MATH 121 - COLLEGE ALGEBRA -
Equations and properties of conic
sections
Question Bank - Set 4
Liberty University
Question 1
Question
Determine the standard form of the equation of the hyperbola with vertices at
(-4,3) and (-4,-3) and a distance between the foci of 10 units.
Solution
Step 1: Determine the center of the hyperbola by finding the midpoint of the
vertices. The center Cis given by the midpoint formula: C(x1+x2
2,y1+y2
2).
Given vertices: V1(−4,3) and V2(−4,−3). Calculating the midpoint:
C(−4+(−4)
2,3+(−3)
2)=C(−4,0)
Step 2: Find the distance between the vertices which is 2a. The distance for-
mula between two points is √(x2−x1)2+ (y2−y1)2. Given vertices: V1(−4,3)
and V2(−4,−3). Calculating the distance:
2a=√(−4 + 4)2+ (3 −(−3))2=√0 + 36 = 6
So, a= 3.
Step 3: Determine the distance between the foci which is 2c. Given that the
distance between the foci is 10 units, we have 2c= 10. So, c= 5.
Step 4: Use the relationship between a,b, and cin a hyperbola: c2=a2+b2.
Substitute the known values: 52= 32+b2. Solving for b:25 = 9 + b2b2= 16
b= 4
Step 5: Write the equation of the hyperbola in standard form using the
center, values of aand b: The standard form of a hyperbola with center (h, k)
is (x−h)2
a2−(y−k)2
b2= 1. Substitute the known values:
(x+ 4)2
9−y2
16 = 1
Therefore, the standard form of the equation of the hyperbola is (x+4)2
9−y2
16 =
1.
Question 2
Question
Solve the following system of equations:
{y2−x2= 16
x2+y2
4= 5
Solution
Step 1: Rewrite the system of equations by rearranging the terms:
{y2−x2= 16
4x2+y2= 20
Step 2: Add the two equations together to eliminate x2:
5y2= 36
Step 3: Solve for y:
y2=36
5=⇒y=±√36
5=±6
√5=±6√5
5
Step 4: Substitute the value of yback into one of the original equations to
solve for x: From the first equation:
(±6√5
5)2−x2= 16
36 ×5
25 −x2= 16
180
25 −x2= 16
36
5−x2= 16
2
x2=36
5−16
x2=36
5−80
5
x2=36 −80
5
x2=−44
5=⇒x=±√−44
5=±2√55i
√5=±2√11i
√5
Therefore, the solution to the system of equations is:
(x, y) = (±2√11i
√5,±6√5
5)
Question 3
Question
Find the equation of the hyperbola with vertices at (−5,0) and (5,0) and foci
at (−7,0) and (7,0).
Solution
Step 1: Recall that the standard form of the equation of a hyperbola centered
at the origin with vertices at (±a, 0) and foci at (±c, 0) is
x2
a2−y2
b2= 1,
where c2=a2+b2.
Step 2: From the given information, we have a= 5,c= 7.
Step 3: Use the relationship c2=a2+b2to find b:
72= 52+b2
49 = 25 + b2
b2= 24
Step 4: Substitute the values of aand binto the standard form of the
equation of a hyperbola to find the equation:
x2
52−y2
√242= 1
x2
25 −y2
24 = 1
x2
25 −y2
24 = 1
Step 5: Therefore, the equation of the hyperbola with vertices at (−5,0) and
(5,0) and foci at (−7,0) and (7,0) is x2
25 −y2
24 = 1.
3
Question 4
Question
Solve the system of equations:
{x2+y2= 25
x−y= 3
Solution
Step 1: Rewrite the second equation in terms of one variable by solving for x.
x−y= 3
x=y+ 3
Step 2: Substitute xin the first equation with y+ 3.
(y+ 3)2+y2= 25
y2+ 6y+9+y2= 25
2y2+ 6y−16 = 0
y2+ 3y−8 = 0
Step 3: Solve the quadratic equation y2+ 3y−8 = 0 by factoring.
(y+ 4)(y−2) = 0
y=−4or y= 2
Step 4: Substitute yback into x=y+ 3 to find the corresponding values of
x.For y=−4 : x=−4 + 3 = −1
For y= 2 : x= 2 + 3 = 5
Therefore, the solution to the system of equations is (x, y) = (−1,−4),(5,2) .
Question 5
Question
Given the equation of a hyperbola in standard form: (x−1)2
9−(y+2)2
4= 1,
determine the center, vertices, foci, and asymptotes of the hyperbola.
4
Solution
Step 1: Identify the center of the hyperbola by matching the form of the given
equation with the standard form of a hyperbola:
(x−h)2
a2−(y−k)2
b2= 1.
Here, the center of the hyperbola is at (h, k) = (1,−2).
Step 2: Calculate the values of a2and b2:
a2= 9 =⇒a= 3,
b2= 4 =⇒b= 2.
Step 3: Find the vertices of the hyperbola. Since the vertices are located ±a
units from the center in the horizontal direction, the vertices are at (h±a, k).
Therefore, the vertices are at (1 + 3,−2) = (4,−2) and (1 −3,−2) = (−2,−2).
Step 4: Calculate the distance from the center to the foci using the relation-
ship c2=a2+b2:
c2= 9 + 4 =⇒c=√13 ≈3.61.
Step 5: Determine the foci of the hyperbola. The foci are located ±cunits
from the center in the horizontal direction, so the foci are at (h±c, k). Therefore,
the foci are at (1 + √13,−2) ≈(4.61,−2) and (1 −√13,−2) ≈(−1.61,−2).
Step 6: Write the equations of the asymptotes using the formula y=
±(b
a)(x−h) + k. The slope of the asymptotes is ±b
a=±2
3. Therefore, the
equations of the asymptotes are y=±2
3(x−1) −2.
Question 6
Question
Given the equation of a hyperbola in standard form: (x−3)2
16 −(y+1)2
9= 1. Find
the coordinates of the center, vertices, and foci.
Solution
Step 1: To find the coordinates of the center, we identify the values hand kin
the standard form of a hyperbola: (x−h)2
a2−(y−k)2
b2= 1. In this case, h= 3 and
k=−1, so the center of the hyperbola is at (3,−1).
Step 2: To find the vertices, we need to determine the distance from the
center to the vertices along the transverse axis. The distance from the center to
the vertices is a= 4. Therefore, the vertices are located at (3 ±4,−1), which
gives us vertices at (7,−1) and (−1,−1).
Step 3: The distance between the center and the foci is given by c, where
c2=a2+b2. In this case, a2= 16 and b2= 9, so c2= 25 and c= 5. Therefore,
the foci are located at (3 ±5,−1), which gives us foci at (8,−1) and (−2,−1).
5
Question 7
Question
Solve the following system of equations:
{x2+y2= 25
4x−3y= 7
Solution
Step 1: We can start by solving the second equation for y:
y=4x−7
3
Step 2: Substitute this expression for yinto the first equation:
x2+(4x−7
3)2
= 25
Step 3: Simplify the equation by expanding and collecting like terms:
x2+(16x2−56x+ 49
9)= 25
Step 4: Multiply through by 9 to get rid of the fraction:
9x2+ 16x2−56x+ 49 = 225
Step 5: Combine like terms:
25x2−56x+ 49 = 225
Step 6: Rearrange the equation into standard quadratic form:
25x2−56x−176 = 0
Step 7: Solve this quadratic equation using the quadratic formula:
x=56 ±√(−56)2−4(25)(−176)
2(25)
Step 8: Simplify the expression inside the square root:
x=56 ±√3136 + 17600
50
Step 9: Further simplify to find xvalues:
x=56 ±√20736
50
6
Step 10: Simplify the square root:
x=56 ±144
50
Step 11: Find the two possible values of x:
x=56 + 144
50 or x=56 −144
50
Step 12: Solve for x:
x=200
50 = 4 or x=−88
50 =−44
25
Step 13: Use these values to find the corresponding yvalues using the equa-
tion y=4x−7
3. For x= 4, we have y=4(4)−7
3=16−7
3=9
3= 3. For x=−44
25 ,
we have y=4(−44
25 )−7
3=−176
25 −7
3=−
176+175
25
3=−351
25
3=−351
25 ·1
3=−351
75 =−117
25
Step 14: Therefore, the solutions to the system of equations are (4,3) and
(−44
25 ,−117
25 ).
Question 8
Question
Solve the following equation for yin terms of x:2x2−y2+ 4y+ 9 = 0.
Solution
Step 1: First, rearrange the equation by grouping the y-terms together:
(2x2−y2)+4y+ 9 = 0
Step 2: Factor out −1from the y2term:
(2x2−y2−4y) + 9 = 0
Step 3: Complete the square for the y-terms by adding and subtracting
(4/2)2= 4 inside the parenthesis:
(2x2−(y2+ 4y+ 4 −4)) + 9 = 0
Step 4: Simplify the expression inside the parenthesis:
(2x2−(y+ 2)2+ 4) + 9 = 0
Step 5: Move the constant term to the right side:
(2x2−(y+ 2)2=−13
7
Step 6: Divide by 2 on both sides:
x2−(y+ 2)2
2=−13
2
Step 7: Isolate the squared term:
−(y+ 2)2
2=−x2−13
2
Step 8: Multiply by −1to simplify the equation:
(y+ 2)2
2=x2+13
2
Step 9: Take the square root of both sides to solve for y:
y+ 2 = ±√2√x2+13
2
Step 10: Subtract 2 from both sides to isolate y:
y=−2±√2√x2+13
2
Therefore, the solution to the equation 2x2−y2+ 4y+ 9 = 0 in terms of y
is y=−2±√2√x2+13
2.
Question 9
Question
Find the standard form equation of the ellipse with foci F1(−3,0) and F2(3,0),
and major axis length 10.
Solution
Step 1: Find the center of the ellipse. The center of the ellipse is the midpoint of
the segment connecting the foci. Let C(h, k)be the center of the ellipse. Then
the center is given by:
h=−3+3
2= 0
k=0+0
2= 0
Therefore, the center of the ellipse is C(0,0).
