MATH 121 - COLLEGE ALGEBRA Equations and properties of conic sections Question Bank Set 10

MATH 121 - COLLEGE ALGEBRA -

Equations and properties of conic

sections

Question Bank - Set 10

Liberty University

Question 1

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 8.

Solution

Step 1: Find the center of the ellipse. Since the foci lie on the same horizontal

line, the center of the ellipse is the midpoint of the line segment joining the foci.

Let C=(−1+3

2,1+1

2)= (1,1).

Step 2: Find the distance between the two foci. The distance between the

two foci is 2a= 8, where ais the semi-major axis. So, a= 4.

Step 3: Find the distance between the center and each focus. The distance

between the center and each focus is c= 1 (half the distance between the foci).

Step 4: Use the relationship c2=a2−b2for an ellipse. Substitute c= 1 and

a= 4 into the equation to solve for b.

1 = 16 −b2

b2= 15

Step 5: Write the standard form of the equation of the ellipse. The standard

form of the equation of an ellipse centered at (h, k)is:

(x−h)2

a2+(y−k)2

b2= 1

So, for our ellipse with center at (1,1),a= 4, and b=√15, the equation is:

(x−1)2

16 +(y−1)2

15 = 1

Therefore, the standard form equation of the ellipse is (x−1)2

16 +(y−1)2

15 = 1 .

Question 2

Question

Find the equation of the hyperbola with vertices at (−5,0) and (5,0), and foci

at (−4,0) and (4,0).

Solution

Step 1: Find the center of the hyperbola by finding the midpoint of the segment

connecting the vertices:

Midpoint =(−5+5

2,0+0

2)= (0,0)

So, the center of the hyperbola is at (0,0).

Step 2: Find the distance between the center and the foci to determine c:

c= 4 −0 = 4

Step 3: Find the distance between the center and either vertex to determine

a= 5 −0 = 5

Step 4: Use the relationship between a,b(distance from center to vertex

along the transverse axis), and cto find b:

c2=a2+b2=⇒42= 52+b2=⇒16 = 25 + b2=⇒b2= 16 −25 = −9

Since b2is negative, this hyperbola opens left and right.

Step 5: Assembling the equation of the hyperbola: Since the hyperbola opens

horizontally, the equation is:

a2−y2

b2= 1 =⇒x2

25 −y2

−9= 1 =⇒x2

25 +y2

9= 1

Question 3

Question

Solve the following system of equations:

{x2+y2= 25

x+ 2y= 10

Solution

Step 1: We can start by solving the second equation for xin terms of y:

x= 10 −2y

Step 2: Substitute this expression for xinto the first equation:

(10 −2y)2+y2= 25

Step 3: Expand and simplify the left side of the equation:

100 −40y+ 4y2+y2= 25

Step 4: Combine like terms and rearrange the equation:

5y2−40y+ 75 = 0

Step 5: Divide the entire equation by 5 to simplify:

y2−8y+ 15 = 0

Step 6: Factor the quadratic equation:

(y−3)(y−5) = 0

Step 7: Use the zero product property to find the possible values of y:

y= 3 or y= 5

Step 8: Substitute these values back into the equation x= 10 −2yto find

the corresponding values of x: For y= 3:

x= 10 −2(3) = 10 −6 = 4

So, when y= 3,x= 4.

For y= 5:

x= 10 −2(5) = 10 −10 = 0

So, when y= 5,x= 0.

Step 9: Therefore, the solutions to the system of equations are:

(x, y) = (4,3) and (x, y) = (0,5)

Question 4

Question

Solve the following system of equations:

{x2+y2= 25

x−y= 1

Solution

Step 1: Subtract the second equation from the first to eliminate y.

x2+y2−(x−y) = 25

Step 2: Simplify the equation obtained in Step 1.

x2+y2−x+y= 25

Step 3: Since x2+y2= 25 according to the first equation, substitute 25 for

x2+y2in the equation from Step 2.

25 −x+y= 25

Step 4: Subtract 25 from both sides of the equation in Step 3 and simplify.

−x+y= 0

Step 5: Add xto both sides of the equation from step 4 to solve for y.

y=x

Step 6: Substitute ywith xin the second equation ( x−y= 1 ) to solve for

x−x= 1

Step 7: Combine like terms in the equation from Step 6.

0 = 1

Step 8: Since the equation 0 = 1 is false, there is no solution to the original

system of equations. Thus, the system of equations is inconsistent and has no

solution.

Question 5

Question

Find the standard form of the equation of the ellipse with foci at (−3,0) and

(3,0) and a major axis of length 10 units.

Solution

Step 1: The distance between the foci is the length of the major axis, which is

given as 10 units. Therefore, the distance between the foci is 2c= 10, which

implies c= 5.

Step 2: The center of the ellipse is the midpoint between the foci. The x-

coordinate of the center is the average of the x-coordinates of the foci, which

is (3 + (−3))/2=0. And the y-coordinate of the center is the average of the

y-coordinates of the foci, which is (0 + 0)/2 = 0. Thus, the center of the ellipse

is at the point (0,0).

Step 3: The distance between the center and each focus is c= 5, and the

distance between the center and each vertex is a, where ais half the length of

the major axis. Using the relationship c2=a2−b2, where a > b, we can find a.

Step 4: Since a= 5 and c= 5, we can substitute these values into the

equation c2=a2−b2to find b. Thus, 52= 52−b2⇒25 = 25 −b2⇒b2= 0 ⇒

b= 0.

Step 5: Now that we have a= 5 and b= 0, we can write the equation of the

ellipse in standard form. The standard form of an ellipse with center at (h, k),

major axis of length 2a, and minor axis of length 2bis (x−h)2

a2+(y−k)2

b2= 1.

Substituting h= 0,k= 0,a= 5, and b= 0, we have x2

25 +y2

0= 1.

Step 6: Since b= 0, the minor axis is essentially non-existent, making this

ellipse a circle with radius 5 units. So, the standard form of the equation for

the given ellipse is x2

25 = 1 .

Question 6

Question

Consider the equation of a hyperbola in standard form: (x−3)2

16 −(y+2)2

9= 1.

Determine the center, vertices, foci, and asymptotes of the hyperbola.

