MATH 121 - COLLEGE ALGEBRA -
Equations and properties of conic
sections
Question Bank - Set 8
Liberty University
Question 1
Question
Find the standard form of the equation of the hyperbola with vertices at (−5,2)
and (3,2) and with foci at (−7,2) and (5,2).
Solution
Step 1: Find the center of the hyperbola by averaging the coordinates of the
vertices.
Center =(−5+3
2,2+2
2)= (−1,2)
Step 2: Find the distance between the center and one of the vertices to
determine a(distance from center to vertex).
a= 3 −(−1) = 4
Step 3: Find the distance between the center and one of the foci to determine
c(distance from center to focus).
c= 5 −(−1) = 6
Step 4: Use the relationship c2=a2+b2to find b.
b2=c2−a2= 62−42= 36 −16 = 20
Step 5: The standard form of the equation of a hyperbola centered at (h, k)
with vertices on the x-axis is (x−h)2
a2−(y−k)2
b2= 1. Thus, the equation of the
hyperbola is
(x+ 1)2
16 −(y−2)2
20 = 1
Question 2
Question
Find the standard form of the equation of the ellipse with foci at (−1,1) and
(3,1) and a major axis of length 6 units.
Solution
Step 1: Find the center of the ellipse. The center of the ellipse is the midpoint
between the foci. Let’s calculate it using the midpoint formula:
(−1+3
2,1+1
2) = (1,1)
Step 2: Find the distance between the foci. Let 2cbe the distance between
the foci of the ellipse. The major axis is 2a= 6 units, so a= 3 units. Using the
distance formula:
2c=√(3 −(−1))2+ (1 −1)2=√16 = 4
Thus, c= 2.
Step 3: Find the value of b. The formula relating a,b, and cfor an ellipse is
a2=b2+c2. Substituting the values we know:
32=b2+ 22
9 = b2+ 4
b2= 5
b=√5
Step 4: Write the equation of the ellipse. The standard form of the equation
of an ellipse centered at (h, k)is:
(x−h
a)2+ (y−k
b)2= 1
Plugging in the values of a,b,h, and k:
(x−1
3)2+ (y−1
√5)2= 1
Therefore, the standard form of the equation of the ellipse is (x−1
3)2+ (y−1
√5)2= 1 .
Question 3
Question
Solve the system of equations for xand y:
{9x2−4y2= 36
3x+ 4y= 6
2
Solution
Step 1: Let’s solve the second equation for x:
3x+ 4y= 6
3x= 6 −4y
x= 2 −4
3y
Step 2: Substitute the expression for xinto the first equation:
9(2 −4
3y)2−4y2= 36
9(4 −16
3y+16
9y2)−4y2= 36
36 −48y+ 16y2= 36 −4y2
Step 3: Simplify the equation by expanding and combining like terms:
36 −48y+ 16y2= 36 −4y2
16y2−48y+ 36 = 36 −4y2
20y2−48y= 0
Step 4: Factor out 4y:
4y(5y−12) = 0
Step 5: Solve for y:
4y= 0 or 5y−12 = 0
y= 0 or y=12
5
Step 6: Substitute the values of yback into the equation for xto find the
corresponding values: For y= 0, substituting into x= 2 −4
3ygives x= 2.
For y=12
5, substituting into x= 2 −4
3ygives x=−4
5.
Therefore, the solutions to the system of equations are (2,0) and (−4
5,12
5).
Question 4
Question
Solve the following system of equations:
{x2+y2= 16
2x−y= 3
3
Solution
Step 1: Solve the second equation for y in terms of x.
2x−y= 3
y= 2x−3
Step 2: Substitute yfrom the second equation into the first equation.
x2+ (2x−3)2= 16
x2+ 4x2−12x+ 9 = 16
5x2−12x−7 = 0
Step 3: Solve the quadratic equation 5x2−12x−7 = 0.
x=−(−12) ±√(−12)2−4(5)(−7)
2(5)
x=12 ±√144 + 140
10
x=12 ±√284
10
x=12 ±2√71
10
x=6±√71
5
Step 4: Substitute the values of xback into the equation y= 2x−3to find
the corresponding yvalues.
For x=6 + √71
5:y= 2 (6 + √71
5)−3 = 12 + 2√71
5−15
5=−3+2√71
5
For x=6−√71
5:y= 2 (6−√71
5)−3 = 12 −2√71
5−15
5=−3−2√71
5
Therefore, the solutions to the system of equations are:
(6 + √71
5,−3+2√71
5)and (6−√71
5,−3−2√71
5)
Question 5
Question
Given the equation of the hyperbola: (x−2)2
16 −(y+3)2
9= 1, find the standard
form of the equation of this hyperbola, the center, the vertices, the asymptotes,
and the foci.
4
Solution
Step 1: Identify the standard form of the equation of a hyperbola. Step 2: Use
properties of the given hyperbola to determine the values needed to represent
the equation in standard form. Step 3: Find the center of the hyperbola. Step
4: Calculate the vertices. Step 5: Determine the asymptotes. Step 6: Find the
foci.
Step 1: The standard form of the equation of a hyperbola is (x−h)2
a2−
(y−k)2
b2= 1, where (h, k)is the center of the hyperbola, ais the distance from
the center to the vertices along the x-axis, and bis the distance from the center
to the vertices along the y-axis.
Step 2: By comparing the given equation to the standard form, we have
h= 2,k=−3,a2= 16, and b2= 9.
Step 3: The center of the hyperbola is (h, k) = (2,−3).
Step 4: The vertices are given by (h±a, k), so the vertices are (2 + 4,−3)
and (2 −4,−3), which simplifies to (6,−3) and (−2,−3).
Step 5: The slopes of the asymptotes are given by ±b
a, so the slopes are
±3
4. Using the center (2,−3), we can find the equations of the asymptotes as
y+ 3 = ±3
4(x−2).
Step 6: To find the foci, we use the relationship c2=a2+b2. Solving
for c, we find c=√a2+b2=√16 + 9 = √25 = 5. The foci are located at
(h±c, k), so the foci are (2 + 5,−3) and (2 −5,−3), which simplifies to (7,−3)
and (−3,−3).
Question 6
Question
Solve the following equation for x:
9x2+ 16y2+ 12x−8y+ 4 = 0
Solution
Step 1: Rearrange the given equation by grouping the xterms and completing
the square for x.
9x2+ 16y2+ 12x−8y+ 4 = 0
9x2+ 12x+ 16y2−8y+ 4 = 0
9(x2+4
3x) + 16(y2−1
2y) = −4
Step 2: Complete the square for xby adding (4
3)2=16
9inside the parenthe-
5
ses.
9(x2+4
3x+16
9) + 16(y2−1
2y) = −4 + 9( 16
9)
9(x+4
3)2+ 16(y−1
4)2=−4 + 16
9(x+4
3)2+ 16(y−1
4)2= 12
Step 3: Divide both sides by the constant term on the right to obtain the
standard form of the equation for an ellipse.
9(x+4
3)2
12 +16(y−1
4)2
12 = 1
(x+4
3)2
(2√3
3)2+(y−1
4)2
(2√3
4)2= 1
Therefore, the equation represents an ellipse centered at (−4
3,1
4)with major
axis 2√3along the xaxis and minor axis 2√3along the yaxis.
Question 7
Question
Let y= 3x2−4be the equation of a parabola.
1. Find the coordinates of the vertex of the parabola.
2. Determine whether the parabola opens upwards or downwards.
3. Find the equation of the axis of symmetry.
Solution
1. To find the coordinates of the vertex of the parabola, we need to first rewrite
the given equation in vertex form. The vertex form of a parabola is y=a(x−
h)2+k, where (h, k)is the vertex of the parabola. Given y= 3x2−4, we can
rewrite it by completing the square:
y= 3(x2)−4
y= 3(x2−0x)−4
y= 3(x2−0x+ 02)−4−3(02)
y= 3(x−0)2−4
Comparing this with the vertex form, we see that a= 3,h= 0, and k=−4.
Therefore, the vertex of the parabola is (0,−4).
2. The parabola opens upwards if a > 0, and downwards if a < 0. Since
a= 3 in this case, the parabola opens upwards.
6
3. The equation of the axis of symmetry for a parabola in vertex form
y=a(x−h)2+kis given by x=h. Therefore, in this case, the equation of the
axis of symmetry is x= 0.
Question 8
Question
Solve the system of equations:
{x2+ 4y2= 16
3x2−2y2= 9
Solution
Step 1: Let’s write the given system of equations in standard form for conic
sections. We can rewrite the first equation as x2
16 +y2
4= 1 and the second
equation as x2
3−y2
9= 1.
Step 2: Next, notice that the first equation represents an ellipse centered at
the origin with major axis along the x-axis, while the second equation represents
a hyperbola centered at the origin with main axes along the x and y axes.
Step 3: Let’s solve the system by finding the points where the ellipse and
hyperbola intersect. Since each equation represents a curve, the solutions will
be the points where their graphs intersect.
Step 4: Substituting the first equation into the second equation, we get:
16
16 +y2
4= 1 =⇒1 + y2
4= 1 =⇒y2
4= 0 =⇒y= 0
Step 5: Substitute y= 0 back into the first equation to find the correspond-
ing x-values:
x2+ 4(0)2= 16 =⇒x2= 16 =⇒x=±4
Step 6: Therefore, the points of intersection are (4,0) and (−4,0). Thus,
the solution to the system of equations is (x, y) = (4,0),(−4,0) .
Question 9
Question
Simplify the following equation of a hyperbola in standard form:
(x−2)2
9−(y+ 1)2
25 = 1
7
Solution
Step 1: Identify the center of the hyperbola.
The standard form of the equation of a hyperbola is (x−h)2
a2−(y−k)2
b2= 1,
where (h, k)is the center of the hyperbola.
Comparing the given equation to the standard form, we can see that the
center of the hyperbola is at (2,−1).
Step 2: Determine the vertices of the hyperbola.
The vertices of the hyperbola are located at (h±a, k). In this case, the
vertices are at (2 ±3,−1), so the vertices are at (5,−1) and (−1,−1).
Step 3: Find the foci of the hyperbola.
The distance from the center to the foci is given by c=√a2+b2. Plugging
in the values a= 3 and b= 5, we find c=√9 + 25 = √34.
Therefore, the foci of the hyperbola are at (2 ±√34,−1), which are approx-
imately (6.83,−1) and (−2.83,−1).
Step 4: Determine the equations of the asymptotes of the hyperbola.
The equations of the asymptotes for a hyperbola in standard form can be
written as y−k=±b
a(x−h). Substituting the center and the values of aand b
into this equation, we get the equations of the asymptotes as y+ 1 = ±5
3(x−2).
Step 5: Sketch the hyperbola with the help of the center, vertices, foci, and
asymptotes.