Step 2: Find the distance between the foci. The distance between the foci
is equal to 2c, where cis the distance from the center to a focus. Since the
distance between the foci is 6((−3,0) to (3,0)), we can calculate c:
2c= 6
8
c= 3
Step 3: Find the distance from the center to a vertex. The distance from
the center to a vertex is given by a, the semi-major axis. Since the major axis
length is given as 10, we have:
2a= 10
a= 5
Step 4: Determine the minor axis length. The minor axis length 2bcan be
found using the relation c2=a2−b2. Given that a= 5 and c= 3, we can solve
for b:
32= 52−b2
9 = 25 −b2
b2= 16
b= 4
Step 5: Write the standard form equation of the ellipse. The standard form
equation of an ellipse with center (h, k), major axis length 2a, and minor axis
length 2bis:
(x−h)2
a2+(y−k)2
b2= 1
Plugging in the values we found, the equation of the ellipse is:
x2
25 +y2
16 = 1
Question 10
Question
Solve the equation of the hyperbola (x−2)2
9−(y+1)2
4= 1 for y.
Solution
Step 1: Rearrange the given equation to isolate the fraction containing y. Step
2: Multiply both sides by 4 to eliminate the fraction in the equation. Step 3:
Expand the equation. Step 4: Rearrange the terms to gather all terms with y
on one side. Step 5: Divide by 4 and take the square root to solve for y. Step
6: Simplify the result to find the final solution.
Step 1: Rewrite the equation of the hyperbola as:
(x−2)2−9(y+ 1)2= 4
Step 2: Multiply both sides by 4:
4(x−2)2−36(y+ 1)2= 16
9
Step 3: Expand the equation:
4(x2−4x+ 4) −36(y2+ 2y+ 1) = 16
Step 4: Rearrange the terms:
4x2−16x+ 16 −36y2−72y−36 = 16
Step 5: Divide by 4 and take the square root to solve for y:
−9y2−18y= 4x2−16x−36
9y2+ 18y= 16x−4x2+ 36
y2+ 2y=16x
9−4x2
9+ 4
(y+ 1)2=16x
9−4x2
9+ 4
Step 6: Simplify the result to find the final solution:
y+ 1 = ±√16x
9−4x2
9+ 4
y=−1±√16x
9−4x2
9+ 4
Therefore, the solution to the equation of the hyperbola for yis y=−1±
√16x
9−4x2
9+ 4.
Question 11
Question
Find the standard form equation of a parabola with a focus at (−2,4) and a
directrix at y= 6.
Solution
Step 1: Recall that the standard form of the equation of a parabola with a focus
at (h, k +p)and a directrix at y=k−pis given by
(x−h)2= 4p(y−k)
Step 2: From the given information, we can identify h=−2,k= 4, and
k−p= 6. This allows us to solve for p.
4 = 6 −p
p= 2
Step 3: Now substituting h,k, and pinto the standard form equation of the
parabola gives us:
(x+ 2)2= 8(y−4)
Therefore, the standard form equation of the parabola is (x+ 2)2= 8(y−4).
10
Question 12
Question
Solve the following system of equations:
{x2−y2= 1
xy = 2
Solution
Step 1: We have the system of equations:
{x2−y2= 1
xy = 2
Step 2: From the second equation, we can express yin terms of x:
y=2
x
Step 3: Substitute y=2
xinto the first equation:
x2−(2
x)2
= 1
Step 4: Simplify the equation:
x2−4
x2= 1 ⇒x4−x2−4 = 0
Step 5: Let u=x2:
u2−u−4 = 0
Step 6: Solve for uusing the quadratic formula:
u=1±√12−4·1·(−4)
2=1±√17
2
Step 7: Substitute back x2:
x2=1 + √17
2or x2=1−√17
2
Step 8: Solve for x:
x=±√1 + √17
2or x=±√1−√17
2
Step 9: Substitute the values of xback into y=2
xto find the corresponding
yvalues.
Therefore, the solutions to the system of equations are x=±√1+√17
2,
y=±√1−√17
2and x=±√1−√17
2,y=±√1+√17
2
11
Question 13
Question
Solve the equation of the hyperbola: (x−3)2
16 −(y+1)2
9= 1.
Solution
Step 1: Identify the center and vertices of the hyperbola. The equation of a
hyperbola in standard form is (x−h)2
a2−(y−k)2
b2= 1, where the center is at (h, k).
Comparing with the given equation (x−3)2
16 −(y+1)2
9= 1, we have: Center:
(h, k) = (3,−1)
The distance from the center to the vertices along the x-axis is a, and along
the y-axis is b.
Vertices: Along x-axis: 3±a= 3 ±4=7and −1So, vertices are (7,−1)
and (−1,−1).
Step 2: Identify the foci of the hyperbola. The distance from the center to
the foci along the x-axis is c, where c2=a2+b2.
Foci: c2=a2+b2c2= 16 + 9 c2= 25 c= 5
Foci lie on the major axis from the center in the positive and negative direc-
tions: (3 + 5,−1) and (3 −5,−1). So, foci are (8,−1) and (−2,−1).
Step 3: Sketch the hyperbola. Now, we can sketch the hyperbola with
the center at (3,−1), vertices at (7,−1) and (−1,−1), and foci at (8,−1) and
(−2,−1).
Question 14
Question
Find the standard form of the equation of the hyperbola with foci at (−1,2)
and (5,2), and a horizontal axis.
Solution
Step 1: Identify the center of the hyperbola. The center of the hyperbola is the
midpoint of the segment connecting the foci. Let C(−1+5
2,2+2
2)be the center
of the hyperbola. Therefore, C(2,2) is the center of the hyperbola.
Step 2: Determine the equation of the transverse axis. Since the foci are
on the same horizontal line, the transverse axis is horizontal. Therefore, the
equation of the transverse axis is y= 2.
Step 3: Find a, the distance from the center to a vertex. The distance
between the center and one of the foci is the value of c.c= 5 −2=3The
distance between the center and a vertex is given by a=√c2+b2, where bis the
distance from the center to a co-vertex. Since this is a hyperbola, a2=b2+c2.
12
By substituting c= 3 into the equation above, we have: a2=b2+ 32Since
b= 2, we now have: a2= 22+ 32= 13
Step 4: Write the standard form of the equation. The standard form of the
equation of a hyperbola centered at (h, k)with vertices at (h±a, k)and foci at
(h±c, k)is given by
(x−h)2
a2−(y−k)2
b2= 1
Plugging in the values of a,b,h, and k, the equation becomes:
(x−2)2
13 −(y−2)2
4= 1
Thus, the standard form of the equation of the hyperbola with foci at (−1,2)
and (5,2) and a horizontal axis is (x−2)2
13 −(y−2)2
4= 1.
Question 15
Question
Find the standard form of the equation of the hyperbola with vertices at (−2,3)
and (4,3), and foci at (6,3) and (−8,3).
Solution
Step 1: Find the center of the hyperbola by finding the midpoint of the line
segment between the vertices.
The midpoint formula is given by
(x1+x2
2,y1+y2
2)
where (x1, y1)and (x2, y2)are the coordinates of the two points. Given the
vertices (−2,3) and (4,3), the center of the hyperbola is
(−2+4
2,3+3
2)= (1,3)
Step 2: Find the distance between the center and one of the vertices to
determine a.
The distance formula is given by
√(x2−x1)2+ (y2−y1)2
where (x1, y1)is the center of the hyperbola and (x2, y2)is one of the vertices.
Let’s use (−2,3) as the vertex. The distance ais
√(1 −(−2))2+ (3 −3)2=√32+ 0 = 3
13
Step 3: Find the distance between the center and one of the foci to determine
c.
From the information given, we can see that the distance between the center
and either of the foci is
c= 6 −1 = 5
Step 4: Use the formula c2=a2+b2to solve for b.
Substitute the values of aand cinto the formula to find b:
52= 32+b2
25 = 9 + b2
b2= 16
b= 4
Step 5: Write the equation of the hyperbola in standard form.
The equation of a hyperbola centered at (h, k)with vertices on the transverse
axis is given by
(x−h)2
a2−(y−k)2
b2= 1
Substitute h= 1,k= 3,a= 3, and b= 4 into the equation:
(x−1)2
32−(y−3)2
42= 1
(x−1)2
9−(y−3)2
16 = 1
Therefore, the standard form of the equation of the hyperbola is
(x−1)2
9−(y−3)2
16 = 1
Question 16
Question
Solve the following system of equations:
{x2+y2= 10
x−2y= 3
14
Solution
Step 1: Rearrange the second equation to solve for x:
x= 3 + 2y
Step 2: Substitute this expression for xinto the first equation:
(3 + 2y)2+y2= 10
Step 3: Expand and simplify the equation:
(3 + 2y)(3 + 2y) + y2= 10
9+6y+ 6y+ 4y2+y2= 10
9 + 12y+ 5y2= 10
Step 4: Rearrange the equation and set it equal to zero:
5y2+ 12y−1 = 0
Step 5: Solve for yusing the quadratic formula:
y=−b±√b2−4ac
2a
Substitute a= 5,b= 12, and c=−1:
y=−12 ±√122−4(5)(−1)
2(5)
y=−12 ±√144 + 20
10
y=−12 ±√164
10
y=−12 ±2√41
10
y=−6±√41
5
Step 6: Find the corresponding xvalues for each yvalue: For y=−6+√41
5:
x= 3 + 2 (−6 + √41
5)= 3 −12
5+2√41
5=−12 + 2√41
5
For y=−6−√41
5:
x= 3 + 2 (−6−√41
5)= 3 −12
5−2√41
5=−12 −2√41
5
15
Hence, the solutions to the system of equations are:
(x, y) = (−12 + 2√41
5,−6 + √41
5)
(x, y) = (−12 −2√41
5,−6−√41
5)
Question 17
Question
Find the standard form equation of the ellipse that satisfies the following con-
ditions: The major axis is along the y-axis, the center is at (-2,3), one vertex is
at (-2,6), and the co-vertex on the right side is at (-5,3).
Solution
Step 1: The standard form equation of an ellipse centered at (h,k) with major
axis along the y-axis is given by:
(y−k)2
b2+(x−h)2
a2= 1
where ais the length of the semi-major axis (half the length of the major axis),
and bis the length of the semi-minor axis (half the length of the minor axis).