Solution

Step 1: First, identify the center of the hyperbola from the standard form. The

center of the hyperbola is given by (h, k), where his the x-coordinate and kis

the y-coordinate. In this case, the center is (3,−2).

Step 2: Next, we can determine the vertices of the hyperbola. For a hyper-

bola with equation (x−h)2

a2−(y−k)2

b2= 1, the vertices are located at (h±a, k).

Therefore, the vertices are at (3 + 4,−2) = (7,−2) and (3 −4,−2) = (−1,−2).

Step 3: To find the foci of the hyperbola, we use the formula c=√a2+b2.

The distance between the center and each focus is cunits. In this case, a2= 16

and b2= 9. So, c=√16 + 9 = √25 = 5. Therefore, the foci are located at

(3 + 5,−2) = (8,−2) and (3 −5,−2) = (−2,−2).

Step 4: The slopes of the asymptotes of the hyperbola are given by ±b

So in this case, the slopes are ±3

4. The equations of the asymptotes passing

through the center (3,−2) are y+ 2 = ±3

4(x−3).

Therefore, the center of the hyperbola is (3,−2), the vertices are at (7,−2)

and (−1,−2), the foci are at (8,−2) and (−2,−2), and the asymptotes are

y+ 2 = 3

4(x−3) and y+ 2 = −3

4(x−3).

Question 7

Question

Solve the equation 3x2+ 2xy −y2= 8 for yin terms of x.

Solution

To solve the equation 3x2+ 2xy −y2= 8 for yin terms of x, we will complete

the square.

Step 1: Rearrange the terms to prepare for completing the square:

3x2+ 2xy −y2= 8

3x2+ 2xy +(2x

2)2

−(2x

2)2

−y2= 8

3x2+ 2xy +x2−4x2

4−y2= 8

Step 2: Factor the perfect square trinomial:

(x+y)2−(2x)2−y2= 8

(x+y)2−4x2−y2= 8

Step 3: Rewrite the equation in the form of a difference of squares:

(x+y)2−(2x)2−y2= 8

(x+y+ 2x)(x+y−2x) = 8

(3x+y)(−x+y) = 8

Step 4: Solve for y:

(3x+y)(−x+y) = 8

−3x2+ 3xy −x2+xy = 8

−4x2+ 4xy = 8

4xy = 4x2+ 8

y=4x2+ 8

y=x+ 2

Therefore, the solution to the equation 3x2+ 2xy −y2= 8 for yin terms of

xis y=x+ 2.

Question 8

Question

Solve the following system of equations:

{2x2−y2= 1

x2+y2= 5

Solution

Step 1: Let’s isolate y2in the first equation:

2x2−y2= 1 =⇒y2= 2x2−1

Step 2: Now, substitute y2from the first equation into the second equation:

x2+ (2x2−1) = 5

Step 3: Simplify the equation:

3x2−1 = 5

Step 4: Add 1 to both sides to isolate the x2term:

3x2= 6

Step 5: Divide by 3 to solve for x:

x2= 2 =⇒x=±√2

Step 6: Plug the values of xback into the equation we found earlier to find

the corresponding values of y: For x=√2:

y2= 2(√2)2−1 =⇒y2= 3 =⇒y=±√3

For x=−√2:

y2= 2(−√2)2−1 =⇒y2= 3 =⇒y=±√3

Step 7: Therefore, the solutions to the system of equations are:

(x, y) = (√2,√3),(√2,−√3),(−√2,√3),(−√2,−√3)

Question 9

Question

Find the equation of the ellipse that satisfies the following conditions: the major

axis is parallel to the x-axis, the center is at (-2,3), passing through the points

(-4,3) and (-1,6).

Solution

Step 1: Find the length of the major axis. Since the major axis is parallel to the

x-axis, the length of the major axis is the difference between the x-coordinates

of the two given points.

Major axis length =| − 1−(−4)|= 3

Step 2: Find the length of the minor axis. Since an ellipse is symmetric, the

length of the minor axis is the distance between the y-coordinates of the two

given points.

Minor axis length =|6−3|= 3

Step 3: Determine the coordinates of the endpoints of the major axis. Since

the center is at (-2,3), the endpoints of the major axis are (-2,3 ± 1.5). So the

endpoints are (-2,1.5) and (-2,4.5).

Step 4: Find the semi-major axis (a) and semi-minor axis (b). The semi-

major axis (a) is half the major axis length: a=3

2= 1.5The semi-minor axis

(b) is half the minor axis length: b=3

2= 1.5

Step 5: Write the equation of the ellipse. The equation of an ellipse centered

at (h,k) with semi-major axis aand semi-minor axis bis:

(x−h)2

a2+(y−k)2

b2= 1

Substitute h=−2,k= 3,a= 1.5, and b= 1.5into the equation:

(x+ 2)2

2.25 +(y−3)2

2.25 = 1

Therefore, the equation of the ellipse is (x+ 2)2

2.25 +(y−3)2

2.25 = 1 .

Question 10

Question

Solve the following equation for y:3x2−2xy + 3y2−4x−8y+ 8 = 0.

Solution

Step 1: Rearrange the equation to write it in the form of a general conic section,

i.e., Ax2+Bxy +Cy2+Dx +Ey +F= 0. Step 2: Once the equation is in

the proper form, determine the discriminant B2−4AC to classify the type of

conic section. Step 3: Solve for yby using the quadratic formula and simplify

the result.

Question 11

Question

Solve for x:2x2+ 5x+ 3 = 0.

Solution

Step 1: We first recognize that the quadratic equation is in the form ax2+bx +

c= 0, where a= 2,b= 5, and c= 3.

Step 2: To solve for x, we can use the quadratic formula: x=−b±√b2−4ac

2a.

Step 3: Plugging in the values of a,b, and c, we get: x=−5±√52−4(2)(3)

2(2)

Step 4: Simplifying inside the square root: x=−5±√25−24

Step 5: Further simplifying: x=−5±√1

Step 6: Since √1 = 1, we have two possible solutions: x1=−5+1

4=−4

−1x2=−5−1

4=−6

4=−3

Step 7: Thus, the solutions to the equation 2x2+ 5x+ 3 = 0 are x=−1

and x=−3

Question 12

Question

Solve the equation of the ellipse: (x−3)2

16 +(y+2)2

9= 1.