Now that we have all the key points and information about the hyperbola,
we can sketch it on the coordinate plane.
This completes the solution.
Question 10
Question
Solve the following system of equations for xand y:
{2x2+ 5y2= 13
4x2−y2= 3
Solution
To solve the system of equations for xand y, we first need to isolate one variable
in one of the equations and substitute that into the other equation to find the
solution.
8
Step 1: Solve for yin terms of xin the first equation.
2x2+ 5y2= 13
5y2= 13 −2x2
y2=13 −2x2
5
y=±√13 −2x2
5
Step 2: Substitute yinto the second equation and solve for x.
4x2−(13 −2x2
5)= 3
20x2−13 + 2x2= 15
22x2= 28
x2=28
22
x=±√14
11
Step 3: Substitute xback into the equation to find the correspond-
ing yvalues.
y=±√13 −2(14
11 )
5
y=±√13 −28
11
5
y=±√143 −28
55
y=±√115
55
y=±√2
Therefore, the solutions to the system of equations are x=±√14
11 and
y=±√2.
Question 11
Question
Consider the equation of a conic section given by 5x2−12xy +5y2−18x−18y−
18 = 0. Determine the type of conic section represented by this equation and
sketch its graph.
9
Solution
Step 1: To determine the type of conic section represented by the given equation,
we need to consider the discriminant B2−4AC of the general conic section
equation Ax2+Bxy +Cy2+Dx +Ey +F= 0. In this case, A= 5,B=−12,
and C= 5.
Step 2: Calculate the discriminant:
B2−4AC = (−12)2−4(5)(5) = 144 −100 = 44
Step 3: Since the discriminant B2−4AC = 44 >0, the conic section is an
ellipse.
Step 4: Next, we can rewrite the given equation in standard form to deter-
mine the center, vertices, and major and minor axes of the ellipse. We start by
completing the square.
Step 5: Completing the square for 5x2−12xy + 5y2−18x−18y−18 = 0,
we get:
5(x2−2x)−12xy + 5(y2−2y) = 18
Step 6: Continuing to complete the square:
5(x2−2x+ 1) −12xy + 5(y2−2y+ 1) = 18 + 5 + 5
Step 7: Simplifying further:
5(x−1)2−12xy + 5(y−1)2= 28
Step 8: So, the standard form of the given ellipse equation is 5(x−1)2−
12xy + 5(y−1)2= 28.
Step 9: From the standard equation, we can identify the center of the ellipse
as (1,1).
Step 10: The major axis of the ellipse is along the line of symmetry defined
by the major axis direction. The minor axis is perpendicular to the major axis.
Step 11: Now, let’s sketch the graph of the ellipse with the center at (1,1)
and with major and minor axes determined by the equation.
Question 12
Question
Solve the following system of equations for xand y:
2x2+ 5xy + 3y2= 29
3x2−4xy −8y2=−7
10
Solution
Step 1: Let’s rewrite the system of equations as a matrix equation AX =B,
where A,X, and Bare as follows:
A=[2 5
3−4], X =[x
y], B =[29
−7]
Step 2: To solve for X, we need to find the inverse of matrix A. The inverse
of a 2×2matrix [a b
c d]is given by:
A−1=1
ad −bc [d−b
−c a ]
Step 3: Compute the determinant of matrix A:
|A|= (2 × −4) −(5 ×3) = −8−15 = −23
Step 4: Find the inverse of matrix A:
A−1=1
−23 [−4−5
−3 2 ]=[4
23
5
23
3
23 −2
23 ]
Step 5: Multiply A−1by matrix Bto solve for X:
X=A−1×B=[4
23
5
23
3
23 −2
23 ]×[29
−7]
X=[3
2]
Therefore, the solution to the system of equations is x= 3 and y= 2.
Question 13
Question
Solve the system of equations:
{x2+y2= 25
2x−y= 1
Solution
Step 1: Begin by solving the second equation for y in terms of x.
2x−y= 1 =⇒y= 2x−1
11
Step 2: Substitute the expression for y into the first equation.
x2+ (2x−1)2= 25
Step 3: Expand and simplify the equation.
x2+ 4x2−4x+ 1 = 25
Step 4: Combine like terms.
5x2−4x+ 1 = 25
Step 5: Subtract 25 from both sides.
5x2−4x−24 = 0
Step 6: Factor the quadratic equation.
(5x+ 12)(x−2) = 0
Step 7: Set each factor to zero and solve for x.
5x+ 12 = 0 or x−2 = 0
Step 8: Solve for x.
x=−12
5or x= 2
Step 9: Substitute the values of x back into the equation y= 2x−1to find
the corresponding y-values. For x=−12
5:
y= 2 ·(−12
5)−1 = −24
5−5
5=−29
5
For x= 2:
y= 2 ·2−1 = 4 −1 = 3
Step 10: The solutions to the system of equations are (−12
5,−29
5)and (2,3).
Question 14
Question
Determine the standard form equation of the hyperbola with vertices at (−3,0)
and (3,0) and foci at (−5,0) and (5,0).
12
Solution
Step 1: Determine the center of the hyperbola. The center of the hyperbola is
the midpoint of the segment connecting the vertices. The center of the hyperbola
is (−3+3
2,0+0
2)= (0,0).
Step 2: Determine the distance between the center and one of the vertices
to find a, the distance from the center to a vertex. 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 a focus. c= 5 −0 = 5.
Step 4: Use the relationship for a hyperbola, c2=a2+b2, to find b, where b
is the distance from the center to a co-vertex. 52= 32+b225 = 9 + b2b2= 16
b= 4 or b=−4
Step 5: Since the hyperbola is centered at the origin, the standard form
equation of the hyperbola is x2
a2−y2
b2= 1. Substituting the values of aand b,
we can write the equation as:
x2
32−y2
42= 1
x2
9−y2
16 = 1
Question 15
Question
Solve the following system of equations for xand y:
{4x2−y2= 16
x−y= 2
Solution
Step 1: We will first isolate yin the second equation to express yin terms of
x. Step 2: We will then substitute this expression for yinto the first equation
to solve for x. Step 3: Once we have found the value of x, we will substitute it
back into the second equation to find the corresponding value of y.
Step 1: Solve the second equation for y.
x−y= 2
y=x−2
Step 2: Substitute y=x−2into the first equation.
4x2−(x−2)2= 16
13
4x2−(x2−4x+ 4) = 16
4x2−x2+ 4x−4 = 16
3x2+ 4x−20 = 0
Step 3: Solve the quadratic equation 3x2+ 4x−20 = 0.
(3x−6)(x+ 5) = 0
x= 2,−5
3
Now, substitute x= 2 and x=−5
3back into y=x−2to find the corre-
sponding values of y.
If x= 2:
y= 2 −2 = 0
∴x= 2, y = 0
If x=−5
3:
y=−5
3−2 = −11
3
∴x=−5
3, y =−11
3
Therefore, the solutions to the system of equations are (2,0) and (−5
3,−11
3).
Question 16
Question
Determine the standard form of the equation of the parabola that satisfies the
given conditions: vertex at (1,−3) and directrix x= 5.
Solution
Step 1: The standard form of the equation of a vertical parabola with vertex
(h, k)and directrix x=ais given by (x−h)2= 4p(y−k), where pis the
distance between the vertex and the focus.
Step 2: Since the directrix is to the right of the vertex, pis positive. The
distance between the vertex (h, k)and the directrix x=ais p.
Step 3: In this case, the vertex is at (1,−3) and the directrix is x= 5. Since
pis the distance between the vertex and the directrix, we have p= 5 −1 = 4.
Step 4: The equation of the parabola is:
(x−1)2= 4(4)(y+ 3)
Step 5:
(x−1)2= 16(y+ 3)
Therefore, the standard form of the equation of the parabola is (x−1)2=
16(y+ 3).
14
Question 17
Question
Solve the equation 4x2−9y2−16x+ 18y+ 56 = 0 and determine the type of
conic section it represents.
Solution
Step 1: Rearrange the equation by grouping like terms:
(4x2−16x)−(9y2−18y) = −56
Step 2: Complete the square for xby adding and subtracting (16/2)2= 42=
16 inside the parentheses:
4(x2−4x+ 4) −(9y2−18y) = −56 + 4(4)
Step 3: Complete the square for yby adding and subtracting (18/2)2= 92=
81 inside the parentheses:
4(x2−4x+ 4) −9(y2−2y+ 1) = −56 + 16 −81
Step 4: Simplify the equation:
4(x−2)2−9(y−1)2=−121
Step 5: Divide by −121 to get the equation in the standard form:
(x−2)2
9−(y−1)2
(−121)/9= 1
Step 6: Compare the equation to the standard form of a conic section to
determine its type: The equation represents a hyperbola since the terms have
opposite signs in the standard form equation.
Question 18
Question
Find the standard form of the equation of the hyperbola that satisfies the given
conditions: vertices at (−5,1) and (1,1), and passes through the point (3,4).
Solution
Step 1: Find the center of the hyperbola.
The center of the hyperbola is the midpoint of the segment connecting the
vertices. Use the midpoint formula M=(x1+x2
2,y1+y2
2).
15
Given vertices: (−5,1) and (1,1)
Center M=(−5+1
2,1+1
2)
Center M= (−2,1)
Step 2: Find the equation of the hyperbola with center (−2,1).
Since the hyperbola has a horizontal transverse axis, the standard form of
the equation of the hyperbola is
(x−h)2/a2−(y−k)2/b2= 1
where (h, k)is the center of the hyperbola, ais the distance from the center
to a vertex, and bis the distance from the center to a co-vertex.
Given h=−2,k= 1, the vertices are at (−5,1) and (1,1).
ais the distance from the center to a vertex, so a=| − 2−(−5)|= 3.
So, we have
(x+ 2)2/32−(y−1)2/b2= 1
Step 3: Find the value of b.
The hyperbola passes through the point (3,4). Substitute x= 3 and y= 4
into the equation:
(3+2
3)2
−(4 −1)2/b2= 1
25/9−9/b2= 1
Solve for b2:
b2= 9/(25/9−1) = 9/(16/9) = 81/16
b=√81/16 = 9/4
Thus, the equation of the hyperbola is:
(x+ 2)2
9−(y−1)2
(9
4)2= 1
Question 19
Question
Find the standard form equation of the conic section with the following prop-
erties: a parabola with a focus at F(3,1) and a directrix y=−1.
Solution
Step 1: Recall that the standard form equation for a parabola with a vertical
axis of symmetry is given by (x−h)2= 4p(y−k).
Step 2: The vertex of the parabola is equidistant from the focus and the
directrix. Since the directrix is y=−1, the vertex, V, must have a y-coordinate
16
equal to the average of the y-coordinate of the focus and the y-coordinate of a
point on the directrix. Therefore, Vis at (3,0).