Step 2: We are given that the center of the ellipse is at (-2,3). So, h=−2
and k= 3.
Step 3: The distance between the center (-2,3) and one of the vertices (-2,6)
gives us the length of the semi-major axis, which is 3: a= 3.
Step 4: The distance between the center (-2,3) and the co-vertex (-5,3) gives
us the length of the semi-minor axis, which is 3: b= 3.
Step 5: Plug in the values of h,k,a, and binto the standard form equation
of the ellipse:
(y−3)2
32+(x+ 2)2
32= 1
Step 6: Simplify the equation:
(y−3)2
9+(x+ 2)2
9= 1
Step 7: Multiply through by 9 to get the standard form equation of the
ellipse:
(y−3)2+ (x+ 2)2= 9
Step 8: The standard form equation of the ellipse is (y−3)2+ (x+ 2)2= 9.
16
Step 5: Write the equation of the hyperbola in standard form using the
center, values of aand b: The standard form of a hyperbola with center (h, k)
is (x−h)2
a2−(y−k)2
b2= 1. Substitute the known values:
(x+ 4)2
9−y2
16 = 1
Therefore, the standard form of the equation of the hyperbola is (x+4)2
9−y2
16 =
1.
Question 2
Question
Solve the following system of equations:
{y2−x2= 16
x2+y2
4= 5
Solution
Step 1: Rewrite the system of equations by rearranging the terms:
{y2−x2= 16
4x2+y2= 20
Step 2: Add the two equations together to eliminate x2:
5y2= 36
Step 3: Solve for y:
y2=36
5=⇒y=±√36
5=±6
√5=±6√5
5
Step 4: Substitute the value of yback into one of the original equations to
solve for x: From the first equation:
(±6√5
5)2−x2= 16
36 ×5
25 −x2= 16
180
25 −x2= 16
36
5−x2= 16
2
x2=36
5−16
x2=36
5−80
5
x2=36 −80
5
x2=−44
5=⇒x=±√−44
5=±2√55i
√5=±2√11i
√5
Therefore, the solution to the system of equations is:
(x, y) = (±2√11i
√5,±6√5
5)
Question 3
Question
Find the equation of the hyperbola with vertices at (−5,0) and (5,0) and foci
at (−7,0) and (7,0).
Solution
Step 1: Recall that the standard form of the equation of a hyperbola centered
at the origin with vertices at (±a, 0) and foci at (±c, 0) is
x2
a2−y2
b2= 1,
where c2=a2+b2.
Step 2: From the given information, we have a= 5,c= 7.
Step 3: Use the relationship c2=a2+b2to find b:
72= 52+b2
49 = 25 + b2
b2= 24
Step 4: Substitute the values of aand binto the standard form of the
equation of a hyperbola to find the equation:
x2
52−y2
√242= 1
x2
25 −y2
24 = 1
x2
25 −y2
24 = 1
Step 5: Therefore, the equation of the hyperbola with vertices at (−5,0) and
(5,0) and foci at (−7,0) and (7,0) is x2
25 −y2
24 = 1.
3
Question 4
Question
Solve the system of equations:
{x2+y2= 25
x−y= 3
Solution
Step 1: Rewrite the second equation in terms of one variable by solving for x.
x−y= 3
x=y+ 3
Step 2: Substitute xin the first equation with y+ 3.
(y+ 3)2+y2= 25
y2+ 6y+9+y2= 25
2y2+ 6y−16 = 0
y2+ 3y−8 = 0
Step 3: Solve the quadratic equation y2+ 3y−8 = 0 by factoring.
(y+ 4)(y−2) = 0
y=−4or y= 2
Step 4: Substitute yback into x=y+ 3 to find the corresponding values of
x.For y=−4 : x=−4 + 3 = −1
For y= 2 : x= 2 + 3 = 5
Therefore, the solution to the system of equations is (x, y) = (−1,−4),(5,2) .
Question 5
Question
Given the equation of a hyperbola in standard form: (x−1)2
9−(y+2)2
4= 1,
determine the center, vertices, foci, and asymptotes of the hyperbola.
4
Solution
Step 1: Identify the center of the hyperbola by matching the form of the given
equation with the standard form of a hyperbola:
(x−h)2
a2−(y−k)2
b2= 1.
Here, the center of the hyperbola is at (h, k) = (1,−2).
Step 2: Calculate the values of a2and b2:
a2= 9 =⇒a= 3,
b2= 4 =⇒b= 2.
Step 3: Find the vertices of the hyperbola. Since the vertices are located ±a
units from the center in the horizontal direction, the vertices are at (h±a, k).
Therefore, the vertices are at (1 + 3,−2) = (4,−2) and (1 −3,−2) = (−2,−2).
Step 4: Calculate the distance from the center to the foci using the relation-
ship c2=a2+b2:
c2= 9 + 4 =⇒c=√13 ≈3.61.
Step 5: Determine the foci of the hyperbola. The foci are located ±cunits
from the center in the horizontal direction, so the foci are at (h±c, k). Therefore,
the foci are at (1 + √13,−2) ≈(4.61,−2) and (1 −√13,−2) ≈(−1.61,−2).
Step 6: Write the equations of the asymptotes using the formula y=
±(b
a)(x−h) + k. The slope of the asymptotes is ±b
a=±2
3. Therefore, the
equations of the asymptotes are y=±2
3(x−1) −2.
Question 6
Question
Given the equation of a hyperbola in standard form: (x−3)2
16 −(y+1)2
9= 1. Find
the coordinates of the center, vertices, and foci.
Solution
Step 1: To find the coordinates of the center, we identify the values hand kin
the standard form of a hyperbola: (x−h)2
a2−(y−k)2
b2= 1. In this case, h= 3 and
k=−1, so the center of the hyperbola is at (3,−1).
Step 2: To find the vertices, we need to determine the distance from the
center to the vertices along the transverse axis. The distance from the center to
the vertices is a= 4. Therefore, the vertices are located at (3 ±4,−1), which
gives us vertices at (7,−1) and (−1,−1).
Step 3: The distance between the center and the foci is given by c, where
c2=a2+b2. In this case, a2= 16 and b2= 9, so c2= 25 and c= 5. Therefore,
the foci are located at (3 ±5,−1), which gives us foci at (8,−1) and (−2,−1).
5
Question 7
Question
Solve the following system of equations:
{x2+y2= 25
4x−3y= 7
Solution
Step 1: We can start by solving the second equation for y:
y=4x−7
3
Step 2: Substitute this expression for yinto the first equation:
x2+(4x−7
3)2
= 25
Step 3: Simplify the equation by expanding and collecting like terms:
x2+(16x2−56x+ 49
9)= 25
Step 4: Multiply through by 9 to get rid of the fraction:
9x2+ 16x2−56x+ 49 = 225
Step 5: Combine like terms:
25x2−56x+ 49 = 225
Step 6: Rearrange the equation into standard quadratic form:
25x2−56x−176 = 0
Step 7: Solve this quadratic equation using the quadratic formula:
x=56 ±√(−56)2−4(25)(−176)
2(25)
Step 8: Simplify the expression inside the square root:
x=56 ±√3136 + 17600
50
Step 9: Further simplify to find xvalues:
x=56 ±√20736
50
6
Step 10: Simplify the square root:
x=56 ±144
50
Step 11: Find the two possible values of x:
x=56 + 144
50 or x=56 −144
50
Step 12: Solve for x:
x=200
50 = 4 or x=−88
50 =−44
25
Step 13: Use these values to find the corresponding yvalues using the equa-
tion y=4x−7
3. For x= 4, we have y=4(4)−7
3=16−7
3=9
3= 3. For x=−44
25 ,
we have y=4(−44
25 )−7
3=−176
25 −7
3=−
176+175
25
3=−351
25
3=−351
25 ·1
3=−351
75 =−117
25
Step 14: Therefore, the solutions to the system of equations are (4,3) and
(−44
25 ,−117
25 ).
Question 8
Question
Solve the following equation for yin terms of x:2x2−y2+ 4y+ 9 = 0.
Solution
Step 1: First, rearrange the equation by grouping the y-terms together:
(2x2−y2)+4y+ 9 = 0
Step 2: Factor out −1from the y2term:
(2x2−y2−4y) + 9 = 0
Step 3: Complete the square for the y-terms by adding and subtracting
(4/2)2= 4 inside the parenthesis:
(2x2−(y2+ 4y+ 4 −4)) + 9 = 0
Step 4: Simplify the expression inside the parenthesis:
(2x2−(y+ 2)2+ 4) + 9 = 0
Step 5: Move the constant term to the right side:
(2x2−(y+ 2)2=−13
7
Step 6: Divide by 2 on both sides:
x2−(y+ 2)2
2=−13
2
Step 7: Isolate the squared term:
−(y+ 2)2
2=−x2−13
2
Step 8: Multiply by −1to simplify the equation:
(y+ 2)2
2=x2+13
2
Step 9: Take the square root of both sides to solve for y:
y+ 2 = ±√2√x2+13
2
Step 10: Subtract 2 from both sides to isolate y:
y=−2±√2√x2+13
2
Therefore, the solution to the equation 2x2−y2+ 4y+ 9 = 0 in terms of y
is y=−2±√2√x2+13
2.
Question 9
Question
Find the standard form equation of the ellipse with foci F1(−3,0) and F2(3,0),
and major axis length 10.
Solution
Step 1: Find the center of the ellipse. The center of the ellipse is the midpoint of
the segment connecting the foci. Let C(h, k)be the center of the ellipse. Then
the center is given by:
h=−3+3
2= 0
k=0+0
2= 0
Therefore, the center of the ellipse is C(0,0).