Solution

Step 1: First, identify the center and major/minor axes of the ellipse. The

center is given by (h, k) = (3,−2), where his the x-coordinate of the center and

kis the y-coordinate of the center. The length of the major axis is 2a= 8, so

a= 4, and the length of the minor axis is 2b= 6, so b= 3.

Step 2: Now, determine which axis is the major axis and which is the minor

axis. Given the equation (x−3)2

16 +(y+2)2

9= 1, the term with the denominator

of 16 corresponds to the x-axis, while the term with the denominator of 9

corresponds to the y-axis. Therefore, the major axis is vertical.

Step 3: Next, find the foci of the ellipse. The distance from the center to

each focus is given by c=√a2−b2=√16 −9 = √7. Therefore, the foci are

located at (h, k ±c) = (3,−2±√7).

Step 4: Finally, write the equation of the ellipse in standard form. Since

the major axis is vertical, the standard form of the equation of the ellipse is

(y+2)2

9−(x−3)2

16 = 1.

Question 13

Question

Solve the system of equations:

{x2+ 4y2= 16

x+y= 3

Solution

Step 1: Let’s solve the second equation for xin terms of y: We can rewrite the

second equation as x= 3 −y.

Step 2: Substitute x= 3 −yinto the first equation:

(3 −y)2+ 4y2= 16

Step 3: Expand and simplify the equation:

9−6y+y2+ 4y2= 16

5y2−6y−7 = 0

Step 4: Solve the quadratic equation for y: We can solve the above quadratic

equation by factoring or using the quadratic formula. Factoring, we get:

(5y+ 3)(y−1) = 0

Step 5: Set each factor to zero and solve for y: Setting 5y+ 3 = 0 gives

y=−3

5. Setting y−1 = 0 gives y= 1.

Step 6: Substitute the solutions for yback into the equation x= 3−yto find

the corresponding values of x: When y=−3

5, then x= 3 −(−3

5) = 3 + 3

5=18

When y= 1, then x= 3 −1 = 2.

Step 7: Therefore, the solutions to the system of equations are:

{x=18

5, y =−3

x= 2, y = 1

Question 14

Question

Find the standard form of the equation of the parabola with a focus at (−1,3)

and a directrix given by y= 7.

Solution

Step 1: Recall that the standard form of the equation of a parabola with its

vertex at (h, k), focus F(h, k +p), and directrix y=k−popening vertically is

given by:

(y−k)2= 4p(x−h).

Step 2: From the given information, we can identify the vertex V(h, k)as

(−1,5) since it lies halfway between the focus and the directrix.

Step 3: The distance between the vertex and the focus is equal to the distance

between the vertex and the directrix. Using this property, we can find the value

of p:

p= 7 −5 = 2.

Step 4: Now that we have the vertex at (−1,5) and p= 2, we can write the

standard form of the equation of the parabola:

(y−5)2= 8(x+ 1).

Therefore, the standard form of the equation of the parabola is (y−5)2=

8(x+ 1).

Question 15

Question

Solve the following system of equations:

{2x2+ 5y2= 18

4x−y= 1

Solution

To solve the system of equations, we can use the substitution method.

• Start by solving the second equation for y:

Step 1: 4x−y= 1

⇒y= 4x−1

• Now substitute yin terms of xinto the first equation:

Step 2: 2x2+ 5(4x−1)2= 18

⇒2x2+ 5(16x2−8x+ 1) = 18

⇒2x2+ 80x2−40x+ 5 = 18

⇒82x2−40x−13 = 0

• Now solve the quadratic equation for x:

Step 3: x=−(−40) ±√(−40)2−4(82)(−13)

2(82)

⇒x=40 ±√1600 + 4264

164

⇒x=40 ±√5864

164

⇒x=40 ±2√1466

164

Simplify further to get the values of x, then substitute back into the equation

to solve for y.

Question 16

Question

Find the standard form equation of the hyperbola that satisfies the given con-

ditions: Foci at (-5,0) and (5,0), and passes through the point (6,4).

Solution

Step 1: Find the center of the hyperbola. Since the foci are at (-5,0) and (5,0),

the center is at the midpoint of the foci, which is at (−5+5

2,0+0

2) = (0,0).

Step 2: Find the distance from the center to a focus. This distance is the

value of c, the distance from the center to a focus, which is 5 units in this case.

Step 3: Use the distance formula to find the value of a, the distance from

the center to a vertex. Since the point (6,4) lies on the hyperbola, the distance

from the center to this point is a. We have:

√(6 −0)2+ (4 −0)2=a

a=√36 + 16 = √52

Step 4: The standard form equation for a hyperbola centered at the origin

is x2

a2−y2

b2= 1, where aand bare the distances from the center to the vertices

along the x-axis and y-axis respectively. Since the hyperbola is horizontally

oriented, a2=c2+b2.

Step 5: Substitute the values of aand cinto the equation from Step 4 and

solve for b2.√522= 52+b2

52 = 25 + b2

b2= 27

Step 6: The standard form equation for the hyperbola is:

52 −y2

27 = 1

Question 17

Question

Find the standard form of the equation of the hyperbola that satisfies the given

conditions: - Foci at (−4,0) and (4,0) - Vertices 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 foci on the x-axis is given by x2

a2−y2

b2= 1, where arepresents the

distance from the center to each vertex, and crepresents the distance from the

center to each focus. We can find the values of aand cbased on the information

given.

Step 2: The distance between the foci is 2c= 8, so c= 4. Therefore, the

value of acan be found using the relationship c2=a2+b2.

Step 3: Since the vertices are at (−7,0) and (7,0), the value of ais 7. Now,

we can solve for busing the relationship c2=a2+b2.

Step 4: Substituting a= 7 and c= 4 into c2=a2+b2, we get 42= 72+b2.

Solving for b, we find b=√16 −49 = √−33.

Step 5: Since b2is negative, this indicates that the hyperbola is horizontal.

Thus, the equation of the hyperbola is x2

49 −y2

−33 = 1.

Step 6: Simplifying, we obtain x2

49 +y2

33 = 1. Therefore, the standard form

of the equation of the hyperbola is x2

49 +y2

33 = 1 .

Question 18

Question

Find the equation of the circle with center at (2,−3) and passing through the

point (−1,4).

Solution

Step 1: Use the formula for the equation of a circle with center (h, k)and radius

r:(x−h)2+ (y−k)2=r2.