Step 3: The distance between the focus Fand the vertex Vgives the value
of pin the equation. Using the distance formula, we find that p= 1.
Step 4: Substituting the values of h,k, and pinto the standard form equa-
tion, we get (x−3)2= 4(y).
Hence, the standard form equation of the parabola is (x−3)2= 4y.
Question 20
Question
Find the equation of the hyperbola with foci at (−5,0) and (5,0) and vertices
at (−8,0) and (8,0).
Solution
Step 1: Find the center of the hyperbola. Since the center is the midpoint of
the vertices, we have:
(h, k) = (−8+8
2,0)= (0,0)
Step 2: Find the distance between the center and the foci. The distance
between the foci and the center, which is the value of c, is:
c= 5 −(−5) = 10
Step 3: Find the value of a(distance from the center to the vertex). The
value of ais the distance between the center and one of the vertices:
a= 8 −0 = 8
Step 4: Determine the equation of the hyperbola. The equation of a hyper-
bola centered at (h, k)with vertices along the x-axis is:
(x−h)2
a2−(y−k)2
b2= 1
Step 5: Find the value of busing the relationship c2=a2+b2:
b=√c2−a2=√102−82=√100 −64 = √36 = 6
Step 6: Substitute the known values into the equation of the hyperbola:
x2
64 −y2
36 = 1
Therefore, the equation of the hyperbola is x2
64 −y2
36 = 1.
17
Question 21
Question
Solve the following system of equations:
{x2+y2= 25
x−2y= 4
Solution
Step 1: Solve the second equation for xin terms of y:
x−2y= 4
x= 4 + 2y
Step 2: Substitute xfrom the second equation into the first equation:
(4 + 2y)2+y2= 25
16 + 16y+ 4y2+y2= 25
4y2+ 17y−9 = 0
Step 3: Solve the quadratic equation 4y2+ 17y−9=0using the quadratic
formula:
y=−b±√b2−4ac
2a
y=−17 ±√172−4(4)(−9)
2(4)
y=−17 ±√289 + 144
8
y=−17 ±√433
8
Step 4: Calculate the values of y:
y=−17 + √433
8or y=−17 −√433
8
Step 5: Substitute the values of yback into the equation x= 4 + 2yto find
the corresponding values of x:
For y=−17 + √433
8:x= 4 + 2 (−17 + √433
8)
For y=−17 −√433
8:x= 4 + 2 (−17 −√433
8)
Therefore, the solution to the system of equations is:
(−17 + √433
8,−17 + √433
4)and (−17 −√433
8,−17 −√433
4)
18
Question 22
Question
Solve the following system of equations:
{4x2+y2= 25
x−y= 1
Solution
Step 1: Solve the second equation for xin terms of y. Step 2: Substitute xin
terms of yinto the first equation. Step 3: Solve the resulting equation to find
the possible values of y. Step 4: Substitute the values of yback into the second
equation to find the corresponding values of x.
Step 1: Solve the second equation for xin terms of y. From x−y= 1, we
have x=y+ 1.
Step 2: Substitute xin terms of yinto the first equation. Substitute x=
y+ 1 into 4x2+y2= 25:
4(y+ 1)2+y2= 25
Step 3: Solve the resulting equation to find the possible values of y. Ex-
panding the left side of the equation:
4(y2+ 2y+ 1) + y2= 25
4y2+ 8y+4+y2= 25
5y2+ 8y−21 = 0
Using the quadratic formula to solve for y:
y=−8±√82−4(5)(−21)
2(5)
y=−8±√64 + 420
10
y=−8±√484
10
y=−8±22
10
Therefore, y=−3.4or y= 3.0.
Step 4: Substitute the values of yback into the second equation to find the
corresponding values of x. For y=−3.4:x=−3.4 + 1 = −2.4
For y= 3.0:x= 3.0 + 1 = 4.0
Thus, the solutions to the system of equations are (−2.4,−3.4) and (4.0,3.0).
19
Question 23
Question
Solve the following system of equations:
{x2+y2−6x+ 4y−12 = 0
x−2y= 0
Solution
Step 1: We can solve the second equation for x:
x= 2y
Step 2: Substitute xin terms of yinto the first equation:
(2y)2+y2−6(2y)+4y−12 = 0
Step 3: Simplify the equation:
4y2+y2−12y+ 4y−12 = 0
5y2−8y−12 = 0
Step 4: Solve the quadratic equation by factoring or using the quadratic
formula: The equation 5y2−8y−12 = 0 can be factored as (5y+ 6)(y−2) = 0.
This gives us two possible solutions: y=−6
5or y= 2.
Step 5: Substitute the values of yback into the equation x= 2yto find the
corresponding values of x. For y=−6
5:
x= 2 (−6
5)=−12
5
For y= 2:
x= 2(2) = 4
So, the solutions to the system of equations are: (x, y) = (−12
5,−6
5)and
(x, y) = (4,2).
Question 24
Question
Find the standard form of the equation of the parabola with focus F(2,3) and
directrix x= 1.
20
Solution
Step 1: Find the vertex of the parabola. The vertex of the parabola is the
midpoint between the focus and the directrix. We have the focus F(2,3) and
the directrix x= 1. Using the formula for the midpoint between two points, the
vertex Vis:
V=(1+2
2,3+3
2)=(3
2,3).
Step 2: Find the equation of the axis of symmetry. The axis of symmetry is
a vertical line passing through the vertex. Therefore, the equation of the axis
of symmetry is x=3
2.
Step 3: Determine whether the parabola opens to the left or right. Since
the directrix is to the left of the focus, the parabola opens to the right.
Step 4: Find the distance between the vertex and the focus. The distance
between the vertex and the focus is the same as the distance between the vertex
and the directrix. This distance is called the focal length, denoted as |p|. In
this case, |p|=|1−3
2|=1
2.
Step 5: Write the standard form of the equation. Since the parabola opens
to the right, the standard form of the equation is:
(x−h)2= 4p(y−k),
where (h, k)is the vertex. Plugging in the values of the vertex (3
2,3) and |p|=1
2,
we have:
(x−3
2)2= 4(1
2)(y−3).
Simplifying, we get:
(x−3
2)2= 2(y−3).
Therefore, the standard form of the equation of the parabola is (x−3
2)2=
2(y−3).
Question 25
Question
Find the standard form equation of the ellipse that satisfies the given conditions:
The major axis is along the line 3x−4y= 2 and passes through the points (−1,3)
and (5,1).
Solution
Step 1: Find the center of the ellipse using the midpoint formula. The midpoint
of the line segment connecting the points (−1,3) and (5,1) is the center of the
ellipse. Let the center be (h, k). The midpoint formula is given by:
(x1+x2
2,y1+y2
2)
21
Applying this formula to (−1,3) and (5,1):
(−1+5
2,3+1
2)= (2,2)
So, the center of the ellipse is at (h, k) = (2,2).
Step 2: Find the distance between the two given points which is the length
of the major axis. The distance formula is given by:
d=√(x2−x1)2+ (y2−y1)2
Calculating the distance between (−1,3) and (5,1):
d=√(5 −(−1))2+ (1 −3)2=√36 + 4 = √40 = 2√10
So, the length of the major axis is 2√10.
Step 3: Determine the major axis direction. We have the equation of the
major axis as 3x−4y= 2. Rewrite it in slope-intercept form as y=3
4x−1
2.
This means the slope of the major axis is 3
4. Therefore, the major axis of the
ellipse is in the direction of the line y=3
4x.
Step 4: Find the equation of the ellipse in standard form. The standard
form of the equation of an ellipse with major axis along the x-axis is:
(x−h)2
a2+(y−k)2
b2= 1
where (h, k)is the center of the ellipse, ais half the length of the major axis,
and bis half the length of the minor axis. In this case, since the major axis is
along the line y=3
4x, the equation becomes:
(x−2)2
10 +(y−2)2
2= 1
Therefore, the standard form equation of the ellipse is
(x−2)2
10 +(y−2)2
2= 1
.
22
Question 2
Question
Find the standard form of the equation of the ellipse with foci at (−1,1) and
(3,1) and a major axis of length 6 units.
Solution
Step 1: Find the center of the ellipse. The center of the ellipse is the midpoint
between the foci. Let’s calculate it using the midpoint formula:
(−1+3
2,1+1
2) = (1,1)
Step 2: Find the distance between the foci. Let 2cbe the distance between
the foci of the ellipse. The major axis is 2a= 6 units, so a= 3 units. Using the
distance formula:
2c=√(3 −(−1))2+ (1 −1)2=√16 = 4
Thus, c= 2.
Step 3: Find the value of b. The formula relating a,b, and cfor an ellipse is
a2=b2+c2. Substituting the values we know:
32=b2+ 22
9 = b2+ 4
b2= 5
b=√5
Step 4: Write the equation of the ellipse. The standard form of the equation
of an ellipse centered at (h, k)is:
(x−h
a)2+ (y−k
b)2= 1
Plugging in the values of a,b,h, and k:
(x−1
3)2+ (y−1
√5)2= 1
Therefore, the standard form of the equation of the ellipse is (x−1
3)2+ (y−1
√5)2= 1 .
Question 3
Question
Solve the system of equations for xand y:
{9x2−4y2= 36
3x+ 4y= 6
2
Solution
Step 1: Let’s solve the second equation for x:
3x+ 4y= 6
3x= 6 −4y
x= 2 −4
3y
Step 2: Substitute the expression for xinto the first equation:
9(2 −4
3y)2−4y2= 36
9(4 −16
3y+16
9y2)−4y2= 36
36 −48y+ 16y2= 36 −4y2
Step 3: Simplify the equation by expanding and combining like terms:
36 −48y+ 16y2= 36 −4y2
16y2−48y+ 36 = 36 −4y2
20y2−48y= 0
Step 4: Factor out 4y:
4y(5y−12) = 0
Step 5: Solve for y:
4y= 0 or 5y−12 = 0
y= 0 or y=12
5
Step 6: Substitute the values of yback into the equation for xto find the
corresponding values: For y= 0, substituting into x= 2 −4
3ygives x= 2.
For y=12
5, substituting into x= 2 −4
3ygives x=−4
5.
Therefore, the solutions to the system of equations are (2,0) and (−4
5,12
5).
Question 4
Question
Solve the following system of equations:
{x2+y2= 16
2x−y= 3
3
Solution
Step 1: Solve the second equation for y in terms of x.
2x−y= 3
y= 2x−3
Step 2: Substitute yfrom the second equation into the first equation.
x2+ (2x−3)2= 16
x2+ 4x2−12x+ 9 = 16
5x2−12x−7 = 0
Step 3: Solve the quadratic equation 5x2−12x−7 = 0.
x=−(−12) ±√(−12)2−4(5)(−7)
2(5)
x=12 ±√144 + 140
10
x=12 ±√284
10
x=12 ±2√71
10
x=6±√71
5
Step 4: Substitute the values of xback into the equation y= 2x−3to find
the corresponding yvalues.