Step 2: Find the distance between the foci. The distance between the foci
is equal to 2c, where cis the distance from the center to a focus. Since the
distance between the foci is 6((−3,0) to (3,0)), we can calculate c:
2c= 6
8
c= 3
Step 3: Find the distance from the center to a vertex. The distance from
the center to a vertex is given by a, the semi-major axis. Since the major axis
length is given as 10, we have:
2a= 10
a= 5
Step 4: Determine the minor axis length. The minor axis length 2bcan be
found using the relation c2=a2−b2. Given that a= 5 and c= 3, we can solve
for b:
32= 52−b2
9 = 25 −b2
b2= 16
b= 4
Step 5: Write the standard form equation of the ellipse. The standard form
equation of an ellipse with center (h, k), major axis length 2a, and minor axis
length 2bis:
(x−h)2
a2+(y−k)2
b2= 1
Plugging in the values we found, the equation of the ellipse is:
x2
25 +y2
16 = 1
Question 10
Question
Solve the equation of the hyperbola (x−2)2
9−(y+1)2
4= 1 for y.
Solution
Step 1: Rearrange the given equation to isolate the fraction containing y. Step
2: Multiply both sides by 4 to eliminate the fraction in the equation. Step 3:
Expand the equation. Step 4: Rearrange the terms to gather all terms with y
on one side. Step 5: Divide by 4 and take the square root to solve for y. Step
6: Simplify the result to find the final solution.
Step 1: Rewrite the equation of the hyperbola as:
(x−2)2−9(y+ 1)2= 4
Step 2: Multiply both sides by 4:
4(x−2)2−36(y+ 1)2= 16
9
Step 3: Expand the equation:
4(x2−4x+ 4) −36(y2+ 2y+ 1) = 16
Step 4: Rearrange the terms:
4x2−16x+ 16 −36y2−72y−36 = 16
Step 5: Divide by 4 and take the square root to solve for y:
−9y2−18y= 4x2−16x−36
9y2+ 18y= 16x−4x2+ 36
y2+ 2y=16x
9−4x2
9+ 4
(y+ 1)2=16x
9−4x2
9+ 4
Step 6: Simplify the result to find the final solution:
y+ 1 = ±√16x
9−4x2
9+ 4
y=−1±√16x
9−4x2
9+ 4
Therefore, the solution to the equation of the hyperbola for yis y=−1±
√16x
9−4x2
9+ 4.
Question 11
Question
Find the standard form equation of a parabola with a focus at (−2,4) and a
directrix at y= 6.
Solution
Step 1: Recall that the standard form of the equation of a parabola with a focus
at (h, k +p)and a directrix at y=k−pis given by
(x−h)2= 4p(y−k)
Step 2: From the given information, we can identify h=−2,k= 4, and
k−p= 6. This allows us to solve for p.
4 = 6 −p
p= 2
Step 3: Now substituting h,k, and pinto the standard form equation of the
parabola gives us:
(x+ 2)2= 8(y−4)
Therefore, the standard form equation of the parabola is (x+ 2)2= 8(y−4).
10
Question 12
Question
Solve the following system of equations:
{x2−y2= 1
xy = 2
Solution
Step 1: We have the system of equations:
{x2−y2= 1
xy = 2
Step 2: From the second equation, we can express yin terms of x:
y=2
x
Step 3: Substitute y=2
xinto the first equation:
x2−(2
x)2
= 1
Step 4: Simplify the equation:
x2−4
x2= 1 ⇒x4−x2−4 = 0
Step 5: Let u=x2:
u2−u−4 = 0
Step 6: Solve for uusing the quadratic formula:
u=1±√12−4·1·(−4)
2=1±√17
2
Step 7: Substitute back x2:
x2=1 + √17
2or x2=1−√17
2
Step 8: Solve for x:
x=±√1 + √17
2or x=±√1−√17
2
Step 9: Substitute the values of xback into y=2
xto find the corresponding
yvalues.
Therefore, the solutions to the system of equations are x=±√1+√17
2,
y=±√1−√17
2and x=±√1−√17
2,y=±√1+√17
2
11
Question 13
Question
Solve the equation of the hyperbola: (x−3)2
16 −(y+1)2
9= 1.
Solution
Step 1: Identify the center and vertices of the hyperbola. The equation of a
hyperbola in standard form is (x−h)2
a2−(y−k)2
b2= 1, where the center is at (h, k).
Comparing with the given equation (x−3)2
16 −(y+1)2
9= 1, we have: Center:
(h, k) = (3,−1)
The distance from the center to the vertices along the x-axis is a, and along
the y-axis is b.
Vertices: Along x-axis: 3±a= 3 ±4=7and −1So, vertices are (7,−1)
and (−1,−1).
Step 2: Identify the foci of the hyperbola. The distance from the center to
the foci along the x-axis is c, where c2=a2+b2.
Foci: c2=a2+b2c2= 16 + 9 c2= 25 c= 5
Foci lie on the major axis from the center in the positive and negative direc-
tions: (3 + 5,−1) and (3 −5,−1). So, foci are (8,−1) and (−2,−1).
Step 3: Sketch the hyperbola. Now, we can sketch the hyperbola with
the center at (3,−1), vertices at (7,−1) and (−1,−1), and foci at (8,−1) and
(−2,−1).
Question 14
Question
Find the standard form of the equation of the hyperbola with foci at (−1,2)
and (5,2), and a horizontal axis.
Solution
Step 1: Identify the center of the hyperbola. The center of the hyperbola is the
midpoint of the segment connecting the foci. Let C(−1+5
2,2+2
2)be the center
of the hyperbola. Therefore, C(2,2) is the center of the hyperbola.
Step 2: Determine the equation of the transverse axis. Since the foci are
on the same horizontal line, the transverse axis is horizontal. Therefore, the
equation of the transverse axis is y= 2.
Step 3: Find a, the distance from the center to a vertex. The distance
between the center and one of the foci is the value of c.c= 5 −2=3The
distance between the center and a vertex is given by a=√c2+b2, where bis the
distance from the center to a co-vertex. Since this is a hyperbola, a2=b2+c2.
12
By substituting c= 3 into the equation above, we have: a2=b2+ 32Since
b= 2, we now have: a2= 22+ 32= 13
Step 4: Write the standard form of the equation. The standard form of the
equation of a hyperbola centered at (h, k)with vertices at (h±a, k)and foci at
(h±c, k)is given by
(x−h)2
a2−(y−k)2
b2= 1
Plugging in the values of a,b,h, and k, the equation becomes:
(x−2)2
13 −(y−2)2
4= 1
Thus, the standard form of the equation of the hyperbola with foci at (−1,2)
and (5,2) and a horizontal axis is (x−2)2
13 −(y−2)2
4= 1.
Question 15
Question
Find the standard form of the equation of the hyperbola with vertices at (−2,3)
and (4,3), and foci at (6,3) and (−8,3).
Solution
Step 1: Find the center of the hyperbola by finding the midpoint of the line
segment between the vertices.
The midpoint formula is given by
(x1+x2
2,y1+y2
2)
where (x1, y1)and (x2, y2)are the coordinates of the two points. Given the
vertices (−2,3) and (4,3), the center of the hyperbola is
(−2+4
2,3+3
2)= (1,3)
Step 2: Find the distance between the center and one of the vertices to
determine a.
The distance formula is given by
√(x2−x1)2+ (y2−y1)2
where (x1, y1)is the center of the hyperbola and (x2, y2)is one of the vertices.
Let’s use (−2,3) as the vertex. The distance ais
√(1 −(−2))2+ (3 −3)2=√32+ 0 = 3
13
Step 3: Find the distance between the center and one of the foci to determine
c.
From the information given, we can see that the distance between the center
and either of the foci is
c= 6 −1 = 5
Step 4: Use the formula c2=a2+b2to solve for b.
Substitute the values of aand cinto the formula to find b:
52= 32+b2
25 = 9 + b2
b2= 16
b= 4
Step 5: Write the equation of the hyperbola in standard form.
The equation of a hyperbola centered at (h, k)with vertices on the transverse
axis is given by
(x−h)2
a2−(y−k)2
b2= 1
Substitute h= 1,k= 3,a= 3, and b= 4 into the equation:
(x−1)2
32−(y−3)2
42= 1
(x−1)2
9−(y−3)2
16 = 1
Therefore, the standard form of the equation of the hyperbola is
(x−1)2
9−(y−3)2
16 = 1
Question 16
Question
Solve the following system of equations:
{x2+y2= 10
x−2y= 3
14
Solution
Step 1: Rearrange the second equation to solve for x:
x= 3 + 2y
Step 2: Substitute this expression for xinto the first equation:
(3 + 2y)2+y2= 10
Step 3: Expand and simplify the equation:
(3 + 2y)(3 + 2y) + y2= 10
9+6y+ 6y+ 4y2+y2= 10
9 + 12y+ 5y2= 10
Step 4: Rearrange the equation and set it equal to zero:
5y2+ 12y−1 = 0
Step 5: Solve for yusing the quadratic formula:
y=−b±√b2−4ac
2a
Substitute a= 5,b= 12, and c=−1:
y=−12 ±√122−4(5)(−1)
2(5)
y=−12 ±√144 + 20
10
y=−12 ±√164
10
y=−12 ±2√41
10
y=−6±√41
5
Step 6: Find the corresponding xvalues for each yvalue: For y=−6+√41
5:
x= 3 + 2 (−6 + √41
5)= 3 −12
5+2√41
5=−12 + 2√41
5
For y=−6−√41
5:
x= 3 + 2 (−6−√41
5)= 3 −12
5−2√41
5=−12 −2√41
5
15
Hence, the solutions to the system of equations are:
(x, y) = (−12 + 2√41
5,−6 + √41
5)
(x, y) = (−12 −2√41
5,−6−√41
5)
Question 17
Question
Find the standard form equation of the ellipse that satisfies the following con-
ditions: The major axis is along the y-axis, the center is at (-2,3), one vertex is
at (-2,6), and the co-vertex on the right side is at (-5,3).
Solution
Step 1: The standard form equation of an ellipse centered at (h,k) with major
axis along the y-axis is given by:
(y−k)2
b2+(x−h)2
a2= 1
where ais the length of the semi-major axis (half the length of the major axis),
and bis the length of the semi-minor axis (half the length of the minor axis).
Step 2: We are given that the center of the ellipse is at (-2,3). So, h=−2
and k= 3.
Step 3: The distance between the center (-2,3) and one of the vertices (-2,6)
gives us the length of the semi-major axis, which is 3: a= 3.