Step 2: Plug in the values h= 2,k=−3, and (−1,4) for another point on

the circle to find the radius.

(x−2)2+ (y+ 3)2=r2

(−1−2)2+ (4 + 3)2=r2

(−3)2+ (7)2=r2

9 + 49 = r2

58 = r2

r=√58

Step 3: Substitute back the values of h= 2,k=−3, and r=√58 to get

the final equation of the circle.

(x−2)2+ (y+ 3)2= 58

Therefore, the equation of the circle with center at (2,−3) and passing

through the point (−1,4) is (x−2)2+ (y+ 3)2= 58.

Question 19

Question

Solve the system of equations:

{x2−y2= 9

xy =−2

Solution

Step 1: We can solve the second equation for yin terms of xby dividing both

sides by x:

y=−2

Step 2: Now, substitute y=−2

xinto the first equation:

x2−(−2

x)2

= 9

Step 3: Simplify the equation:

x2−4

x2= 9

Step 4: Multiply through by x2to clear the fraction:

x4−4 = 9x2

Step 5: Rearrange the terms to form a quadratic equation:

x4−9x2−4 = 0

Step 6: Let u=x2, then the equation becomes:

u2−9u−4 = 0

Step 7: Solve the quadratic equation for u:

u=9±√81 + 16

u=9±√97

Step 8: Since u=x2, there are two possible values for x:

x=±√9±√97

Step 9: Substitute the found values of xback into y=−2

xto find the

corresponding yvalues for each x.

Therefore, the solutions to the system of equations are:



√9 + √97

2,−2

√9+√97



and 

−√9 + √97

2,2

√9+√97





Question 20

Question

Find the standard form of the equation of the ellipse with foci at (2,1) and

(−2,1) passing through the point (−3,4).

Solution

Step 1: Find the center of the ellipse. The center is the midpoint of the line

segment joining the foci, which is ((−2 + 2)/2,(1 + 1)/2) = (0,1).

Step 2: Find the distance from the center to one of the foci. This distance

is the distance from the origin to the foci, which is 2.

Step 3: Since the ellipse passes through the point (−3,4), the sum of the

distances from this point to each of the foci must be constant and equal to

twice the distance from the center to a focus. Let (x, y)be a point on the

ellipse. The distances are √(x−2)2+ (y−1)2and √(x+ 2)2+ (y−1)2. We

have the equation

√(x−2)2+ (y−1)2+√(x+ 2)2+ (y−1)2= 4.

Step 4: Square both sides of the equation from step 3 to eliminate the square

roots. Simplifying, we get

(x−2)2+(y−1)2+2√[(x−2)2+ (y−1)2][(x+ 2)2+ (y−1)2]+(x+2)2+(y−1)2= 16.

Step 5: Expand and simplify the equation from step 4. This results in

2x2+ 4y2−12x−8 = 0.

Therefore, the standard form of the equation of the ellipse is 2x2+ 4y2−

12x−8 = 0.

Question 21

Question

Find the standard form of the equation of the hyperbola with vertices at (−3,0)

and (3,0) and foci at (−5,0) and (5,0).

Solution

Step 1: Find the center of the hyperbola.

The center of the hyperbola is the midpoint between the vertices, which is the

point ((3 + (−3))/2,(0 + 0)/2). So, the center of the hyperbola is (0,0).

Step 2: Find a(distance from the center to a vertex) and c(distance from

the center to a focus).

The distance between the center and a vertex is 3units, so a= 3. The distance

between the center and a focus is 5units, so c= 5.

Step 3: Find b(distance from the center to the transverse axis).

We know that c2=a2+b2for a hyperbola. Substituting the values of aand c:

52= 32+b2

25 = 9 + b2

b2= 16

b= 4

Step 4: Write the equation in standard form.

The equation of a hyperbola with transverse axis along the x-axis is:

(x−h)2

a2−(y−k)2

b2= 1

Plugging in the values of h,k,a, and b:

9−y2

16 = 1

So, the standard form of the equation of the hyperbola is x2

9−y2

16 = 1.

Question 22

Question

Given the equation of a hyperbola in standard form:

(y−3)2

16 −(x+ 1)2

9= 1

Find the center, vertices, and foci of the hyperbola.

Solution

Step 1: Identify the center of the hyperbola.

To find the center of the hyperbola, we compare the standard form to the

general equation for a hyperbola:

(y−k)2

a2−(x−h)2

b2= 1

Here, the center of the hyperbola is at the point (h, k). Therefore, the center

of the given hyperbola is (−1,3).

Step 2: Find the vertices of the hyperbola.

The distance from the center to the vertices along the transverse axis is equal

to a. In this case, a= 4 (since a2= 16).

The vertices are located at (h, k +a)and (h, k −a). Therefore, the vertices

of the hyperbola are (−1,7) and (−1,−1).

Step 3: Determine the foci of the hyperbola.

The distance from the center to the foci along the transverse axis is equal

to c. We can find cusing the relationship:

c2=a2+b2

In this case, a2= 16 and b2= 9, so a2+b2= 25. Hence, c= 5.

The foci are located at (h, k +c)and (h, k −c). Therefore, the foci of the

hyperbola are (−1,8) and (−1,2).

Question 23

Question

Given the equation of a hyperbola: (y+2)2

25 −(x−1)2

16 = 1, identify the center,

vertices, foci, and asymptotes of the hyperbola.

Solution

Step 1: First, identify the center of the hyperbola. From the given equation, we

can see that the center of the hyperbola is at the point (h, k) = (1,−2).

Step 2: Next, let’s find the vertices of the hyperbola. The distance from

the center to the vertices along the transverse axis is a=√25 = 5. Thus, the

vertices are located at (h, k +a) = (1,−2 + 5) = (1,3) and (1,−2−5) = (1,−7).

Step 3: Now, let’s determine the foci of the hyperbola. The distance from the

center to the foci along the transverse axis is c=√a2+b2=√25 + 16 = √41.

Hence, the foci are at (h, k +c) = (1,−2 + √41) and (1,−2−√41).