For x=6 + √71
5:y= 2 (6 + √71
5)−3 = 12 + 2√71
5−15
5=−3+2√71
5
For x=6−√71
5:y= 2 (6−√71
5)−3 = 12 −2√71
5−15
5=−3−2√71
5
Therefore, the solutions to the system of equations are:
(6 + √71
5,−3+2√71
5)and (6−√71
5,−3−2√71
5)
Question 5
Question
Given the equation of the hyperbola: (x−2)2
16 −(y+3)2
9= 1, find the standard
form of the equation of this hyperbola, the center, the vertices, the asymptotes,
and the foci.
4
Solution
Step 1: Identify the standard form of the equation of a hyperbola. Step 2: Use
properties of the given hyperbola to determine the values needed to represent
the equation in standard form. Step 3: Find the center of the hyperbola. Step
4: Calculate the vertices. Step 5: Determine the asymptotes. Step 6: Find the
foci.
Step 1: The standard form of the equation of a hyperbola is (x−h)2
a2−
(y−k)2
b2= 1, where (h, k)is the center of the hyperbola, ais the distance from
the center to the vertices along the x-axis, and bis the distance from the center
to the vertices along the y-axis.
Step 2: By comparing the given equation to the standard form, we have
h= 2,k=−3,a2= 16, and b2= 9.
Step 3: The center of the hyperbola is (h, k) = (2,−3).
Step 4: The vertices are given by (h±a, k), so the vertices are (2 + 4,−3)
and (2 −4,−3), which simplifies to (6,−3) and (−2,−3).
Step 5: The slopes of the asymptotes are given by ±b
a, so the slopes are
±3
4. Using the center (2,−3), we can find the equations of the asymptotes as
y+ 3 = ±3
4(x−2).
Step 6: To find the foci, we use the relationship c2=a2+b2. Solving
for c, we find c=√a2+b2=√16 + 9 = √25 = 5. The foci are located at
(h±c, k), so the foci are (2 + 5,−3) and (2 −5,−3), which simplifies to (7,−3)
and (−3,−3).
Question 6
Question
Solve the following equation for x:
9x2+ 16y2+ 12x−8y+ 4 = 0
Solution
Step 1: Rearrange the given equation by grouping the xterms and completing
the square for x.
9x2+ 16y2+ 12x−8y+ 4 = 0
9x2+ 12x+ 16y2−8y+ 4 = 0
9(x2+4
3x) + 16(y2−1
2y) = −4
Step 2: Complete the square for xby adding (4
3)2=16
9inside the parenthe-
5
ses.
9(x2+4
3x+16
9) + 16(y2−1
2y) = −4 + 9( 16
9)
9(x+4
3)2+ 16(y−1
4)2=−4 + 16
9(x+4
3)2+ 16(y−1
4)2= 12
Step 3: Divide both sides by the constant term on the right to obtain the
standard form of the equation for an ellipse.
9(x+4
3)2
12 +16(y−1
4)2
12 = 1
(x+4
3)2
(2√3
3)2+(y−1
4)2
(2√3
4)2= 1
Therefore, the equation represents an ellipse centered at (−4
3,1
4)with major
axis 2√3along the xaxis and minor axis 2√3along the yaxis.
Question 7
Question
Let y= 3x2−4be the equation of a parabola.
1. Find the coordinates of the vertex of the parabola.
2. Determine whether the parabola opens upwards or downwards.
3. Find the equation of the axis of symmetry.
Solution
1. To find the coordinates of the vertex of the parabola, we need to first rewrite
the given equation in vertex form. The vertex form of a parabola is y=a(x−
h)2+k, where (h, k)is the vertex of the parabola. Given y= 3x2−4, we can
rewrite it by completing the square:
y= 3(x2)−4
y= 3(x2−0x)−4
y= 3(x2−0x+ 02)−4−3(02)
y= 3(x−0)2−4
Comparing this with the vertex form, we see that a= 3,h= 0, and k=−4.
Therefore, the vertex of the parabola is (0,−4).
2. The parabola opens upwards if a > 0, and downwards if a < 0. Since
a= 3 in this case, the parabola opens upwards.
6
3. The equation of the axis of symmetry for a parabola in vertex form
y=a(x−h)2+kis given by x=h. Therefore, in this case, the equation of the
axis of symmetry is x= 0.
Question 8
Question
Solve the system of equations:
{x2+ 4y2= 16
3x2−2y2= 9
Solution
Step 1: Let’s write the given system of equations in standard form for conic
sections. We can rewrite the first equation as x2
16 +y2
4= 1 and the second
equation as x2
3−y2
9= 1.
Step 2: Next, notice that the first equation represents an ellipse centered at
the origin with major axis along the x-axis, while the second equation represents
a hyperbola centered at the origin with main axes along the x and y axes.
Step 3: Let’s solve the system by finding the points where the ellipse and
hyperbola intersect. Since each equation represents a curve, the solutions will
be the points where their graphs intersect.
Step 4: Substituting the first equation into the second equation, we get:
16
16 +y2
4= 1 =⇒1 + y2
4= 1 =⇒y2
4= 0 =⇒y= 0
Step 5: Substitute y= 0 back into the first equation to find the correspond-
ing x-values:
x2+ 4(0)2= 16 =⇒x2= 16 =⇒x=±4
Step 6: Therefore, the points of intersection are (4,0) and (−4,0). Thus,
the solution to the system of equations is (x, y) = (4,0),(−4,0) .
Question 9
Question
Simplify the following equation of a hyperbola in standard form:
(x−2)2
9−(y+ 1)2
25 = 1
7
Solution
Step 1: Identify the center of the hyperbola.
The standard form of the equation of a hyperbola is (x−h)2
a2−(y−k)2
b2= 1,
where (h, k)is the center of the hyperbola.
Comparing the given equation to the standard form, we can see that the
center of the hyperbola is at (2,−1).
Step 2: Determine the vertices of the hyperbola.
The vertices of the hyperbola are located at (h±a, k). In this case, the
vertices are at (2 ±3,−1), so the vertices are at (5,−1) and (−1,−1).
Step 3: Find the foci of the hyperbola.
The distance from the center to the foci is given by c=√a2+b2. Plugging
in the values a= 3 and b= 5, we find c=√9 + 25 = √34.
Therefore, the foci of the hyperbola are at (2 ±√34,−1), which are approx-
imately (6.83,−1) and (−2.83,−1).
Step 4: Determine the equations of the asymptotes of the hyperbola.
The equations of the asymptotes for a hyperbola in standard form can be
written as y−k=±b
a(x−h). Substituting the center and the values of aand b
into this equation, we get the equations of the asymptotes as y+ 1 = ±5
3(x−2).
Step 5: Sketch the hyperbola with the help of the center, vertices, foci, and
asymptotes.
Now that we have all the key points and information about the hyperbola,
we can sketch it on the coordinate plane.
This completes the solution.
Question 10
Question
Solve the following system of equations for xand y:
{2x2+ 5y2= 13
4x2−y2= 3
Solution
To solve the system of equations for xand y, we first need to isolate one variable
in one of the equations and substitute that into the other equation to find the
solution.
8
Step 1: Solve for yin terms of xin the first equation.
2x2+ 5y2= 13
5y2= 13 −2x2
y2=13 −2x2
5
y=±√13 −2x2
5
Step 2: Substitute yinto the second equation and solve for x.
4x2−(13 −2x2
5)= 3
20x2−13 + 2x2= 15
22x2= 28
x2=28
22
x=±√14
11
Step 3: Substitute xback into the equation to find the correspond-
ing yvalues.
y=±√13 −2(14
11 )
5
y=±√13 −28
11
5
y=±√143 −28
55
y=±√115
55
y=±√2
Therefore, the solutions to the system of equations are x=±√14
11 and
y=±√2.
Question 11
Question
Consider the equation of a conic section given by 5x2−12xy +5y2−18x−18y−
18 = 0. Determine the type of conic section represented by this equation and
sketch its graph.
9
Solution
Step 1: To determine the type of conic section represented by the given equation,
we need to consider the discriminant B2−4AC of the general conic section
equation Ax2+Bxy +Cy2+Dx +Ey +F= 0. In this case, A= 5,B=−12,
and C= 5.
Step 2: Calculate the discriminant:
B2−4AC = (−12)2−4(5)(5) = 144 −100 = 44
Step 3: Since the discriminant B2−4AC = 44 >0, the conic section is an
ellipse.
Step 4: Next, we can rewrite the given equation in standard form to deter-
mine the center, vertices, and major and minor axes of the ellipse. We start by
completing the square.
Step 5: Completing the square for 5x2−12xy + 5y2−18x−18y−18 = 0,
we get:
5(x2−2x)−12xy + 5(y2−2y) = 18
Step 6: Continuing to complete the square:
5(x2−2x+ 1) −12xy + 5(y2−2y+ 1) = 18 + 5 + 5
Step 7: Simplifying further:
5(x−1)2−12xy + 5(y−1)2= 28
Step 8: So, the standard form of the given ellipse equation is 5(x−1)2−
12xy + 5(y−1)2= 28.
Step 9: From the standard equation, we can identify the center of the ellipse
as (1,1).
Step 10: The major axis of the ellipse is along the line of symmetry defined
by the major axis direction. The minor axis is perpendicular to the major axis.
Step 11: Now, let’s sketch the graph of the ellipse with the center at (1,1)
and with major and minor axes determined by the equation.
Question 12
Question
Solve the following system of equations for xand y:
2x2+ 5xy + 3y2= 29
3x2−4xy −8y2=−7
10
Solution
Step 1: Let’s rewrite the system of equations as a matrix equation AX =B,
where A,X, and Bare as follows:
A=[2 5
3−4], X =[x
y], B =[29
−7]
Step 2: To solve for X, we need to find the inverse of matrix A. The inverse
of a 2×2matrix [a b
c d]is given by:
A−1=1
ad −bc [d−b
−c a ]
Step 3: Compute the determinant of matrix A:
|A|= (2 × −4) −(5 ×3) = −8−15 = −23
Step 4: Find the inverse of matrix A:
A−1=1
−23 [−4−5
−3 2 ]=[4
23
5
23
3
23 −2
23 ]
Step 5: Multiply A−1by matrix Bto solve for X:
X=A−1×B=[4
23
5
23
3
23 −2
23 ]×[29
−7]
X=[3
2]
Therefore, the solution to the system of equations is x= 3 and y= 2.
Question 13
Question
Solve the system of equations:
{x2+y2= 25
2x−y= 1
Solution
Step 1: Begin by solving the second equation for y in terms of x.