Step 4: The distance between the center (-2,3) and the co-vertex (-5,3) gives
us the length of the semi-minor axis, which is 3: b= 3.
Step 5: Plug in the values of h,k,a, and binto the standard form equation
of the ellipse:
(y−3)2
32+(x+ 2)2
32= 1
Step 6: Simplify the equation:
(y−3)2
9+(x+ 2)2
9= 1
Step 7: Multiply through by 9 to get the standard form equation of the
ellipse:
(y−3)2+ (x+ 2)2= 9
Step 8: The standard form equation of the ellipse is (y−3)2+ (x+ 2)2= 9.
16
Step 5: Write the equation of the hyperbola in standard form using the
center, values of aand b: The standard form of a hyperbola with center (h, k)
is (x−h)2
a2−(y−k)2
b2= 1. Substitute the known values:
(x+ 4)2
9−y2
16 = 1
Therefore, the standard form of the equation of the hyperbola is (x+4)2
9−y2
16 =
1.
Question 2
Question
Solve the following system of equations:
{y2−x2= 16
x2+y2
4= 5
Solution
Step 1: Rewrite the system of equations by rearranging the terms:
{y2−x2= 16
4x2+y2= 20
Step 2: Add the two equations together to eliminate x2:
5y2= 36
Step 3: Solve for y:
y2=36
5=⇒y=±√36
5=±6
√5=±6√5
5
Step 4: Substitute the value of yback into one of the original equations to
solve for x: From the first equation:
(±6√5
5)2−x2= 16
36 ×5
25 −x2= 16
180
25 −x2= 16
36
5−x2= 16
2
x2=36
5−16
x2=36
5−80
5
x2=36 −80
5
x2=−44
5=⇒x=±√−44
5=±2√55i
√5=±2√11i
√5
Therefore, the solution to the system of equations is:
(x, y) = (±2√11i
√5,±6√5
5)
Question 3
Question
Find the equation of the hyperbola with vertices at (−5,0) and (5,0) and foci
at (−7,0) and (7,0).
Solution
Step 1: Recall that the standard form of the equation of a hyperbola centered
at the origin with vertices at (±a, 0) and foci at (±c, 0) is
x2
a2−y2
b2= 1,
where c2=a2+b2.
Step 2: From the given information, we have a= 5,c= 7.
Step 3: Use the relationship c2=a2+b2to find b:
72= 52+b2
49 = 25 + b2
b2= 24
Step 4: Substitute the values of aand binto the standard form of the
equation of a hyperbola to find the equation:
x2
52−y2
√242= 1
x2
25 −y2
24 = 1
x2
25 −y2
24 = 1
Step 5: Therefore, the equation of the hyperbola with vertices at (−5,0) and
(5,0) and foci at (−7,0) and (7,0) is x2
25 −y2
24 = 1.
3
Question 4
Question
Solve the system of equations:
{x2+y2= 25
x−y= 3
Solution
Step 1: Rewrite the second equation in terms of one variable by solving for x.
x−y= 3
x=y+ 3
Step 2: Substitute xin the first equation with y+ 3.
(y+ 3)2+y2= 25
y2+ 6y+9+y2= 25
2y2+ 6y−16 = 0
y2+ 3y−8 = 0
Step 3: Solve the quadratic equation y2+ 3y−8 = 0 by factoring.
(y+ 4)(y−2) = 0
y=−4or y= 2
Step 4: Substitute yback into x=y+ 3 to find the corresponding values of
x.For y=−4 : x=−4 + 3 = −1
For y= 2 : x= 2 + 3 = 5
Therefore, the solution to the system of equations is (x, y) = (−1,−4),(5,2) .
Question 5
Question
Given the equation of a hyperbola in standard form: (x−1)2
9−(y+2)2
4= 1,
determine the center, vertices, foci, and asymptotes of the hyperbola.
4
Solution
Step 1: Identify the center of the hyperbola by matching the form of the given
equation with the standard form of a hyperbola:
(x−h)2
a2−(y−k)2
b2= 1.
Here, the center of the hyperbola is at (h, k) = (1,−2).
Step 2: Calculate the values of a2and b2:
a2= 9 =⇒a= 3,
b2= 4 =⇒b= 2.
Step 3: Find the vertices of the hyperbola. Since the vertices are located ±a
units from the center in the horizontal direction, the vertices are at (h±a, k).
Therefore, the vertices are at (1 + 3,−2) = (4,−2) and (1 −3,−2) = (−2,−2).
Step 4: Calculate the distance from the center to the foci using the relation-
ship c2=a2+b2:
c2= 9 + 4 =⇒c=√13 ≈3.61.
Step 5: Determine the foci of the hyperbola. The foci are located ±cunits
from the center in the horizontal direction, so the foci are at (h±c, k). Therefore,
the foci are at (1 + √13,−2) ≈(4.61,−2) and (1 −√13,−2) ≈(−1.61,−2).
Step 6: Write the equations of the asymptotes using the formula y=
±(b
a)(x−h) + k. The slope of the asymptotes is ±b
a=±2
3. Therefore, the
equations of the asymptotes are y=±2
3(x−1) −2.
Question 6
Question
Given the equation of a hyperbola in standard form: (x−3)2
16 −(y+1)2
9= 1. Find
the coordinates of the center, vertices, and foci.
Solution
Step 1: To find the coordinates of the center, we identify the values hand kin
the standard form of a hyperbola: (x−h)2
a2−(y−k)2
b2= 1. In this case, h= 3 and
k=−1, so the center of the hyperbola is at (3,−1).
Step 2: To find the vertices, we need to determine the distance from the
center to the vertices along the transverse axis. The distance from the center to
the vertices is a= 4. Therefore, the vertices are located at (3 ±4,−1), which
gives us vertices at (7,−1) and (−1,−1).
Step 3: The distance between the center and the foci is given by c, where
c2=a2+b2. In this case, a2= 16 and b2= 9, so c2= 25 and c= 5. Therefore,
the foci are located at (3 ±5,−1), which gives us foci at (8,−1) and (−2,−1).
5
Question 7
Question
Solve the following system of equations:
{x2+y2= 25
4x−3y= 7
Solution
Step 1: We can start by solving the second equation for y:
y=4x−7
3
Step 2: Substitute this expression for yinto the first equation:
x2+(4x−7
3)2
= 25
Step 3: Simplify the equation by expanding and collecting like terms:
x2+(16x2−56x+ 49
9)= 25
Step 4: Multiply through by 9 to get rid of the fraction:
9x2+ 16x2−56x+ 49 = 225
Step 5: Combine like terms:
25x2−56x+ 49 = 225
Step 6: Rearrange the equation into standard quadratic form:
25x2−56x−176 = 0
Step 7: Solve this quadratic equation using the quadratic formula:
x=56 ±√(−56)2−4(25)(−176)
2(25)
Step 8: Simplify the expression inside the square root:
x=56 ±√3136 + 17600
50
Step 9: Further simplify to find xvalues:
x=56 ±√20736
50
6
Step 10: Simplify the square root:
x=56 ±144
50
Step 11: Find the two possible values of x:
x=56 + 144
50 or x=56 −144
50
Step 12: Solve for x:
x=200
50 = 4 or x=−88
50 =−44
25
Step 13: Use these values to find the corresponding yvalues using the equa-
tion y=4x−7
3. For x= 4, we have y=4(4)−7
3=16−7
3=9
3= 3. For x=−44
25 ,
we have y=4(−44
25 )−7
3=−176
25 −7
3=−
176+175
25
3=−351
25
3=−351
25 ·1
3=−351
75 =−117
25
Step 14: Therefore, the solutions to the system of equations are (4,3) and
(−44
25 ,−117
25 ).
Question 8
Question
Solve the following equation for yin terms of x:2x2−y2+ 4y+ 9 = 0.
Solution
Step 1: First, rearrange the equation by grouping the y-terms together:
(2x2−y2)+4y+ 9 = 0
Step 2: Factor out −1from the y2term:
(2x2−y2−4y) + 9 = 0
Step 3: Complete the square for the y-terms by adding and subtracting
(4/2)2= 4 inside the parenthesis:
(2x2−(y2+ 4y+ 4 −4)) + 9 = 0
Step 4: Simplify the expression inside the parenthesis:
(2x2−(y+ 2)2+ 4) + 9 = 0
Step 5: Move the constant term to the right side:
(2x2−(y+ 2)2=−13
7
Step 6: Divide by 2 on both sides:
x2−(y+ 2)2
2=−13
2
Step 7: Isolate the squared term:
−(y+ 2)2
2=−x2−13
2
Step 8: Multiply by −1to simplify the equation:
(y+ 2)2
2=x2+13
2
Step 9: Take the square root of both sides to solve for y:
y+ 2 = ±√2√x2+13
2
Step 10: Subtract 2 from both sides to isolate y:
y=−2±√2√x2+13
2
Therefore, the solution to the equation 2x2−y2+ 4y+ 9 = 0 in terms of y
is y=−2±√2√x2+13
2.
Question 9
Question
Find the standard form equation of the ellipse with foci F1(−3,0) and F2(3,0),
and major axis length 10.
Solution
Step 1: Find the center of the ellipse. The center of the ellipse is the midpoint of
the segment connecting the foci. Let C(h, k)be the center of the ellipse. Then
the center is given by:
h=−3+3
2= 0
k=0+0
2= 0
Therefore, the center of the ellipse is C(0,0).
Step 2: Find the distance between the foci. The distance between the foci
is equal to 2c, where cis the distance from the center to a focus. Since the
distance between the foci is 6((−3,0) to (3,0)), we can calculate c:
2c= 6
8
c= 3
Step 3: Find the distance from the center to a vertex. The distance from
the center to a vertex is given by a, the semi-major axis. Since the major axis
length is given as 10, we have:
2a= 10
a= 5
Step 4: Determine the minor axis length. The minor axis length 2bcan be
found using the relation c2=a2−b2. Given that a= 5 and c= 3, we can solve
for b:
32= 52−b2
9 = 25 −b2
b2= 16
b= 4
Step 5: Write the standard form equation of the ellipse. The standard form
equation of an ellipse with center (h, k), major axis length 2a, and minor axis
length 2bis:
(x−h)2
a2+(y−k)2
b2= 1
Plugging in the values we found, the equation of the ellipse is:
x2
25 +y2
16 = 1
Question 10
Question
Solve the equation of the hyperbola (x−2)2
9−(y+1)2
4= 1 for y.