Step 4: Determine the slopes of the asymptotes. The slopes of the asymp-

totes for a hyperbola of the form (y−k)2

a2−(x−h)2

b2= 1 are given by ±b

a. In this

case, the slope of the asymptotes is ±4

Step 5: Next, we can find the equations of the asymptotes. Using the point-

slope form y−y1=m(x−x1), where (x1, y1)is a point on the line, we can use

either of the vertices to get the equations. So, the equations of the asymptotes

are y−3 = ±4

5(x−1) and y+ 7 = ±4

5(x−1).

Therefore, the center of the hyperbola is at (1,−2), the vertices are at (1,3)

and (1,−7), the foci are at (1,−2+√41) and (1,−2−√41), and the asymptotes

have slopes ±4

5with equations y−3 = ±4

5(x−1) and y+ 7 = ±4

5(x−1).

Question 24

Question

Determine the standard form of the equation of the conic section that satisfies

the following conditions: The center is at (−3,−4), the major axis is horizontal,

with length 10, and passes through the point (−1,−1).

Solution

Step 1: The standard form of the equation for a conic section with a horizontal

major axis is given by

(x−h)2

a2−(y−k)2

b2= 1

where (h, k)is the center of the conic section, ais the half-length of the major

axis, and bis the half-length of the minor axis.

Step 2: Given that the center is at (−3,−4) and the major axis has a length

of 10, we know that a= 5. Since the major axis is horizontal, acorresponds to

the term under x.

Step 3: Let’s denote the equation we are looking for as

(x+ 3)2

25 −(y+ 4)2

b2= 1

Step 4: To find b2, we can use the point (−1,−1) that the conic section

passes through. Substitute x=−1and y=−1into the equation to find b2.

Step 5: Substituting x=−1and y=−1into the equation, we get

(−1 + 3)2

25 −(−1 + 4)2

b2= 1

Step 6: Simplifying, we have

25 −9

b2= 1

Step 7: Multiplying through by 25b2, we get

4b2−225 = 25b2

Step 8: Rearranging terms, we find

21b2= 225

Step 9: Solving for b2, we get

b2=225

21 =75

Step 10: Therefore, the standard form of the equation of the conic section is

(x+ 3)2

25 −(y+ 4)2

= 1

Question 25

Question

Solve the equation 4x2−9y2+ 16x−18y−71 = 0 and determine the type of

conic section it represents.

Solution

Step 1: Rearrange the equation to put it in standard form:

4x2+ 16x−9y2−18y= 71

Step 2: Complete the square for xterms:

4(x2+ 4x)−9y2−18y= 71

4(x2+ 4x+ 4) −4(4) −9y2−18y= 71 + 16 −4

4(x+ 2)2−9y2−18y= 83

Step 3: Group yterms together:

4(x+ 2)2−9(y2+ 2y) = 83

So, for our ellipse with center at (1,1),a= 4, and b=√15, the equation is:

(x−1)2

16 +(y−1)2

15 = 1

Therefore, the standard form equation of the ellipse is (x−1)2

16 +(y−1)2

15 = 1 .

Question 2

Question

Find the equation of the hyperbola with vertices at (−5,0) and (5,0), and foci

at (−4,0) and (4,0).

Solution

Step 1: Find the center of the hyperbola by finding the midpoint of the segment

connecting the vertices:

Midpoint =(−5+5

2,0+0

2)= (0,0)

So, the center of the hyperbola is at (0,0).

Step 2: Find the distance between the center and the foci to determine c:

c= 4 −0 = 4

Step 3: Find the distance between the center and either vertex to determine

a= 5 −0 = 5

Step 4: Use the relationship between a,b(distance from center to vertex

along the transverse axis), and cto find b:

c2=a2+b2=⇒42= 52+b2=⇒16 = 25 + b2=⇒b2= 16 −25 = −9

Since b2is negative, this hyperbola opens left and right.

Step 5: Assembling the equation of the hyperbola: Since the hyperbola opens

horizontally, the equation is:

a2−y2

b2= 1 =⇒x2

25 −y2

−9= 1 =⇒x2

25 +y2

9= 1

Question 3

Question

Solve the following system of equations:

{x2+y2= 25

x+ 2y= 10

Solution

Step 1: We can start by solving the second equation for xin terms of y:

x= 10 −2y

Step 2: Substitute this expression for xinto the first equation:

(10 −2y)2+y2= 25

Step 3: Expand and simplify the left side of the equation:

100 −40y+ 4y2+y2= 25

Step 4: Combine like terms and rearrange the equation:

5y2−40y+ 75 = 0

Step 5: Divide the entire equation by 5 to simplify:

y2−8y+ 15 = 0

Step 6: Factor the quadratic equation:

(y−3)(y−5) = 0

Step 7: Use the zero product property to find the possible values of y:

y= 3 or y= 5

Step 8: Substitute these values back into the equation x= 10 −2yto find

the corresponding values of x: For y= 3:

x= 10 −2(3) = 10 −6 = 4

So, when y= 3,x= 4.

For y= 5:

x= 10 −2(5) = 10 −10 = 0

So, when y= 5,x= 0.

Step 9: Therefore, the solutions to the system of equations are:

(x, y) = (4,3) and (x, y) = (0,5)

Question 4

Question

Solve the following system of equations:

{x2+y2= 25

x−y= 1

Solution

Step 1: Subtract the second equation from the first to eliminate y.

x2+y2−(x−y) = 25

Step 2: Simplify the equation obtained in Step 1.

x2+y2−x+y= 25

Step 3: Since x2+y2= 25 according to the first equation, substitute 25 for

x2+y2in the equation from Step 2.

25 −x+y= 25

Step 4: Subtract 25 from both sides of the equation in Step 3 and simplify.

−x+y= 0

Step 5: Add xto both sides of the equation from step 4 to solve for y.

y=x

Step 6: Substitute ywith xin the second equation ( x−y= 1 ) to solve for

x−x= 1

Step 7: Combine like terms in the equation from Step 6.

0 = 1

Step 8: Since the equation 0 = 1 is false, there is no solution to the original

system of equations. Thus, the system of equations is inconsistent and has no

solution.

Question 5

Question

Find the standard form of the equation of the ellipse with foci at (−3,0) and

(3,0) and a major axis of length 10 units.

Solution

Step 1: The distance between the foci is the length of the major axis, which is

given as 10 units. Therefore, the distance between the foci is 2c= 10, which

implies c= 5.