2x−y= 1 =⇒y= 2x−1
11
Step 2: Substitute the expression for y into the first equation.
x2+ (2x−1)2= 25
Step 3: Expand and simplify the equation.
x2+ 4x2−4x+ 1 = 25
Step 4: Combine like terms.
5x2−4x+ 1 = 25
Step 5: Subtract 25 from both sides.
5x2−4x−24 = 0
Step 6: Factor the quadratic equation.
(5x+ 12)(x−2) = 0
Step 7: Set each factor to zero and solve for x.
5x+ 12 = 0 or x−2 = 0
Step 8: Solve for x.
x=−12
5or x= 2
Step 9: Substitute the values of x back into the equation y= 2x−1to find
the corresponding y-values. For x=−12
5:
y= 2 ·(−12
5)−1 = −24
5−5
5=−29
5
For x= 2:
y= 2 ·2−1 = 4 −1 = 3
Step 10: The solutions to the system of equations are (−12
5,−29
5)and (2,3).
Question 14
Question
Determine the standard form equation of the hyperbola with vertices at (−3,0)
and (3,0) and foci at (−5,0) and (5,0).
12
Solution
Step 1: Determine the center of the hyperbola. The center of the hyperbola is
the midpoint of the segment connecting the vertices. The center of the hyperbola
is (−3+3
2,0+0
2)= (0,0).
Step 2: Determine the distance between the center and one of the vertices
to find a, the distance from the center to a vertex. 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 a focus. c= 5 −0 = 5.
Step 4: Use the relationship for a hyperbola, c2=a2+b2, to find b, where b
is the distance from the center to a co-vertex. 52= 32+b225 = 9 + b2b2= 16
b= 4 or b=−4
Step 5: Since the hyperbola is centered at the origin, the standard form
equation of the hyperbola is x2
a2−y2
b2= 1. Substituting the values of aand b,
we can write the equation as:
x2
32−y2
42= 1
x2
9−y2
16 = 1
Question 15
Question
Solve the following system of equations for xand y:
{4x2−y2= 16
x−y= 2
Solution
Step 1: We will first isolate yin the second equation to express yin terms of
x. Step 2: We will then substitute this expression for yinto the first equation
to solve for x. Step 3: Once we have found the value of x, we will substitute it
back into the second equation to find the corresponding value of y.
Step 1: Solve the second equation for y.
x−y= 2
y=x−2
Step 2: Substitute y=x−2into the first equation.
4x2−(x−2)2= 16
13
4x2−(x2−4x+ 4) = 16
4x2−x2+ 4x−4 = 16
3x2+ 4x−20 = 0
Step 3: Solve the quadratic equation 3x2+ 4x−20 = 0.
(3x−6)(x+ 5) = 0
x= 2,−5
3
Now, substitute x= 2 and x=−5
3back into y=x−2to find the corre-
sponding values of y.
If x= 2:
y= 2 −2 = 0
∴x= 2, y = 0
If x=−5
3:
y=−5
3−2 = −11
3
∴x=−5
3, y =−11
3
Therefore, the solutions to the system of equations are (2,0) and (−5
3,−11
3).
Question 16
Question
Determine the standard form of the equation of the parabola that satisfies the
given conditions: vertex at (1,−3) and directrix x= 5.
Solution
Step 1: The standard form of the equation of a vertical parabola with vertex
(h, k)and directrix x=ais given by (x−h)2= 4p(y−k), where pis the
distance between the vertex and the focus.
Step 2: Since the directrix is to the right of the vertex, pis positive. The
distance between the vertex (h, k)and the directrix x=ais p.
Step 3: In this case, the vertex is at (1,−3) and the directrix is x= 5. Since
pis the distance between the vertex and the directrix, we have p= 5 −1 = 4.
Step 4: The equation of the parabola is:
(x−1)2= 4(4)(y+ 3)
Step 5:
(x−1)2= 16(y+ 3)
Therefore, the standard form of the equation of the parabola is (x−1)2=
16(y+ 3).
14
Question 17
Question
Solve the equation 4x2−9y2−16x+ 18y+ 56 = 0 and determine the type of
conic section it represents.
Solution
Step 1: Rearrange the equation by grouping like terms:
(4x2−16x)−(9y2−18y) = −56
Step 2: Complete the square for xby adding and subtracting (16/2)2= 42=
16 inside the parentheses:
4(x2−4x+ 4) −(9y2−18y) = −56 + 4(4)
Step 3: Complete the square for yby adding and subtracting (18/2)2= 92=
81 inside the parentheses:
4(x2−4x+ 4) −9(y2−2y+ 1) = −56 + 16 −81
Step 4: Simplify the equation:
4(x−2)2−9(y−1)2=−121
Step 5: Divide by −121 to get the equation in the standard form:
(x−2)2
9−(y−1)2
(−121)/9= 1
Step 6: Compare the equation to the standard form of a conic section to
determine its type: The equation represents a hyperbola since the terms have
opposite signs in the standard form equation.
Question 18
Question
Find the standard form of the equation of the hyperbola that satisfies the given
conditions: vertices at (−5,1) and (1,1), and passes through the point (3,4).
Solution
Step 1: Find the center of the hyperbola.
The center of the hyperbola is the midpoint of the segment connecting the
vertices. Use the midpoint formula M=(x1+x2
2,y1+y2
2).
15
Given vertices: (−5,1) and (1,1)
Center M=(−5+1
2,1+1
2)
Center M= (−2,1)
Step 2: Find the equation of the hyperbola with center (−2,1).
Since the hyperbola has a horizontal transverse axis, the standard form of
the equation of the hyperbola is
(x−h)2/a2−(y−k)2/b2= 1
where (h, k)is the center of the hyperbola, ais the distance from the center
to a vertex, and bis the distance from the center to a co-vertex.
Given h=−2,k= 1, the vertices are at (−5,1) and (1,1).
ais the distance from the center to a vertex, so a=| − 2−(−5)|= 3.
So, we have
(x+ 2)2/32−(y−1)2/b2= 1
Step 3: Find the value of b.
The hyperbola passes through the point (3,4). Substitute x= 3 and y= 4
into the equation:
(3+2
3)2
−(4 −1)2/b2= 1
25/9−9/b2= 1
Solve for b2:
b2= 9/(25/9−1) = 9/(16/9) = 81/16
b=√81/16 = 9/4
Thus, the equation of the hyperbola is:
(x+ 2)2
9−(y−1)2
(9
4)2= 1
Question 19
Question
Find the standard form equation of the conic section with the following prop-
erties: a parabola with a focus at F(3,1) and a directrix y=−1.
Solution
Step 1: Recall that the standard form equation for a parabola with a vertical
axis of symmetry is given by (x−h)2= 4p(y−k).
Step 2: The vertex of the parabola is equidistant from the focus and the
directrix. Since the directrix is y=−1, the vertex, V, must have a y-coordinate
16
equal to the average of the y-coordinate of the focus and the y-coordinate of a
point on the directrix. Therefore, Vis at (3,0).
Step 3: The distance between the focus Fand the vertex Vgives the value
of pin the equation. Using the distance formula, we find that p= 1.
Step 4: Substituting the values of h,k, and pinto the standard form equa-
tion, we get (x−3)2= 4(y).
Hence, the standard form equation of the parabola is (x−3)2= 4y.
Question 20
Question
Find the equation of the hyperbola with foci at (−5,0) and (5,0) and vertices
at (−8,0) and (8,0).
Solution
Step 1: Find the center of the hyperbola. Since the center is the midpoint of
the vertices, we have:
(h, k) = (−8+8
2,0)= (0,0)
Step 2: Find the distance between the center and the foci. The distance
between the foci and the center, which is the value of c, is:
c= 5 −(−5) = 10
Step 3: Find the value of a(distance from the center to the vertex). The
value of ais the distance between the center and one of the vertices:
a= 8 −0 = 8
Step 4: Determine the equation of the hyperbola. The equation of a hyper-
bola centered at (h, k)with vertices along the x-axis is:
(x−h)2
a2−(y−k)2
b2= 1
Step 5: Find the value of busing the relationship c2=a2+b2:
b=√c2−a2=√102−82=√100 −64 = √36 = 6
Step 6: Substitute the known values into the equation of the hyperbola:
x2
64 −y2
36 = 1
Therefore, the equation of the hyperbola is x2
64 −y2
36 = 1.
17
Question 21
Question
Solve the following system of equations:
{x2+y2= 25
x−2y= 4
Solution
Step 1: Solve the second equation for xin terms of y:
x−2y= 4
x= 4 + 2y
Step 2: Substitute xfrom the second equation into the first equation:
(4 + 2y)2+y2= 25
16 + 16y+ 4y2+y2= 25
4y2+ 17y−9 = 0
Step 3: Solve the quadratic equation 4y2+ 17y−9=0using the quadratic
formula:
y=−b±√b2−4ac
2a
y=−17 ±√172−4(4)(−9)
2(4)
y=−17 ±√289 + 144
8
y=−17 ±√433
8
Step 4: Calculate the values of y:
y=−17 + √433
8or y=−17 −√433
8
Step 5: Substitute the values of yback into the equation x= 4 + 2yto find
the corresponding values of x:
For y=−17 + √433
8:x= 4 + 2 (−17 + √433
8)
For y=−17 −√433
8:x= 4 + 2 (−17 −√433
8)
Therefore, the solution to the system of equations is:
(−17 + √433
8,−17 + √433
4)and (−17 −√433
8,−17 −√433
4)
18
Question 22
Question
Solve the following system of equations:
{4x2+y2= 25
x−y= 1
Solution
Step 1: Solve the second equation for xin terms of y. Step 2: Substitute xin
terms of yinto the first equation. Step 3: Solve the resulting equation to find
the possible values of y. Step 4: Substitute the values of yback into the second
equation to find the corresponding values of x.
Step 1: Solve the second equation for xin terms of y. From x−y= 1, we
have x=y+ 1.
Step 2: Substitute xin terms of yinto the first equation. Substitute x=
y+ 1 into 4x2+y2= 25:
4(y+ 1)2+y2= 25
Step 3: Solve the resulting equation to find the possible values of y. Ex-
panding the left side of the equation:
4(y2+ 2y+ 1) + y2= 25
4y2+ 8y+4+y2= 25
5y2+ 8y−21 = 0
Using the quadratic formula to solve for y:
y=−8±√82−4(5)(−21)
2(5)
y=−8±√64 + 420
10
y=−8±√484
10
y=−8±22
10
Therefore, y=−3.4or y= 3.0.
Step 4: Substitute the values of yback into the second equation to find the
corresponding values of x. For y=−3.4:x=−3.4 + 1 = −2.4
For y= 3.0:x= 3.0 + 1 = 4.0
Thus, the solutions to the system of equations are (−2.4,−3.4) and (4.0,3.0).