Solution
Step 1: Rearrange the given equation to isolate the fraction containing y. Step
2: Multiply both sides by 4 to eliminate the fraction in the equation. Step 3:
Expand the equation. Step 4: Rearrange the terms to gather all terms with y
on one side. Step 5: Divide by 4 and take the square root to solve for y. Step
6: Simplify the result to find the final solution.
Step 1: Rewrite the equation of the hyperbola as:
(x−2)2−9(y+ 1)2= 4
Step 2: Multiply both sides by 4:
4(x−2)2−36(y+ 1)2= 16
9
Step 3: Expand the equation:
4(x2−4x+ 4) −36(y2+ 2y+ 1) = 16
Step 4: Rearrange the terms:
4x2−16x+ 16 −36y2−72y−36 = 16
Step 5: Divide by 4 and take the square root to solve for y:
−9y2−18y= 4x2−16x−36
9y2+ 18y= 16x−4x2+ 36
y2+ 2y=16x
9−4x2
9+ 4
(y+ 1)2=16x
9−4x2
9+ 4
Step 6: Simplify the result to find the final solution:
y+ 1 = ±√16x
9−4x2
9+ 4
y=−1±√16x
9−4x2
9+ 4
Therefore, the solution to the equation of the hyperbola for yis y=−1±
√16x
9−4x2
9+ 4.
Question 11
Question
Find the standard form equation of a parabola with a focus at (−2,4) and a
directrix at y= 6.
Solution
Step 1: Recall that the standard form of the equation of a parabola with a focus
at (h, k +p)and a directrix at y=k−pis given by
(x−h)2= 4p(y−k)
Step 2: From the given information, we can identify h=−2,k= 4, and
k−p= 6. This allows us to solve for p.
4 = 6 −p
p= 2
Step 3: Now substituting h,k, and pinto the standard form equation of the
parabola gives us:
(x+ 2)2= 8(y−4)
Therefore, the standard form equation of the parabola is (x+ 2)2= 8(y−4).
10
Question 12
Question
Solve the following system of equations:
{x2−y2= 1
xy = 2
Solution
Step 1: We have the system of equations:
{x2−y2= 1
xy = 2
Step 2: From the second equation, we can express yin terms of x:
y=2
x
Step 3: Substitute y=2
xinto the first equation:
x2−(2
x)2
= 1
Step 4: Simplify the equation:
x2−4
x2= 1 ⇒x4−x2−4 = 0
Step 5: Let u=x2:
u2−u−4 = 0
Step 6: Solve for uusing the quadratic formula:
u=1±√12−4·1·(−4)
2=1±√17
2
Step 7: Substitute back x2:
x2=1 + √17
2or x2=1−√17
2
Step 8: Solve for x:
x=±√1 + √17
2or x=±√1−√17
2
Step 9: Substitute the values of xback into y=2
xto find the corresponding
yvalues.
Therefore, the solutions to the system of equations are x=±√1+√17
2,
y=±√1−√17
2and x=±√1−√17
2,y=±√1+√17
2
11
Question 13
Question
Solve the equation of the hyperbola: (x−3)2
16 −(y+1)2
9= 1.
Solution
Step 1: Identify the center and vertices of the hyperbola. The equation of a
hyperbola in standard form is (x−h)2
a2−(y−k)2
b2= 1, where the center is at (h, k).
Comparing with the given equation (x−3)2
16 −(y+1)2
9= 1, we have: Center:
(h, k) = (3,−1)
The distance from the center to the vertices along the x-axis is a, and along
the y-axis is b.
Vertices: Along x-axis: 3±a= 3 ±4=7and −1So, vertices are (7,−1)
and (−1,−1).
Step 2: Identify the foci of the hyperbola. The distance from the center to
the foci along the x-axis is c, where c2=a2+b2.
Foci: c2=a2+b2c2= 16 + 9 c2= 25 c= 5
Foci lie on the major axis from the center in the positive and negative direc-
tions: (3 + 5,−1) and (3 −5,−1). So, foci are (8,−1) and (−2,−1).
Step 3: Sketch the hyperbola. Now, we can sketch the hyperbola with
the center at (3,−1), vertices at (7,−1) and (−1,−1), and foci at (8,−1) and
(−2,−1).
Question 14
Question
Find the standard form of the equation of the hyperbola with foci at (−1,2)
and (5,2), and a horizontal axis.
Solution
Step 1: Identify the center of the hyperbola. The center of the hyperbola is the
midpoint of the segment connecting the foci. Let C(−1+5
2,2+2
2)be the center
of the hyperbola. Therefore, C(2,2) is the center of the hyperbola.
Step 2: Determine the equation of the transverse axis. Since the foci are
on the same horizontal line, the transverse axis is horizontal. Therefore, the
equation of the transverse axis is y= 2.
Step 3: Find a, the distance from the center to a vertex. The distance
between the center and one of the foci is the value of c.c= 5 −2=3The
distance between the center and a vertex is given by a=√c2+b2, where bis the
distance from the center to a co-vertex. Since this is a hyperbola, a2=b2+c2.
12
By substituting c= 3 into the equation above, we have: a2=b2+ 32Since
b= 2, we now have: a2= 22+ 32= 13
Step 4: Write the standard form of the equation. The standard form of the
equation of a hyperbola centered at (h, k)with vertices at (h±a, k)and foci at
(h±c, k)is given by
(x−h)2
a2−(y−k)2
b2= 1
Plugging in the values of a,b,h, and k, the equation becomes:
(x−2)2
13 −(y−2)2
4= 1
Thus, the standard form of the equation of the hyperbola with foci at (−1,2)
and (5,2) and a horizontal axis is (x−2)2
13 −(y−2)2
4= 1.
Question 15
Question
Find the standard form of the equation of the hyperbola with vertices at (−2,3)
and (4,3), and foci at (6,3) and (−8,3).
Solution
Step 1: Find the center of the hyperbola by finding the midpoint of the line
segment between the vertices.
The midpoint formula is given by
(x1+x2
2,y1+y2
2)
where (x1, y1)and (x2, y2)are the coordinates of the two points. Given the
vertices (−2,3) and (4,3), the center of the hyperbola is
(−2+4
2,3+3
2)= (1,3)
Step 2: Find the distance between the center and one of the vertices to
determine a.
The distance formula is given by
√(x2−x1)2+ (y2−y1)2
where (x1, y1)is the center of the hyperbola and (x2, y2)is one of the vertices.
Let’s use (−2,3) as the vertex. The distance ais
√(1 −(−2))2+ (3 −3)2=√32+ 0 = 3
13
Step 3: Find the distance between the center and one of the foci to determine
c.
From the information given, we can see that the distance between the center
and either of the foci is
c= 6 −1 = 5
Step 4: Use the formula c2=a2+b2to solve for b.
Substitute the values of aand cinto the formula to find b:
52= 32+b2
25 = 9 + b2
b2= 16
b= 4
Step 5: Write the equation of the hyperbola in standard form.
The equation of a hyperbola centered at (h, k)with vertices on the transverse
axis is given by
(x−h)2
a2−(y−k)2
b2= 1
Substitute h= 1,k= 3,a= 3, and b= 4 into the equation:
(x−1)2
32−(y−3)2
42= 1
(x−1)2
9−(y−3)2
16 = 1
Therefore, the standard form of the equation of the hyperbola is
(x−1)2
9−(y−3)2
16 = 1
Question 16
Question
Solve the following system of equations:
{x2+y2= 10
x−2y= 3
14
Solution
Step 1: Rearrange the second equation to solve for x:
x= 3 + 2y
Step 2: Substitute this expression for xinto the first equation:
(3 + 2y)2+y2= 10
Step 3: Expand and simplify the equation:
(3 + 2y)(3 + 2y) + y2= 10
9+6y+ 6y+ 4y2+y2= 10
9 + 12y+ 5y2= 10
Step 4: Rearrange the equation and set it equal to zero:
5y2+ 12y−1 = 0
Step 5: Solve for yusing the quadratic formula:
y=−b±√b2−4ac
2a
Substitute a= 5,b= 12, and c=−1:
y=−12 ±√122−4(5)(−1)
2(5)
y=−12 ±√144 + 20
10
y=−12 ±√164
10
y=−12 ±2√41
10
y=−6±√41
5
Step 6: Find the corresponding xvalues for each yvalue: For y=−6+√41
5:
x= 3 + 2 (−6 + √41
5)= 3 −12
5+2√41
5=−12 + 2√41
5
For y=−6−√41
5:
x= 3 + 2 (−6−√41
5)= 3 −12
5−2√41
5=−12 −2√41
5
15
Hence, the solutions to the system of equations are:
(x, y) = (−12 + 2√41
5,−6 + √41
5)
(x, y) = (−12 −2√41
5,−6−√41
5)
Question 17
Question
Find the standard form equation of the ellipse that satisfies the following con-
ditions: The major axis is along the y-axis, the center is at (-2,3), one vertex is
at (-2,6), and the co-vertex on the right side is at (-5,3).
Solution
Step 1: The standard form equation of an ellipse centered at (h,k) with major
axis along the y-axis is given by:
(y−k)2
b2+(x−h)2
a2= 1
where ais the length of the semi-major axis (half the length of the major axis),
and bis the length of the semi-minor axis (half the length of the minor axis).
Step 2: We are given that the center of the ellipse is at (-2,3). So, h=−2
and k= 3.
Step 3: The distance between the center (-2,3) and one of the vertices (-2,6)
gives us the length of the semi-major axis, which is 3: a= 3.
Step 4: The distance between the center (-2,3) and the co-vertex (-5,3) gives
us the length of the semi-minor axis, which is 3: b= 3.