Step 2: The center of the ellipse is the midpoint between the foci. The x-

coordinate of the center is the average of the x-coordinates of the foci, which

is (3 + (−3))/2=0. And the y-coordinate of the center is the average of the

y-coordinates of the foci, which is (0 + 0)/2 = 0. Thus, the center of the ellipse

is at the point (0,0).

Step 3: The distance between the center and each focus is c= 5, and the

distance between the center and each vertex is a, where ais half the length of

the major axis. Using the relationship c2=a2−b2, where a > b, we can find a.

Step 4: Since a= 5 and c= 5, we can substitute these values into the

equation c2=a2−b2to find b. Thus, 52= 52−b2⇒25 = 25 −b2⇒b2= 0 ⇒

b= 0.

Step 5: Now that we have a= 5 and b= 0, we can write the equation of the

ellipse in standard form. The standard form of an ellipse with center at (h, k),

major axis of length 2a, and minor axis of length 2bis (x−h)2

a2+(y−k)2

b2= 1.

Substituting h= 0,k= 0,a= 5, and b= 0, we have x2

25 +y2

0= 1.

Step 6: Since b= 0, the minor axis is essentially non-existent, making this

ellipse a circle with radius 5 units. So, the standard form of the equation for

the given ellipse is x2

25 = 1 .

Question 6

Question

Consider the equation of a hyperbola in standard form: (x−3)2

16 −(y+2)2

9= 1.

Determine the center, vertices, foci, and asymptotes of the hyperbola.

Solution

Step 1: First, identify the center of the hyperbola from the standard form. The

center of the hyperbola is given by (h, k), where his the x-coordinate and kis

the y-coordinate. In this case, the center is (3,−2).

Step 2: Next, we can determine the vertices of the hyperbola. For a hyper-

bola with equation (x−h)2

a2−(y−k)2

b2= 1, the vertices are located at (h±a, k).

Therefore, the vertices are at (3 + 4,−2) = (7,−2) and (3 −4,−2) = (−1,−2).

Step 3: To find the foci of the hyperbola, we use the formula c=√a2+b2.

The distance between the center and each focus is cunits. In this case, a2= 16

and b2= 9. So, c=√16 + 9 = √25 = 5. Therefore, the foci are located at

(3 + 5,−2) = (8,−2) and (3 −5,−2) = (−2,−2).

Step 4: The slopes of the asymptotes of the hyperbola are given by ±b

So in this case, the slopes are ±3

4. The equations of the asymptotes passing

through the center (3,−2) are y+ 2 = ±3

4(x−3).

Therefore, the center of the hyperbola is (3,−2), the vertices are at (7,−2)

and (−1,−2), the foci are at (8,−2) and (−2,−2), and the asymptotes are

y+ 2 = 3

4(x−3) and y+ 2 = −3

4(x−3).

Question 7

Question

Solve the equation 3x2+ 2xy −y2= 8 for yin terms of x.

Solution

To solve the equation 3x2+ 2xy −y2= 8 for yin terms of x, we will complete

the square.

Step 1: Rearrange the terms to prepare for completing the square:

3x2+ 2xy −y2= 8

3x2+ 2xy +(2x

2)2

−(2x

2)2

−y2= 8

3x2+ 2xy +x2−4x2

4−y2= 8

Step 2: Factor the perfect square trinomial:

(x+y)2−(2x)2−y2= 8

(x+y)2−4x2−y2= 8

Step 3: Rewrite the equation in the form of a difference of squares:

(x+y)2−(2x)2−y2= 8

(x+y+ 2x)(x+y−2x) = 8

(3x+y)(−x+y) = 8

Step 4: Solve for y:

(3x+y)(−x+y) = 8

−3x2+ 3xy −x2+xy = 8

−4x2+ 4xy = 8

4xy = 4x2+ 8

y=4x2+ 8

y=x+ 2

Therefore, the solution to the equation 3x2+ 2xy −y2= 8 for yin terms of

xis y=x+ 2.

Question 8

Question

Solve the following system of equations:

{2x2−y2= 1

x2+y2= 5

Solution

Step 1: Let’s isolate y2in the first equation:

2x2−y2= 1 =⇒y2= 2x2−1

Step 2: Now, substitute y2from the first equation into the second equation:

x2+ (2x2−1) = 5

Step 3: Simplify the equation:

3x2−1 = 5

Step 4: Add 1 to both sides to isolate the x2term:

3x2= 6

Step 5: Divide by 3 to solve for x:

x2= 2 =⇒x=±√2

Step 6: Plug the values of xback into the equation we found earlier to find

the corresponding values of y: For x=√2:

y2= 2(√2)2−1 =⇒y2= 3 =⇒y=±√3

For x=−√2:

y2= 2(−√2)2−1 =⇒y2= 3 =⇒y=±√3

Step 7: Therefore, the solutions to the system of equations are:

(x, y) = (√2,√3),(√2,−√3),(−√2,√3),(−√2,−√3)

Question 9

Question

Find the equation of the ellipse that satisfies the following conditions: the major

axis is parallel to the x-axis, the center is at (-2,3), passing through the points

(-4,3) and (-1,6).

Solution

Step 1: Find the length of the major axis. Since the major axis is parallel to the

x-axis, the length of the major axis is the difference between the x-coordinates

of the two given points.

Major axis length =| − 1−(−4)|= 3

Step 2: Find the length of the minor axis. Since an ellipse is symmetric, the

length of the minor axis is the distance between the y-coordinates of the two

given points.

Minor axis length =|6−3|= 3

Step 3: Determine the coordinates of the endpoints of the major axis. Since

the center is at (-2,3), the endpoints of the major axis are (-2,3 ± 1.5). So the

endpoints are (-2,1.5) and (-2,4.5).

Step 4: Find the semi-major axis (a) and semi-minor axis (b). The semi-

major axis (a) is half the major axis length: a=3

2= 1.5The semi-minor axis

(b) is half the minor axis length: b=3

2= 1.5

Step 5: Write the equation of the ellipse. The equation of an ellipse centered

at (h,k) with semi-major axis aand semi-minor axis bis:

(x−h)2

a2+(y−k)2

b2= 1

Substitute h=−2,k= 3,a= 1.5, and b= 1.5into the equation:

(x+ 2)2

2.25 +(y−3)2

2.25 = 1

Therefore, the equation of the ellipse is (x+ 2)2

2.25 +(y−3)2

2.25 = 1 .