19
Question 23
Question
Solve the following system of equations:
{x2+y2−6x+ 4y−12 = 0
x−2y= 0
Solution
Step 1: We can solve the second equation for x:
x= 2y
Step 2: Substitute xin terms of yinto the first equation:
(2y)2+y2−6(2y)+4y−12 = 0
Step 3: Simplify the equation:
4y2+y2−12y+ 4y−12 = 0
5y2−8y−12 = 0
Step 4: Solve the quadratic equation by factoring or using the quadratic
formula: The equation 5y2−8y−12 = 0 can be factored as (5y+ 6)(y−2) = 0.
This gives us two possible solutions: y=−6
5or y= 2.
Step 5: Substitute the values of yback into the equation x= 2yto find the
corresponding values of x. For y=−6
5:
x= 2 (−6
5)=−12
5
For y= 2:
x= 2(2) = 4
So, the solutions to the system of equations are: (x, y) = (−12
5,−6
5)and
(x, y) = (4,2).
Question 24
Question
Find the standard form of the equation of the parabola with focus F(2,3) and
directrix x= 1.
20
Solution
Step 1: Find the vertex of the parabola. The vertex of the parabola is the
midpoint between the focus and the directrix. We have the focus F(2,3) and
the directrix x= 1. Using the formula for the midpoint between two points, the
vertex Vis:
V=(1+2
2,3+3
2)=(3
2,3).
Step 2: Find the equation of the axis of symmetry. The axis of symmetry is
a vertical line passing through the vertex. Therefore, the equation of the axis
of symmetry is x=3
2.
Step 3: Determine whether the parabola opens to the left or right. Since
the directrix is to the left of the focus, the parabola opens to the right.
Step 4: Find the distance between the vertex and the focus. The distance
between the vertex and the focus is the same as the distance between the vertex
and the directrix. This distance is called the focal length, denoted as |p|. In
this case, |p|=|1−3
2|=1
2.
Step 5: Write the standard form of the equation. Since the parabola opens
to the right, the standard form of the equation is:
(x−h)2= 4p(y−k),
where (h, k)is the vertex. Plugging in the values of the vertex (3
2,3) and |p|=1
2,
we have:
(x−3
2)2= 4(1
2)(y−3).
Simplifying, we get:
(x−3
2)2= 2(y−3).
Therefore, the standard form of the equation of the parabola is (x−3
2)2=
2(y−3).
Question 25
Question
Find the standard form equation of the ellipse that satisfies the given conditions:
The major axis is along the line 3x−4y= 2 and passes through the points (−1,3)
and (5,1).
Solution
Step 1: Find the center of the ellipse using the midpoint formula. The midpoint
of the line segment connecting the points (−1,3) and (5,1) is the center of the
ellipse. Let the center be (h, k). The midpoint formula is given by:
(x1+x2
2,y1+y2
2)
21
Applying this formula to (−1,3) and (5,1):
(−1+5
2,3+1
2)= (2,2)
So, the center of the ellipse is at (h, k) = (2,2).
Step 2: Find the distance between the two given points which is the length
of the major axis. The distance formula is given by:
d=√(x2−x1)2+ (y2−y1)2
Calculating the distance between (−1,3) and (5,1):
d=√(5 −(−1))2+ (1 −3)2=√36 + 4 = √40 = 2√10
So, the length of the major axis is 2√10.
Step 3: Determine the major axis direction. We have the equation of the
major axis as 3x−4y= 2. Rewrite it in slope-intercept form as y=3
4x−1
2.
This means the slope of the major axis is 3
4. Therefore, the major axis of the
ellipse is in the direction of the line y=3
4x.
Step 4: Find the equation of the ellipse in standard form. The standard
form of the equation of an ellipse with major axis along the x-axis is:
(x−h)2
a2+(y−k)2
b2= 1
where (h, k)is the center of the ellipse, ais half the length of the major axis,
and bis half the length of the minor axis. In this case, since the major axis is
along the line y=3
4x, the equation becomes:
(x−2)2
10 +(y−2)2
2= 1
Therefore, the standard form equation of the ellipse is
(x−2)2
10 +(y−2)2
2= 1
.
22
Question 2
Question
Find the standard form of the equation of the ellipse with foci at (−1,1) and
(3,1) and a major axis of length 6 units.
Solution
Step 1: Find the center of the ellipse. The center of the ellipse is the midpoint
between the foci. Let’s calculate it using the midpoint formula:
(−1+3
2,1+1
2) = (1,1)
Step 2: Find the distance between the foci. Let 2cbe the distance between
the foci of the ellipse. The major axis is 2a= 6 units, so a= 3 units. Using the
distance formula:
2c=√(3 −(−1))2+ (1 −1)2=√16 = 4
Thus, c= 2.
Step 3: Find the value of b. The formula relating a,b, and cfor an ellipse is
a2=b2+c2. Substituting the values we know:
32=b2+ 22
9 = b2+ 4
b2= 5
b=√5
Step 4: Write the equation of the ellipse. The standard form of the equation
of an ellipse centered at (h, k)is:
(x−h
a)2+ (y−k
b)2= 1
Plugging in the values of a,b,h, and k:
(x−1
3)2+ (y−1
√5)2= 1
Therefore, the standard form of the equation of the ellipse is (x−1
3)2+ (y−1
√5)2= 1 .
Question 3
Question
Solve the system of equations for xand y:
{9x2−4y2= 36
3x+ 4y= 6
2
Solution
Step 1: Let’s solve the second equation for x:
3x+ 4y= 6
3x= 6 −4y
x= 2 −4
3y
Step 2: Substitute the expression for xinto the first equation:
9(2 −4
3y)2−4y2= 36
9(4 −16
3y+16
9y2)−4y2= 36
36 −48y+ 16y2= 36 −4y2
Step 3: Simplify the equation by expanding and combining like terms:
36 −48y+ 16y2= 36 −4y2
16y2−48y+ 36 = 36 −4y2
20y2−48y= 0
Step 4: Factor out 4y:
4y(5y−12) = 0
Step 5: Solve for y:
4y= 0 or 5y−12 = 0
y= 0 or y=12
5
Step 6: Substitute the values of yback into the equation for xto find the
corresponding values: For y= 0, substituting into x= 2 −4
3ygives x= 2.
For y=12
5, substituting into x= 2 −4
3ygives x=−4
5.
Therefore, the solutions to the system of equations are (2,0) and (−4
5,12
5).
Question 4
Question
Solve the following system of equations:
{x2+y2= 16
2x−y= 3
3
Solution
Step 1: Solve the second equation for y in terms of x.
2x−y= 3
y= 2x−3
Step 2: Substitute yfrom the second equation into the first equation.
x2+ (2x−3)2= 16
x2+ 4x2−12x+ 9 = 16
5x2−12x−7 = 0
Step 3: Solve the quadratic equation 5x2−12x−7 = 0.
x=−(−12) ±√(−12)2−4(5)(−7)
2(5)
x=12 ±√144 + 140
10
x=12 ±√284
10
x=12 ±2√71
10
x=6±√71
5
Step 4: Substitute the values of xback into the equation y= 2x−3to find
the corresponding yvalues.
For x=6 + √71
5:y= 2 (6 + √71
5)−3 = 12 + 2√71
5−15
5=−3+2√71
5
For x=6−√71
5:y= 2 (6−√71
5)−3 = 12 −2√71
5−15
5=−3−2√71
5
Therefore, the solutions to the system of equations are:
(6 + √71
5,−3+2√71
5)and (6−√71
5,−3−2√71
5)
Question 5
Question
Given the equation of the hyperbola: (x−2)2
16 −(y+3)2
9= 1, find the standard
form of the equation of this hyperbola, the center, the vertices, the asymptotes,
and the foci.
4
Solution
Step 1: Identify the standard form of the equation of a hyperbola. Step 2: Use
properties of the given hyperbola to determine the values needed to represent
the equation in standard form. Step 3: Find the center of the hyperbola. Step
4: Calculate the vertices. Step 5: Determine the asymptotes. Step 6: Find the
foci.
Step 1: The standard form of the equation of a hyperbola is (x−h)2
a2−
(y−k)2
b2= 1, where (h, k)is the center of the hyperbola, ais the distance from
the center to the vertices along the x-axis, and bis the distance from the center
to the vertices along the y-axis.
Step 2: By comparing the given equation to the standard form, we have
h= 2,k=−3,a2= 16, and b2= 9.
Step 3: The center of the hyperbola is (h, k) = (2,−3).
Step 4: The vertices are given by (h±a, k), so the vertices are (2 + 4,−3)
and (2 −4,−3), which simplifies to (6,−3) and (−2,−3).
Step 5: The slopes of the asymptotes are given by ±b
a, so the slopes are
±3
4. Using the center (2,−3), we can find the equations of the asymptotes as
y+ 3 = ±3
4(x−2).
Step 6: To find the foci, we use the relationship c2=a2+b2. Solving
for c, we find c=√a2+b2=√16 + 9 = √25 = 5. The foci are located at
(h±c, k), so the foci are (2 + 5,−3) and (2 −5,−3), which simplifies to (7,−3)
and (−3,−3).
Question 6
Question
Solve the following equation for x:
9x2+ 16y2+ 12x−8y+ 4 = 0
Solution
Step 1: Rearrange the given equation by grouping the xterms and completing
the square for x.
9x2+ 16y2+ 12x−8y+ 4 = 0
9x2+ 12x+ 16y2−8y+ 4 = 0
9(x2+4
3x) + 16(y2−1
2y) = −4
Step 2: Complete the square for xby adding (4
3)2=16
9inside the parenthe-
5
ses.
9(x2+4
3x+16
9) + 16(y2−1
2y) = −4 + 9( 16
9)
9(x+4
3)2+ 16(y−1
4)2=−4 + 16
9(x+4
3)2+ 16(y−1
4)2= 12
Step 3: Divide both sides by the constant term on the right to obtain the
standard form of the equation for an ellipse.
9(x+4
3)2
12 +16(y−1
4)2
12 = 1
(x+4
3)2
(2√3
3)2+(y−1
4)2
(2√3
4)2= 1
Therefore, the equation represents an ellipse centered at (−4
3,1
4)with major
axis 2√3along the xaxis and minor axis 2√3along the yaxis.
Question 7
Question
Let y= 3x2−4be the equation of a parabola.
1. Find the coordinates of the vertex of the parabola.
2. Determine whether the parabola opens upwards or downwards.
3. Find the equation of the axis of symmetry.
Solution
1. To find the coordinates of the vertex of the parabola, we need to first rewrite
the given equation in vertex form. The vertex form of a parabola is y=a(x−
h)2+k, where (h, k)is the vertex of the parabola. Given y= 3x2−4, we can
rewrite it by completing the square:
y= 3(x2)−4
y= 3(x2−0x)−4
y= 3(x2−0x+ 02)−4−3(02)
y= 3(x−0)2−4
Comparing this with the vertex form, we see that a= 3,h= 0, and k=−4.