Step 5: Plug in the values of h,k,a, and binto the standard form equation
of the ellipse:
(y−3)2
32+(x+ 2)2
32= 1
Step 6: Simplify the equation:
(y−3)2
9+(x+ 2)2
9= 1
Step 7: Multiply through by 9 to get the standard form equation of the
ellipse:
(y−3)2+ (x+ 2)2= 9
Step 8: The standard form equation of the ellipse is (y−3)2+ (x+ 2)2= 9.
16
Step 5: Write the equation of the hyperbola in standard form using the
center, values of aand b: The standard form of a hyperbola with center (h, k)
is (x−h)2
a2−(y−k)2
b2= 1. Substitute the known values:
(x+ 4)2
9−y2
16 = 1
Therefore, the standard form of the equation of the hyperbola is (x+4)2
9−y2
16 =
1.
Question 2
Question
Solve the following system of equations:
{y2−x2= 16
x2+y2
4= 5
Solution
Step 1: Rewrite the system of equations by rearranging the terms:
{y2−x2= 16
4x2+y2= 20
Step 2: Add the two equations together to eliminate x2:
5y2= 36
Step 3: Solve for y:
y2=36
5=⇒y=±√36
5=±6
√5=±6√5
5
Step 4: Substitute the value of yback into one of the original equations to
solve for x: From the first equation:
(±6√5
5)2−x2= 16
36 ×5
25 −x2= 16
180
25 −x2= 16
36
5−x2= 16
2
x2=36
5−16
x2=36
5−80
5
x2=36 −80
5
x2=−44
5=⇒x=±√−44
5=±2√55i
√5=±2√11i
√5
Therefore, the solution to the system of equations is:
(x, y) = (±2√11i
√5,±6√5
5)
Question 3
Question
Find the equation of the hyperbola with vertices at (−5,0) and (5,0) and foci
at (−7,0) and (7,0).
Solution
Step 1: Recall that the standard form of the equation of a hyperbola centered
at the origin with vertices at (±a, 0) and foci at (±c, 0) is
x2
a2−y2
b2= 1,
where c2=a2+b2.
Step 2: From the given information, we have a= 5,c= 7.
Step 3: Use the relationship c2=a2+b2to find b:
72= 52+b2
49 = 25 + b2
b2= 24
Step 4: Substitute the values of aand binto the standard form of the
equation of a hyperbola to find the equation:
x2
52−y2
√242= 1
x2
25 −y2
24 = 1
x2
25 −y2
24 = 1
Step 5: Therefore, the equation of the hyperbola with vertices at (−5,0) and
(5,0) and foci at (−7,0) and (7,0) is x2
25 −y2
24 = 1.
3
Question 4
Question
Solve the system of equations:
{x2+y2= 25
x−y= 3
Solution
Step 1: Rewrite the second equation in terms of one variable by solving for x.
x−y= 3
x=y+ 3
Step 2: Substitute xin the first equation with y+ 3.
(y+ 3)2+y2= 25
y2+ 6y+9+y2= 25
2y2+ 6y−16 = 0
y2+ 3y−8 = 0
Step 3: Solve the quadratic equation y2+ 3y−8 = 0 by factoring.
(y+ 4)(y−2) = 0
y=−4or y= 2
Step 4: Substitute yback into x=y+ 3 to find the corresponding values of
x.For y=−4 : x=−4 + 3 = −1
For y= 2 : x= 2 + 3 = 5
Therefore, the solution to the system of equations is (x, y) = (−1,−4),(5,2) .
Question 5
Question
Given the equation of a hyperbola in standard form: (x−1)2
9−(y+2)2
4= 1,
determine the center, vertices, foci, and asymptotes of the hyperbola.
4
Solution
Step 1: Identify the center of the hyperbola by matching the form of the given
equation with the standard form of a hyperbola:
(x−h)2
a2−(y−k)2
b2= 1.
Here, the center of the hyperbola is at (h, k) = (1,−2).
Step 2: Calculate the values of a2and b2:
a2= 9 =⇒a= 3,
b2= 4 =⇒b= 2.
Step 3: Find the vertices of the hyperbola. Since the vertices are located ±a
units from the center in the horizontal direction, the vertices are at (h±a, k).
Therefore, the vertices are at (1 + 3,−2) = (4,−2) and (1 −3,−2) = (−2,−2).
Step 4: Calculate the distance from the center to the foci using the relation-
ship c2=a2+b2:
c2= 9 + 4 =⇒c=√13 ≈3.61.
Step 5: Determine the foci of the hyperbola. The foci are located ±cunits
from the center in the horizontal direction, so the foci are at (h±c, k). Therefore,
the foci are at (1 + √13,−2) ≈(4.61,−2) and (1 −√13,−2) ≈(−1.61,−2).
Step 6: Write the equations of the asymptotes using the formula y=
±(b
a)(x−h) + k. The slope of the asymptotes is ±b
a=±2
3. Therefore, the
equations of the asymptotes are y=±2
3(x−1) −2.
Question 6
Question
Given the equation of a hyperbola in standard form: (x−3)2
16 −(y+1)2
9= 1. Find
the coordinates of the center, vertices, and foci.
Solution
Step 1: To find the coordinates of the center, we identify the values hand kin
the standard form of a hyperbola: (x−h)2
a2−(y−k)2
b2= 1. In this case, h= 3 and
k=−1, so the center of the hyperbola is at (3,−1).
Step 2: To find the vertices, we need to determine the distance from the
center to the vertices along the transverse axis. The distance from the center to
the vertices is a= 4. Therefore, the vertices are located at (3 ±4,−1), which
gives us vertices at (7,−1) and (−1,−1).
Step 3: The distance between the center and the foci is given by c, where
c2=a2+b2. In this case, a2= 16 and b2= 9, so c2= 25 and c= 5. Therefore,
the foci are located at (3 ±5,−1), which gives us foci at (8,−1) and (−2,−1).
5
Question 7
Question
Solve the following system of equations:
{x2+y2= 25
4x−3y= 7
Solution
Step 1: We can start by solving the second equation for y:
y=4x−7
3
Step 2: Substitute this expression for yinto the first equation:
x2+(4x−7
3)2
= 25
Step 3: Simplify the equation by expanding and collecting like terms:
x2+(16x2−56x+ 49
9)= 25
Step 4: Multiply through by 9 to get rid of the fraction:
9x2+ 16x2−56x+ 49 = 225
Step 5: Combine like terms:
25x2−56x+ 49 = 225
Step 6: Rearrange the equation into standard quadratic form:
25x2−56x−176 = 0
Step 7: Solve this quadratic equation using the quadratic formula:
x=56 ±√(−56)2−4(25)(−176)
2(25)
Step 8: Simplify the expression inside the square root:
x=56 ±√3136 + 17600
50
Step 9: Further simplify to find xvalues:
x=56 ±√20736
50
6
Step 10: Simplify the square root:
x=56 ±144
50
Step 11: Find the two possible values of x:
x=56 + 144
50 or x=56 −144
50
Step 12: Solve for x:
x=200
50 = 4 or x=−88
50 =−44
25
Step 13: Use these values to find the corresponding yvalues using the equa-
tion y=4x−7
3. For x= 4, we have y=4(4)−7
3=16−7
3=9
3= 3. For x=−44
25 ,
we have y=4(−44
25 )−7
3=−176
25 −7
3=−
176+175
25
3=−351
25
3=−351
25 ·1
3=−351
75 =−117
25
Step 14: Therefore, the solutions to the system of equations are (4,3) and
(−44
25 ,−117
25 ).
Question 8
Question
Solve the following equation for yin terms of x:2x2−y2+ 4y+ 9 = 0.
Solution
Step 1: First, rearrange the equation by grouping the y-terms together:
(2x2−y2)+4y+ 9 = 0
Step 2: Factor out −1from the y2term:
(2x2−y2−4y) + 9 = 0
Step 3: Complete the square for the y-terms by adding and subtracting
(4/2)2= 4 inside the parenthesis:
(2x2−(y2+ 4y+ 4 −4)) + 9 = 0
Step 4: Simplify the expression inside the parenthesis:
(2x2−(y+ 2)2+ 4) + 9 = 0
Step 5: Move the constant term to the right side:
(2x2−(y+ 2)2=−13
7
Step 6: Divide by 2 on both sides:
x2−(y+ 2)2
2=−13
2
Step 7: Isolate the squared term:
−(y+ 2)2
2=−x2−13
2
Step 8: Multiply by −1to simplify the equation:
(y+ 2)2
2=x2+13
2
Step 9: Take the square root of both sides to solve for y:
y+ 2 = ±√2√x2+13
2
Step 10: Subtract 2 from both sides to isolate y:
y=−2±√2√x2+13
2
Therefore, the solution to the equation 2x2−y2+ 4y+ 9 = 0 in terms of y
is y=−2±√2√x2+13
2.
Question 9
Question
Find the standard form equation of the ellipse with foci F1(−3,0) and F2(3,0),
and major axis length 10.
Solution
Step 1: Find the center of the ellipse. The center of the ellipse is the midpoint of
the segment connecting the foci. Let C(h, k)be the center of the ellipse. Then
the center is given by:
h=−3+3
2= 0
k=0+0
2= 0
Therefore, the center of the ellipse is C(0,0).
Step 2: Find the distance between the foci. The distance between the foci
is equal to 2c, where cis the distance from the center to a focus. Since the
distance between the foci is 6((−3,0) to (3,0)), we can calculate c:
2c= 6
8
c= 3
Step 3: Find the distance from the center to a vertex. The distance from
the center to a vertex is given by a, the semi-major axis. Since the major axis
length is given as 10, we have:
2a= 10
a= 5
Step 4: Determine the minor axis length. The minor axis length 2bcan be
found using the relation c2=a2−b2. Given that a= 5 and c= 3, we can solve
for b:
32= 52−b2
9 = 25 −b2
b2= 16
b= 4
Step 5: Write the standard form equation of the ellipse. The standard form
equation of an ellipse with center (h, k), major axis length 2a, and minor axis
length 2bis:
(x−h)2
a2+(y−k)2
b2= 1
Plugging in the values we found, the equation of the ellipse is:
x2
25 +y2
16 = 1
Question 10
Question
Solve the equation of the hyperbola (x−2)2
9−(y+1)2
4= 1 for y.
