axis along the x-axis is
(x
a)2−(y
b)2= 1
where ais half the distance between the vertices and bis half the distance
between the foci.
Step 2: We are given that the distance between the vertices, V1and V2, is
14. This means 2a= 14, so a= 7.
Step 3: We are given that the distance between the foci, F1and F2, is 10.
This means 2ae = 10, where eis the eccentricity of the hyperbola.
Step 4: Since one of the asymptotes has the equation y= 2x, we know that
the slopes of the asymptotes are ±b
a, so b
a= 2.
Step 5: From step 4, we have b
7= 2, so b= 14.
Step 6: Substitute a= 7 and b= 14 into the standard form equation of the
hyperbola to get
(x
7)2−(y
14)2= 1
Step 7: The standard equation for a hyperbola centered at the origin but
with the transverse axis along the y-axis is
(y
a)2−(x
b)2= 1
so the equation for the given hyperbola can be rewritten as
(y
14)2−(x
7)2= 1
Therefore, the standard form equation of the hyperbola that satisfies the
given conditions is
(y
14)2−(x
7)2= 1
17
Question 22
Question
Solve the system of equations:
{x2+y2= 25
4x+ 3y= 10
Solution
Step 1: The first equation x2+y2= 25 represents a circle centered at the origin
with radius 5.
Step 2: We can solve the second equation 4x+3y= 10 for xto get x=10−3y
4.
Step 3: Substitute x=10−3y
4into x2+y2= 25:
(10 −3y
4)2
+y2= 25
Step 4: Simplify the equation:
(10 −3y)2
16 +y2= 25
Step 5: Expand and simplify:
100 −60y+ 9y2
16 +y2= 25
Step 6: Multiply both sides by 16 to eliminate the denominator:
100 −60y+ 9y2+ 16y2= 400
Step 7: Combine like terms:
25y2−60y−300 = 0
Step 8: Divide by 25 to simplify the equation:
y2−60
25y−300
25 = 0
Step 9: Simplify further:
y2−2.4y−12 = 0
Step 10: Use the quadratic formula to solve for y:
y=−(−2.4) ±√(−2.4)2−4(1)(−12)
2(1)
Step 11: Calculate the solutions for yand then substitute back to find the
corresponding xvalues.
18
Question 23
Question
Consider the ellipse with the equation x2
4+y2
9= 1. Find the foci of the ellipse.
Solution
Step 1: Recall that the standard form of an ellipse centered at the origin is
x2
a2+y2
b2= 1, where ais the semi-major axis and bis the semi-minor axis.
Step 2: By comparing the given equation x2
4+y2
9= 1 with the standard
form, we can see that a2= 4 and b2= 9.
Step 3: The distance from the center to the foci is given by c=√a2−b2.
Step 4: Plugging in the values of a2= 4 and b2= 9 into the formula for c,
we get c=√4−9 = √−5.
Step 5: Since the square root of a negative number is imaginary, the foci of
the ellipse are at points (0,±i√5).
Question 24
Question
Solve the following system of equations:
{x2+y2= 25
2x−3y= 5
Solution
Step 1: Start by solving the second equation for xin terms of y:
2x−3y= 5
2x= 3y+ 5
x=3y+ 5
2
Step 2: Substitute this expression for xinto the first equation:
(3y+ 5
2)2
+y2= 25
9y2+ 30y+ 25
4+y2= 25
Step 3: Multiply by 4 to get rid of the fractions:
9y2+ 30y+25+4y2= 100
19
13y2+ 30y+ 25 = 100
Step 4: Rearrange the equation into standard form:
13y2+ 30y+ 25 −100 = 0
13y2+ 30y−75 = 0
Step 5: Now, we can solve this quadratic equation for y. The solutions will
be:
y=−30 ±√302−4(13)(−75)
2(13)
y=−30 ±√900 + 3900
26
y=−30 ±√4800
26
y=−30 ±20√3
26
y=−15 ±10√3
13
Step 6: Substitute the values of yback into the equation we found for xin
terms of yto find the corresponding values of x.
Therefore, the solutions to the system of equations are:
(−15 + 10√3
13 ,2(−15 + 10√3) + 5
3)and (−15 −10√3
13 ,2(−15 −10√3) + 5
3)
Question 25
Question
Find the equation of the circle that passes through the points A(3,2) and
B(−1,−4).
Solution
Step 1: Find the midpoint of the line segment connecting points Aand B. Let
this midpoint be point M.
Step 2: Find the slope of the line passing through points Aand B. This will
be the negative reciprocal of the slope of the line passing through the center of
the circle, M, and the midpoint of the chord formed by connecting Aand B
through the center of the circle.
Step 3: Using the midpoint, M, and the slope of the line passing through
points Aand B, find the equation of the circle passing through points Aand B.
Step 4: Write down the equation of the circle in standard form.
20
Let’s complete these steps in detail.
Step 1: Find the midpoint of the line segment connecting points Aand B.
Let this midpoint be point M. The midpoint formula is given by:
M(x1+x2
2,y1+y2
2)
Substitute the coordinates of points A(3,2) and B(−1,−4) into the formula:
M(3+(−1)
2,2+(−4)
2)=M(2
2,−2
2)=M(1,−1)
So, the midpoint Mis M(1,−1).
Step 2: Find the slope of the line passing through points Aand B. The
slope formula is given by:
m=y2−y1
x2−x1
Substitute the coordinates of points A(3,2) and B(−1,−4) into the formula:
m=−4−2
−1−3=−6
−4=3
2
Thus, the slope of the line passing through points Aand Bis 3
2.
Step 3: Using the midpoint, M(1,−1), and the slope m=3
2, find the
equation of the circle passing through points Aand B. The equation of the
circle can be written in the form (x−h)2+ (y−k)2=r2, where (h, k)is the
center of the circle and ris the radius of the circle.
Since the circle passes through point M(1,−1), the coordinates of the center
are (h, k) = (1,−1).
The negative reciprocal of the slope m=3
2is −2
3, which represents the slope
of the perpendicular bisector of the chord AB passing through M(1,−1).
Since the equation of a line passing through point (1,−1) with slope −2
3that
is perpendicular to the line passing through Aand B(m=3
2) has the equation:
y−(−1) = −2
3(x−1)
On simplifying, we get:
y+ 1 = −2
3x+2
3
y=−2
3x−5
3
Let this line intersect the circle at points Dand E. As this line is the
perpendicular bisector of AB, we can find the distance from Ato Mand set it
equal to the distance from Ato the circle to find r.
r=√(1 −3)2+ (−1−2)2=√(−2)2+ (−3)2=√13
Thus, the equation of the circle is (x−1)2+ (y+ 1)2= 13.
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