Question 10

Question

Solve the following equation for y:3x2−2xy + 3y2−4x−8y+ 8 = 0.

Solution

Step 1: Rearrange the equation to write it in the form of a general conic section,

i.e., Ax2+Bxy +Cy2+Dx +Ey +F= 0. Step 2: Once the equation is in

the proper form, determine the discriminant B2−4AC to classify the type of

conic section. Step 3: Solve for yby using the quadratic formula and simplify

the result.

Question 11

Question

Solve for x:2x2+ 5x+ 3 = 0.

Solution

Step 1: We first recognize that the quadratic equation is in the form ax2+bx +

c= 0, where a= 2,b= 5, and c= 3.

Step 2: To solve for x, we can use the quadratic formula: x=−b±√b2−4ac

2a.

Step 3: Plugging in the values of a,b, and c, we get: x=−5±√52−4(2)(3)

2(2)

Step 4: Simplifying inside the square root: x=−5±√25−24

Step 5: Further simplifying: x=−5±√1

Step 6: Since √1 = 1, we have two possible solutions: x1=−5+1

4=−4

−1x2=−5−1

4=−6

4=−3

Step 7: Thus, the solutions to the equation 2x2+ 5x+ 3 = 0 are x=−1

and x=−3

Question 12

Question

Solve the equation of the ellipse: (x−3)2

16 +(y+2)2

9= 1.

Solution

Step 1: First, identify the center and major/minor axes of the ellipse. The

center is given by (h, k) = (3,−2), where his the x-coordinate of the center and

kis the y-coordinate of the center. The length of the major axis is 2a= 8, so

a= 4, and the length of the minor axis is 2b= 6, so b= 3.

Step 2: Now, determine which axis is the major axis and which is the minor

axis. Given the equation (x−3)2

16 +(y+2)2

9= 1, the term with the denominator

of 16 corresponds to the x-axis, while the term with the denominator of 9

corresponds to the y-axis. Therefore, the major axis is vertical.

Step 3: Next, find the foci of the ellipse. The distance from the center to

each focus is given by c=√a2−b2=√16 −9 = √7. Therefore, the foci are

located at (h, k ±c) = (3,−2±√7).

Step 4: Finally, write the equation of the ellipse in standard form. Since

the major axis is vertical, the standard form of the equation of the ellipse is

(y+2)2

9−(x−3)2

16 = 1.

Question 13

Question

Solve the system of equations:

{x2+ 4y2= 16

x+y= 3

Solution

Step 1: Let’s solve the second equation for xin terms of y: We can rewrite the

second equation as x= 3 −y.

Step 2: Substitute x= 3 −yinto the first equation:

(3 −y)2+ 4y2= 16

Step 3: Expand and simplify the equation:

9−6y+y2+ 4y2= 16

5y2−6y−7 = 0

Step 4: Solve the quadratic equation for y: We can solve the above quadratic

equation by factoring or using the quadratic formula. Factoring, we get:

(5y+ 3)(y−1) = 0

Step 5: Set each factor to zero and solve for y: Setting 5y+ 3 = 0 gives

y=−3

5. Setting y−1 = 0 gives y= 1.

Step 6: Substitute the solutions for yback into the equation x= 3−yto find

the corresponding values of x: When y=−3

5, then x= 3 −(−3

5) = 3 + 3

5=18

When y= 1, then x= 3 −1 = 2.

Step 7: Therefore, the solutions to the system of equations are:

{x=18

5, y =−3

x= 2, y = 1

Question 14

Question

Find the standard form of the equation of the parabola with a focus at (−1,3)

and a directrix given by y= 7.

Solution

Step 1: Recall that the standard form of the equation of a parabola with its

vertex at (h, k), focus F(h, k +p), and directrix y=k−popening vertically is

given by:

(y−k)2= 4p(x−h).

Step 2: From the given information, we can identify the vertex V(h, k)as

(−1,5) since it lies halfway between the focus and the directrix.

Step 3: The distance between the vertex and the focus is equal to the distance

between the vertex and the directrix. Using this property, we can find the value

of p:

p= 7 −5 = 2.

Step 4: Now that we have the vertex at (−1,5) and p= 2, we can write the

standard form of the equation of the parabola:

(y−5)2= 8(x+ 1).

Therefore, the standard form of the equation of the parabola is (y−5)2=

8(x+ 1).

Question 15

Question

Solve the following system of equations:

{2x2+ 5y2= 18

4x−y= 1

Solution

To solve the system of equations, we can use the substitution method.

• Start by solving the second equation for y:

Step 1: 4x−y= 1

⇒y= 4x−1

• Now substitute yin terms of xinto the first equation:

Step 2: 2x2+ 5(4x−1)2= 18

⇒2x2+ 5(16x2−8x+ 1) = 18

⇒2x2+ 80x2−40x+ 5 = 18

⇒82x2−40x−13 = 0

• Now solve the quadratic equation for x:

Step 3: x=−(−40) ±√(−40)2−4(82)(−13)

2(82)

⇒x=40 ±√1600 + 4264

164

⇒x=40 ±√5864

164

⇒x=40 ±2√1466

164

Simplify further to get the values of x, then substitute back into the equation

to solve for y.

Question 16

Question

Find the standard form equation of the hyperbola that satisfies the given con-

ditions: Foci at (-5,0) and (5,0), and passes through the point (6,4).

Solution

Step 1: Find the center of the hyperbola. Since the foci are at (-5,0) and (5,0),

the center is at the midpoint of the foci, which is at (−5+5

2,0+0

2) = (0,0).

Step 2: Find the distance from the center to a focus. This distance is the

value of c, the distance from the center to a focus, which is 5 units in this case.

Step 3: Use the distance formula to find the value of a, the distance from

the center to a vertex. Since the point (6,4) lies on the hyperbola, the distance

from the center to this point is a. We have:

√(6 −0)2+ (4 −0)2=a

a=√36 + 16 = √52

Step 4: The standard form equation for a hyperbola centered at the origin

is x2

a2−y2

b2= 1, where aand bare the distances from the center to the vertices

along the x-axis and y-axis respectively. Since the hyperbola is horizontally

oriented, a2=c2+b2.