Solution
Step 1: Rearrange the given equation to isolate the fraction containing y. Step
2: Multiply both sides by 4 to eliminate the fraction in the equation. Step 3:
Expand the equation. Step 4: Rearrange the terms to gather all terms with y
on one side. Step 5: Divide by 4 and take the square root to solve for y. Step
6: Simplify the result to find the final solution.
Step 1: Rewrite the equation of the hyperbola as:
(x−2)2−9(y+ 1)2= 4
Step 2: Multiply both sides by 4:
4(x−2)2−36(y+ 1)2= 16
9
Step 3: Expand the equation:
4(x2−4x+ 4) −36(y2+ 2y+ 1) = 16
Step 4: Rearrange the terms:
4x2−16x+ 16 −36y2−72y−36 = 16
Step 5: Divide by 4 and take the square root to solve for y:
−9y2−18y= 4x2−16x−36
9y2+ 18y= 16x−4x2+ 36
y2+ 2y=16x
9−4x2
9+ 4
(y+ 1)2=16x
9−4x2
9+ 4
Step 6: Simplify the result to find the final solution:
y+ 1 = ±√16x
9−4x2
9+ 4
y=−1±√16x
9−4x2
9+ 4
Therefore, the solution to the equation of the hyperbola for yis y=−1±
√16x
9−4x2
9+ 4.
Question 11
Question
Find the standard form equation of a parabola with a focus at (−2,4) and a
directrix at y= 6.
Solution
Step 1: Recall that the standard form of the equation of a parabola with a focus
at (h, k +p)and a directrix at y=k−pis given by
(x−h)2= 4p(y−k)
Step 2: From the given information, we can identify h=−2,k= 4, and
k−p= 6. This allows us to solve for p.
4 = 6 −p
p= 2
Step 3: Now substituting h,k, and pinto the standard form equation of the
parabola gives us:
(x+ 2)2= 8(y−4)
Therefore, the standard form equation of the parabola is (x+ 2)2= 8(y−4).
10
Question 12
Question
Solve the following system of equations:
{x2−y2= 1
xy = 2
Solution
Step 1: We have the system of equations:
{x2−y2= 1
xy = 2
Step 2: From the second equation, we can express yin terms of x:
y=2
x
Step 3: Substitute y=2
xinto the first equation:
x2−(2
x)2
= 1
Step 4: Simplify the equation:
x2−4
x2= 1 ⇒x4−x2−4 = 0
Step 5: Let u=x2:
u2−u−4 = 0
Step 6: Solve for uusing the quadratic formula:
u=1±√12−4·1·(−4)
2=1±√17
2
Step 7: Substitute back x2:
x2=1 + √17
2or x2=1−√17
2
Step 8: Solve for x:
x=±√1 + √17
2or x=±√1−√17
2
Step 9: Substitute the values of xback into y=2
xto find the corresponding
yvalues.
Therefore, the solutions to the system of equations are x=±√1+√17
2,
y=±√1−√17
2and x=±√1−√17
2,y=±√1+√17
2
11
Question 13
Question
Solve the equation of the hyperbola: (x−3)2
16 −(y+1)2
9= 1.
Solution
Step 1: Identify the center and vertices of the hyperbola. The equation of a
hyperbola in standard form is (x−h)2
a2−(y−k)2
b2= 1, where the center is at (h, k).
Comparing with the given equation (x−3)2
16 −(y+1)2
9= 1, we have: Center:
(h, k) = (3,−1)
The distance from the center to the vertices along the x-axis is a, and along
the y-axis is b.
Vertices: Along x-axis: 3±a= 3 ±4=7and −1So, vertices are (7,−1)
and (−1,−1).
Step 2: Identify the foci of the hyperbola. The distance from the center to
the foci along the x-axis is c, where c2=a2+b2.
Foci: c2=a2+b2c2= 16 + 9 c2= 25 c= 5
Foci lie on the major axis from the center in the positive and negative direc-
tions: (3 + 5,−1) and (3 −5,−1). So, foci are (8,−1) and (−2,−1).
Step 3: Sketch the hyperbola. Now, we can sketch the hyperbola with
the center at (3,−1), vertices at (7,−1) and (−1,−1), and foci at (8,−1) and
(−2,−1).
Question 14
Question
Find the standard form of the equation of the hyperbola with foci at (−1,2)
and (5,2), and a horizontal axis.
Solution
Step 1: Identify the center of the hyperbola. The center of the hyperbola is the
midpoint of the segment connecting the foci. Let C(−1+5
2,2+2
2)be the center
of the hyperbola. Therefore, C(2,2) is the center of the hyperbola.
Step 2: Determine the equation of the transverse axis. Since the foci are
on the same horizontal line, the transverse axis is horizontal. Therefore, the
equation of the transverse axis is y= 2.
Step 3: Find a, the distance from the center to a vertex. The distance
between the center and one of the foci is the value of c.c= 5 −2=3The
distance between the center and a vertex is given by a=√c2+b2, where bis the
distance from the center to a co-vertex. Since this is a hyperbola, a2=b2+c2.
12
By substituting c= 3 into the equation above, we have: a2=b2+ 32Since
b= 2, we now have: a2= 22+ 32= 13
Step 4: Write the standard form of the equation. The standard form of the
equation of a hyperbola centered at (h, k)with vertices at (h±a, k)and foci at
(h±c, k)is given by
(x−h)2
a2−(y−k)2
b2= 1
Plugging in the values of a,b,h, and k, the equation becomes:
(x−2)2
13 −(y−2)2
4= 1
Thus, the standard form of the equation of the hyperbola with foci at (−1,2)
and (5,2) and a horizontal axis is (x−2)2
13 −(y−2)2
4= 1.
Question 15
Question
Find the standard form of the equation of the hyperbola with vertices at (−2,3)
and (4,3), and foci at (6,3) and (−8,3).
Solution
Step 1: Find the center of the hyperbola by finding the midpoint of the line
segment between the vertices.
The midpoint formula is given by
(x1+x2
2,y1+y2
2)
where (x1, y1)and (x2, y2)are the coordinates of the two points. Given the
vertices (−2,3) and (4,3), the center of the hyperbola is
(−2+4
2,3+3
2)= (1,3)
Step 2: Find the distance between the center and one of the vertices to
determine a.
The distance formula is given by
√(x2−x1)2+ (y2−y1)2
where (x1, y1)is the center of the hyperbola and (x2, y2)is one of the vertices.
Let’s use (−2,3) as the vertex. The distance ais
√(1 −(−2))2+ (3 −3)2=√32+ 0 = 3
13
Step 3: Find the distance between the center and one of the foci to determine
c.
From the information given, we can see that the distance between the center
and either of the foci is
c= 6 −1 = 5
Step 4: Use the formula c2=a2+b2to solve for b.
Substitute the values of aand cinto the formula to find b:
52= 32+b2
25 = 9 + b2
b2= 16
b= 4
Step 5: Write the equation of the hyperbola in standard form.
The equation of a hyperbola centered at (h, k)with vertices on the transverse
axis is given by
(x−h)2
a2−(y−k)2
b2= 1
Substitute h= 1,k= 3,a= 3, and b= 4 into the equation:
(x−1)2
32−(y−3)2
42= 1
(x−1)2
9−(y−3)2
16 = 1
Therefore, the standard form of the equation of the hyperbola is
(x−1)2
9−(y−3)2
16 = 1
Question 16
Question
Solve the following system of equations:
{x2+y2= 10
x−2y= 3
14
Solution
Step 1: Rearrange the second equation to solve for x:
x= 3 + 2y
Step 2: Substitute this expression for xinto the first equation:
(3 + 2y)2+y2= 10
Step 3: Expand and simplify the equation:
(3 + 2y)(3 + 2y) + y2= 10
9+6y+ 6y+ 4y2+y2= 10
9 + 12y+ 5y2= 10
Step 4: Rearrange the equation and set it equal to zero:
5y2+ 12y−1 = 0
Step 5: Solve for yusing the quadratic formula:
y=−b±√b2−4ac
2a
Substitute a= 5,b= 12, and c=−1:
y=−12 ±√122−4(5)(−1)
2(5)
y=−12 ±√144 + 20
10
y=−12 ±√164
10
y=−12 ±2√41
10
y=−6±√41
5
Step 6: Find the corresponding xvalues for each yvalue: For y=−6+√41
5:
x= 3 + 2 (−6 + √41
5)= 3 −12
5+2√41
5=−12 + 2√41
5
For y=−6−√41
5:
x= 3 + 2 (−6−√41
5)= 3 −12
5−2√41
5=−12 −2√41
5
15
Hence, the solutions to the system of equations are:
(x, y) = (−12 + 2√41
5,−6 + √41
5)
(x, y) = (−12 −2√41
5,−6−√41
5)
Question 17
Question
Find the standard form equation of the ellipse that satisfies the following con-
ditions: The major axis is along the y-axis, the center is at (-2,3), one vertex is
at (-2,6), and the co-vertex on the right side is at (-5,3).
Solution
Step 1: The standard form equation of an ellipse centered at (h,k) with major
axis along the y-axis is given by:
(y−k)2
b2+(x−h)2
a2= 1
where ais the length of the semi-major axis (half the length of the major axis),
and bis the length of the semi-minor axis (half the length of the minor axis).
Step 2: We are given that the center of the ellipse is at (-2,3). So, h=−2
and k= 3.
Step 3: The distance between the center (-2,3) and one of the vertices (-2,6)
gives us the length of the semi-major axis, which is 3: a= 3.
Step 4: The distance between the center (-2,3) and the co-vertex (-5,3) gives
us the length of the semi-minor axis, which is 3: b= 3.
Step 5: Plug in the values of h,k,a, and binto the standard form equation
of the ellipse:
(y−3)2
32+(x+ 2)2
32= 1
Step 6: Simplify the equation:
(y−3)2
9+(x+ 2)2
9= 1
Step 7: Multiply through by 9 to get the standard form equation of the
ellipse:
(y−3)2+ (x+ 2)2= 9
Step 8: The standard form equation of the ellipse is (y−3)2+ (x+ 2)2= 9.
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