Therefore, the vertex of the parabola is (0,−4).
2. The parabola opens upwards if a > 0, and downwards if a < 0. Since
a= 3 in this case, the parabola opens upwards.
6
3. The equation of the axis of symmetry for a parabola in vertex form
y=a(x−h)2+kis given by x=h. Therefore, in this case, the equation of the
axis of symmetry is x= 0.
Question 8
Question
Solve the system of equations:
{x2+ 4y2= 16
3x2−2y2= 9
Solution
Step 1: Let’s write the given system of equations in standard form for conic
sections. We can rewrite the first equation as x2
16 +y2
4= 1 and the second
equation as x2
3−y2
9= 1.
Step 2: Next, notice that the first equation represents an ellipse centered at
the origin with major axis along the x-axis, while the second equation represents
a hyperbola centered at the origin with main axes along the x and y axes.
Step 3: Let’s solve the system by finding the points where the ellipse and
hyperbola intersect. Since each equation represents a curve, the solutions will
be the points where their graphs intersect.
Step 4: Substituting the first equation into the second equation, we get:
16
16 +y2
4= 1 =⇒1 + y2
4= 1 =⇒y2
4= 0 =⇒y= 0
Step 5: Substitute y= 0 back into the first equation to find the correspond-
ing x-values:
x2+ 4(0)2= 16 =⇒x2= 16 =⇒x=±4
Step 6: Therefore, the points of intersection are (4,0) and (−4,0). Thus,
the solution to the system of equations is (x, y) = (4,0),(−4,0) .
Question 9
Question
Simplify the following equation of a hyperbola in standard form:
(x−2)2
9−(y+ 1)2
25 = 1
7
Solution
Step 1: Identify the center of the hyperbola.
The standard form of the equation of a hyperbola is (x−h)2
a2−(y−k)2
b2= 1,
where (h, k)is the center of the hyperbola.
Comparing the given equation to the standard form, we can see that the
center of the hyperbola is at (2,−1).
Step 2: Determine the vertices of the hyperbola.
The vertices of the hyperbola are located at (h±a, k). In this case, the
vertices are at (2 ±3,−1), so the vertices are at (5,−1) and (−1,−1).
Step 3: Find the foci of the hyperbola.
The distance from the center to the foci is given by c=√a2+b2. Plugging
in the values a= 3 and b= 5, we find c=√9 + 25 = √34.
Therefore, the foci of the hyperbola are at (2 ±√34,−1), which are approx-
imately (6.83,−1) and (−2.83,−1).
Step 4: Determine the equations of the asymptotes of the hyperbola.
The equations of the asymptotes for a hyperbola in standard form can be
written as y−k=±b
a(x−h). Substituting the center and the values of aand b
into this equation, we get the equations of the asymptotes as y+ 1 = ±5
3(x−2).
Step 5: Sketch the hyperbola with the help of the center, vertices, foci, and
asymptotes.
Now that we have all the key points and information about the hyperbola,
we can sketch it on the coordinate plane.
This completes the solution.
Question 10
Question
Solve the following system of equations for xand y:
{2x2+ 5y2= 13
4x2−y2= 3
Solution
To solve the system of equations for xand y, we first need to isolate one variable
in one of the equations and substitute that into the other equation to find the
solution.
8
Step 1: Solve for yin terms of xin the first equation.
2x2+ 5y2= 13
5y2= 13 −2x2
y2=13 −2x2
5
y=±√13 −2x2
5
Step 2: Substitute yinto the second equation and solve for x.
4x2−(13 −2x2
5)= 3
20x2−13 + 2x2= 15
22x2= 28
x2=28
22
x=±√14
11
Step 3: Substitute xback into the equation to find the correspond-
ing yvalues.
y=±√13 −2(14
11 )
5
y=±√13 −28
11
5
y=±√143 −28
55
y=±√115
55
y=±√2
Therefore, the solutions to the system of equations are x=±√14
11 and
y=±√2.
Question 11
Question
Consider the equation of a conic section given by 5x2−12xy +5y2−18x−18y−
18 = 0. Determine the type of conic section represented by this equation and
sketch its graph.
9
Solution
Step 1: To determine the type of conic section represented by the given equation,
we need to consider the discriminant B2−4AC of the general conic section
equation Ax2+Bxy +Cy2+Dx +Ey +F= 0. In this case, A= 5,B=−12,
and C= 5.
Step 2: Calculate the discriminant:
B2−4AC = (−12)2−4(5)(5) = 144 −100 = 44
Step 3: Since the discriminant B2−4AC = 44 >0, the conic section is an
ellipse.
Step 4: Next, we can rewrite the given equation in standard form to deter-
mine the center, vertices, and major and minor axes of the ellipse. We start by
completing the square.
Step 5: Completing the square for 5x2−12xy + 5y2−18x−18y−18 = 0,
we get:
5(x2−2x)−12xy + 5(y2−2y) = 18
Step 6: Continuing to complete the square:
5(x2−2x+ 1) −12xy + 5(y2−2y+ 1) = 18 + 5 + 5
Step 7: Simplifying further:
5(x−1)2−12xy + 5(y−1)2= 28
Step 8: So, the standard form of the given ellipse equation is 5(x−1)2−
12xy + 5(y−1)2= 28.
Step 9: From the standard equation, we can identify the center of the ellipse
as (1,1).
Step 10: The major axis of the ellipse is along the line of symmetry defined
by the major axis direction. The minor axis is perpendicular to the major axis.
Step 11: Now, let’s sketch the graph of the ellipse with the center at (1,1)
and with major and minor axes determined by the equation.
Question 12
Question
Solve the following system of equations for xand y:
2x2+ 5xy + 3y2= 29
3x2−4xy −8y2=−7
10
Solution
Step 1: Let’s rewrite the system of equations as a matrix equation AX =B,
where A,X, and Bare as follows:
A=[2 5
3−4], X =[x
y], B =[29
−7]
Step 2: To solve for X, we need to find the inverse of matrix A. The inverse
of a 2×2matrix [a b
c d]is given by:
A−1=1
ad −bc [d−b
−c a ]
Step 3: Compute the determinant of matrix A:
|A|= (2 × −4) −(5 ×3) = −8−15 = −23
Step 4: Find the inverse of matrix A:
A−1=1
−23 [−4−5
−3 2 ]=[4
23
5
23
3
23 −2
23 ]
Step 5: Multiply A−1by matrix Bto solve for X:
X=A−1×B=[4
23
5
23
3
23 −2
23 ]×[29
−7]
X=[3
2]
Therefore, the solution to the system of equations is x= 3 and y= 2.
Question 13
Question
Solve the system of equations:
{x2+y2= 25
2x−y= 1
Solution
Step 1: Begin by solving the second equation for y in terms of x.
2x−y= 1 =⇒y= 2x−1
11
Step 2: Substitute the expression for y into the first equation.
x2+ (2x−1)2= 25
Step 3: Expand and simplify the equation.
x2+ 4x2−4x+ 1 = 25
Step 4: Combine like terms.
5x2−4x+ 1 = 25
Step 5: Subtract 25 from both sides.
5x2−4x−24 = 0
Step 6: Factor the quadratic equation.
(5x+ 12)(x−2) = 0
Step 7: Set each factor to zero and solve for x.
5x+ 12 = 0 or x−2 = 0
Step 8: Solve for x.
x=−12
5or x= 2
Step 9: Substitute the values of x back into the equation y= 2x−1to find
the corresponding y-values. For x=−12
5:
y= 2 ·(−12
5)−1 = −24
5−5
5=−29
5
For x= 2:
y= 2 ·2−1 = 4 −1 = 3
Step 10: The solutions to the system of equations are (−12
5,−29
5)and (2,3).
Question 14
Question
Determine the standard form equation of the hyperbola with vertices at (−3,0)
and (3,0) and foci at (−5,0) and (5,0).
12
Solution
Step 1: Determine the center of the hyperbola. The center of the hyperbola is
the midpoint of the segment connecting the vertices. The center of the hyperbola
is (−3+3
2,0+0
2)= (0,0).
Step 2: Determine the distance between the center and one of the vertices
to find a, the distance from the center to a vertex. 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 a focus. c= 5 −0 = 5.
Step 4: Use the relationship for a hyperbola, c2=a2+b2, to find b, where b
is the distance from the center to a co-vertex. 52= 32+b225 = 9 + b2b2= 16
b= 4 or b=−4
Step 5: Since the hyperbola is centered at the origin, the standard form
equation of the hyperbola is x2
a2−y2
b2= 1. Substituting the values of aand b,
we can write the equation as:
x2
32−y2
42= 1
x2
9−y2
16 = 1
Question 15
Question
Solve the following system of equations for xand y:
{4x2−y2= 16
x−y= 2
Solution
Step 1: We will first isolate yin the second equation to express yin terms of
x. Step 2: We will then substitute this expression for yinto the first equation
to solve for x. Step 3: Once we have found the value of x, we will substitute it
back into the second equation to find the corresponding value of y.
Step 1: Solve the second equation for y.
x−y= 2
y=x−2
Step 2: Substitute y=x−2into the first equation.
4x2−(x−2)2= 16
13
4x2−(x2−4x+ 4) = 16
4x2−x2+ 4x−4 = 16
3x2+ 4x−20 = 0
Step 3: Solve the quadratic equation 3x2+ 4x−20 = 0.
(3x−6)(x+ 5) = 0
x= 2,−5
3
Now, substitute x= 2 and x=−5
3back into y=x−2to find the corre-
sponding values of y.
If x= 2:
y= 2 −2 = 0
∴x= 2, y = 0
If x=−5
3:
y=−5
3−2 = −11
3
∴x=−5
3, y =−11
3
Therefore, the solutions to the system of equations are (2,0) and (−5
3,−11
3).
Question 16
Question
Determine the standard form of the equation of the parabola that satisfies the
given conditions: vertex at (1,−3) and directrix x= 5.
Solution
Step 1: The standard form of the equation of a vertical parabola with vertex
(h, k)and directrix x=ais given by (x−h)2= 4p(y−k), where pis the
distance between the vertex and the focus.
Step 2: Since the directrix is to the right of the vertex, pis positive. The
distance between the vertex (h, k)and the directrix x=ais p.
Step 3: In this case, the vertex is at (1,−3) and the directrix is x= 5. Since
pis the distance between the vertex and the directrix, we have p= 5 −1 = 4.
Step 4: The equation of the parabola is:
(x−1)2= 4(4)(y+ 3)
Step 5:
(x−1)2= 16(y+ 3)
Therefore, the standard form of the equation of the parabola is (x−1)2=
16(y+ 3).
14
Question 17
Question
Solve the equation 4x2−9y2−16x+ 18y+ 56 = 0 and determine the type of
conic section it represents.
Solution
Step 1: Rearrange the equation by grouping like terms:
(4x2−16x)−(9y2−18y) = −56
Step 2: Complete the square for xby adding and subtracting (16/2)2= 42=
16 inside the parentheses:
4(x2−4x+ 4) −(9y2−18y) = −56 + 4(4)
Step 3: Complete the square for yby adding and subtracting (18/2)2= 92=
81 inside the parentheses:
4(x2−4x+ 4) −9(y2−2y+ 1) = −56 + 16 −81
Step 4: Simplify the equation:
4(x−2)2−9(y−1)2=−121
Step 5: Divide by −121 to get the equation in the standard form:
(x−2)2
9−(y−1)2
(−121)/9= 1
Step 6: Compare the equation to the standard form of a conic section to
determine its type: The equation represents a hyperbola since the terms have
opposite signs in the standard form equation.
Question 18
Question
Find the standard form of the equation of the hyperbola that satisfies the given
conditions: vertices at (−5,1) and (1,1), and passes through the point (3,4).
Solution
Step 1: Find the center of the hyperbola.
The center of the hyperbola is the midpoint of the segment connecting the
vertices. Use the midpoint formula M=(x1+x2
2,y1+y2
2).
15
Given vertices: (−5,1) and (1,1)
Center M=(−5+1
2,1+1
2)
Center M= (−2,1)
Step 2: Find the equation of the hyperbola with center (−2,1).
Since the hyperbola has a horizontal transverse axis, the standard form of
the equation of the hyperbola is
(x−h)2/a2−(y−k)2/b2= 1
where (h, k)is the center of the hyperbola, ais the distance from the center
to a vertex, and bis the distance from the center to a co-vertex.
Given h=−2,k= 1, the vertices are at (−5,1) and (1,1).
ais the distance from the center to a vertex, so a=| − 2−(−5)|= 3.
So, we have
(x+ 2)2/32−(y−1)2/b2= 1
Step 3: Find the value of b.
The hyperbola passes through the point (3,4). Substitute x= 3 and y= 4
into the equation:
(3+2
3)2
−(4 −1)2/b2= 1
25/9−9/b2= 1
Solve for b2:
b2= 9/(25/9−1) = 9/(16/9) = 81/16
b=√81/16 = 9/4
Thus, the equation of the hyperbola is:
(x+ 2)2
9−(y−1)2
(9
4)2= 1
Question 19
Question
Find the standard form equation of the conic section with the following prop-
erties: a parabola with a focus at F(3,1) and a directrix y=−1.
Solution
Step 1: Recall that the standard form equation for a parabola with a vertical
axis of symmetry is given by (x−h)2= 4p(y−k).
Step 2: The vertex of the parabola is equidistant from the focus and the
directrix. Since the directrix is y=−1, the vertex, V, must have a y-coordinate
16
equal to the average of the y-coordinate of the focus and the y-coordinate of a
point on the directrix. Therefore, Vis at (3,0).
Step 3: The distance between the focus Fand the vertex Vgives the value
of pin the equation. Using the distance formula, we find that p= 1.
Step 4: Substituting the values of h,k, and pinto the standard form equa-
tion, we get (x−3)2= 4(y).
Hence, the standard form equation of the parabola is (x−3)2= 4y.
Question 20
Question
Find the equation of the hyperbola with foci at (−5,0) and (5,0) and vertices
at (−8,0) and (8,0).
Solution
Step 1: Find the center of the hyperbola. Since the center is the midpoint of
the vertices, we have:
(h, k) = (−8+8
2,0)= (0,0)
Step 2: Find the distance between the center and the foci. The distance
between the foci and the center, which is the value of c, is:
c= 5 −(−5) = 10
Step 3: Find the value of a(distance from the center to the vertex). The
value of ais the distance between the center and one of the vertices:
a= 8 −0 = 8
Step 4: Determine the equation of the hyperbola. The equation of a hyper-
bola centered at (h, k)with vertices along the x-axis is:
(x−h)2
a2−(y−k)2
b2= 1
Step 5: Find the value of busing the relationship c2=a2+b2:
b=√c2−a2=√102−82=√100 −64 = √36 = 6
Step 6: Substitute the known values into the equation of the hyperbola:
x2
64 −y2
36 = 1
Therefore, the equation of the hyperbola is x2
64 −y2
36 = 1.
17
Question 21
Question
Solve the following system of equations:
{x2+y2= 25
x−2y= 4
Solution
Step 1: Solve the second equation for xin terms of y:
x−2y= 4
x= 4 + 2y
Step 2: Substitute xfrom the second equation into the first equation:
(4 + 2y)2+y2= 25
16 + 16y+ 4y2+y2= 25
4y2+ 17y−9 = 0
Step 3: Solve the quadratic equation 4y2+ 17y−9=0using the quadratic
formula:
y=−b±√b2−4ac
2a
y=−17 ±√172−4(4)(−9)
2(4)
y=−17 ±√289 + 144
8
y=−17 ±√433
8
Step 4: Calculate the values of y:
y=−17 + √433
8or y=−17 −√433
8
Step 5: Substitute the values of yback into the equation x= 4 + 2yto find
the corresponding values of x:
For y=−17 + √433
8:x= 4 + 2 (−17 + √433
8)
For y=−17 −√433
8:x= 4 + 2 (−17 −√433
8)
Therefore, the solution to the system of equations is:
(−17 + √433
8,−17 + √433
4)and (−17 −√433
8,−17 −√433
4)
18
Question 22
Question
Solve the following system of equations:
{4x2+y2= 25
x−y= 1
Solution
Step 1: Solve the second equation for xin terms of y. Step 2: Substitute xin
terms of yinto the first equation. Step 3: Solve the resulting equation to find
the possible values of y. Step 4: Substitute the values of yback into the second
equation to find the corresponding values of x.
Step 1: Solve the second equation for xin terms of y. From x−y= 1, we
have x=y+ 1.
Step 2: Substitute xin terms of yinto the first equation. Substitute x=
y+ 1 into 4x2+y2= 25:
4(y+ 1)2+y2= 25
Step 3: Solve the resulting equation to find the possible values of y. Ex-
panding the left side of the equation:
4(y2+ 2y+ 1) + y2= 25
4y2+ 8y+4+y2= 25
5y2+ 8y−21 = 0
Using the quadratic formula to solve for y:
y=−8±√82−4(5)(−21)
2(5)
y=−8±√64 + 420
10
y=−8±√484
10
y=−8±22
10
Therefore, y=−3.4or y= 3.0.
Step 4: Substitute the values of yback into the second equation to find the
corresponding values of x. For y=−3.4:x=−3.4 + 1 = −2.4
For y= 3.0:x= 3.0 + 1 = 4.0
Thus, the solutions to the system of equations are (−2.4,−3.4) and (4.0,3.0).
19
Question 23
Question
Solve the following system of equations:
{x2+y2−6x+ 4y−12 = 0
x−2y= 0
Solution
Step 1: We can solve the second equation for x:
x= 2y
Step 2: Substitute xin terms of yinto the first equation:
(2y)2+y2−6(2y)+4y−12 = 0
Step 3: Simplify the equation:
4y2+y2−12y+ 4y−12 = 0
5y2−8y−12 = 0
Step 4: Solve the quadratic equation by factoring or using the quadratic
formula: The equation 5y2−8y−12 = 0 can be factored as (5y+ 6)(y−2) = 0.
This gives us two possible solutions: y=−6
5or y= 2.
Step 5: Substitute the values of yback into the equation x= 2yto find the
corresponding values of x. For y=−6
5:
x= 2 (−6
5)=−12
5
For y= 2:
x= 2(2) = 4
So, the solutions to the system of equations are: (x, y) = (−12
5,−6
5)and
(x, y) = (4,2).
Question 24
Question
Find the standard form of the equation of the parabola with focus F(2,3) and
directrix x= 1.
20
Solution
Step 1: Find the vertex of the parabola. The vertex of the parabola is the
midpoint between the focus and the directrix. We have the focus F(2,3) and
the directrix x= 1. Using the formula for the midpoint between two points, the
vertex Vis:
V=(1+2
2,3+3
2)=(3
2,3).
Step 2: Find the equation of the axis of symmetry. The axis of symmetry is
a vertical line passing through the vertex. Therefore, the equation of the axis
of symmetry is x=3
2.
Step 3: Determine whether the parabola opens to the left or right. Since
the directrix is to the left of the focus, the parabola opens to the right.
Step 4: Find the distance between the vertex and the focus. The distance
between the vertex and the focus is the same as the distance between the vertex
and the directrix. This distance is called the focal length, denoted as |p|. In
this case, |p|=|1−3
2|=1
2.
Step 5: Write the standard form of the equation. Since the parabola opens
to the right, the standard form of the equation is:
(x−h)2= 4p(y−k),
where (h, k)is the vertex. Plugging in the values of the vertex (3
2,3) and |p|=1
2,
we have:
(x−3
2)2= 4(1
2)(y−3).
Simplifying, we get:
(x−3
2)2= 2(y−3).
Therefore, the standard form of the equation of the parabola is (x−3
2)2=
2(y−3).
Question 25
Question
Find the standard form equation of the ellipse that satisfies the given conditions:
The major axis is along the line 3x−4y= 2 and passes through the points (−1,3)
and (5,1).
Solution
Step 1: Find the center of the ellipse using the midpoint formula. The midpoint
of the line segment connecting the points (−1,3) and (5,1) is the center of the
ellipse. Let the center be (h, k). The midpoint formula is given by:
(x1+x2
2,y1+y2
2)
21
Applying this formula to (−1,3) and (5,1):
(−1+5
2,3+1
2)= (2,2)
So, the center of the ellipse is at (h, k) = (2,2).
Step 2: Find the distance between the two given points which is the length
of the major axis. The distance formula is given by:
d=√(x2−x1)2+ (y2−y1)2
Calculating the distance between (−1,3) and (5,1):
d=√(5 −(−1))2+ (1 −3)2=√36 + 4 = √40 = 2√10
So, the length of the major axis is 2√10.
Step 3: Determine the major axis direction. We have the equation of the
major axis as 3x−4y= 2. Rewrite it in slope-intercept form as y=3
4x−1
2.
This means the slope of the major axis is 3
4. Therefore, the major axis of the
ellipse is in the direction of the line y=3
4x.
Step 4: Find the equation of the ellipse in standard form. The standard
form of the equation of an ellipse with major axis along the x-axis is:
(x−h)2
a2+(y−k)2
b2= 1
where (h, k)is the center of the ellipse, ais half the length of the major axis,
and bis half the length of the minor axis. In this case, since the major axis is
along the line y=3
4x, the equation becomes:
(x−2)2
10 +(y−2)2
2= 1
Therefore, the standard form equation of the ellipse is
(x−2)2
10 +(y−2)2
2= 1
.
22