Step 5: Substitute the values of aand cinto the equation from Step 4 and

solve for b2.√522= 52+b2

52 = 25 + b2

b2= 27

Step 6: The standard form equation for the hyperbola is:

52 −y2

27 = 1

Question 17

Question

Find the standard form of the equation of the hyperbola that satisfies the given

conditions: - Foci at (−4,0) and (4,0) - Vertices 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 foci on the x-axis is given by x2

a2−y2

b2= 1, where arepresents the

distance from the center to each vertex, and crepresents the distance from the

center to each focus. We can find the values of aand cbased on the information

given.

Step 2: The distance between the foci is 2c= 8, so c= 4. Therefore, the

value of acan be found using the relationship c2=a2+b2.

Step 3: Since the vertices are at (−7,0) and (7,0), the value of ais 7. Now,

we can solve for busing the relationship c2=a2+b2.

Step 4: Substituting a= 7 and c= 4 into c2=a2+b2, we get 42= 72+b2.

Solving for b, we find b=√16 −49 = √−33.

Step 5: Since b2is negative, this indicates that the hyperbola is horizontal.

Thus, the equation of the hyperbola is x2

49 −y2

−33 = 1.

Step 6: Simplifying, we obtain x2

49 +y2

33 = 1. Therefore, the standard form

of the equation of the hyperbola is x2

49 +y2

33 = 1 .

Question 18

Question

Find the equation of the circle with center at (2,−3) and passing through the

point (−1,4).

Solution

Step 1: Use the formula for the equation of a circle with center (h, k)and radius

r:(x−h)2+ (y−k)2=r2.

Step 2: Plug in the values h= 2,k=−3, and (−1,4) for another point on

the circle to find the radius.

(x−2)2+ (y+ 3)2=r2

(−1−2)2+ (4 + 3)2=r2

(−3)2+ (7)2=r2

9 + 49 = r2

58 = r2

r=√58

Step 3: Substitute back the values of h= 2,k=−3, and r=√58 to get

the final equation of the circle.

(x−2)2+ (y+ 3)2= 58

Therefore, the equation of the circle with center at (2,−3) and passing

through the point (−1,4) is (x−2)2+ (y+ 3)2= 58.

Question 19

Question

Solve the system of equations:

{x2−y2= 9

xy =−2

Solution

Step 1: We can solve the second equation for yin terms of xby dividing both

sides by x:

y=−2

Step 2: Now, substitute y=−2

xinto the first equation:

x2−(−2

x)2

= 9

Step 3: Simplify the equation:

x2−4

x2= 9

Step 4: Multiply through by x2to clear the fraction:

x4−4 = 9x2

Step 5: Rearrange the terms to form a quadratic equation:

x4−9x2−4 = 0

Step 6: Let u=x2, then the equation becomes:

u2−9u−4 = 0

Step 7: Solve the quadratic equation for u:

u=9±√81 + 16

u=9±√97

Step 8: Since u=x2, there are two possible values for x:

x=±√9±√97

Step 9: Substitute the found values of xback into y=−2

xto find the

corresponding yvalues for each x.

Therefore, the solutions to the system of equations are:



√9 + √97

2,−2

√9+√97



and 

−√9 + √97

2,2

√9+√97





Question 20

Question

Find the standard form of the equation of the ellipse with foci at (2,1) and

(−2,1) passing through the point (−3,4).

Solution

Step 1: Find the center of the ellipse. The center is the midpoint of the line

segment joining the foci, which is ((−2 + 2)/2,(1 + 1)/2) = (0,1).

Step 2: Find the distance from the center to one of the foci. This distance

is the distance from the origin to the foci, which is 2.

Step 3: Since the ellipse passes through the point (−3,4), the sum of the

distances from this point to each of the foci must be constant and equal to

twice the distance from the center to a focus. Let (x, y)be a point on the

ellipse. The distances are √(x−2)2+ (y−1)2and √(x+ 2)2+ (y−1)2. We

have the equation

√(x−2)2+ (y−1)2+√(x+ 2)2+ (y−1)2= 4.

Step 4: Square both sides of the equation from step 3 to eliminate the square

roots. Simplifying, we get

(x−2)2+(y−1)2+2√[(x−2)2+ (y−1)2][(x+ 2)2+ (y−1)2]+(x+2)2+(y−1)2= 16.

Step 5: Expand and simplify the equation from step 4. This results in

2x2+ 4y2−12x−8 = 0.

Therefore, the standard form of the equation of the ellipse is 2x2+ 4y2−

12x−8 = 0.

Question 21

Question

Find the standard form of the equation of the hyperbola with vertices at (−3,0)

and (3,0) and foci at (−5,0) and (5,0).

Solution

Step 1: Find the center of the hyperbola.

The center of the hyperbola is the midpoint between the vertices, which is the

point ((3 + (−3))/2,(0 + 0)/2). So, the center of the hyperbola is (0,0).

Step 2: Find a(distance from the center to a vertex) and c(distance from

the center to a focus).

The distance between the center and a vertex is 3units, so a= 3. The distance

between the center and a focus is 5units, so c= 5.

Step 3: Find b(distance from the center to the transverse axis).

We know that c2=a2+b2for a hyperbola. Substituting the values of aand c:

52= 32+b2

25 = 9 + b2

b2= 16

b= 4

Step 4: Write the equation in standard form.

The equation of a hyperbola with transverse axis along the x-axis is:

(x−h)2

a2−(y−k)2

b2= 1

Plugging in the values of h,k,a, and b:

9−y2

16 = 1

So, the standard form of the equation of the hyperbola is x2

9−y2

16 = 1.

Question 22

Question

Given the equation of a hyperbola in standard form:

(y−3)2

16 −(x+ 1)2

9= 1

Find the center, vertices, and foci of the hyperbola.

Solution

Step 1: Identify the center of the hyperbola.

To find the center of the hyperbola, we compare the standard form to the

general equation for a hyperbola:

(y−k)2

a2−(x−h)2

b2= 1

Here, the center of the hyperbola is at the point (h, k). Therefore, the center

of the given hyperbola is (−1,3).

Step 2: Find the vertices of the hyperbola.

The distance from the center to the vertices along the transverse axis is equal

to a. In this case, a= 4 (since a2= 16).

The vertices are located at (h, k +a)and (h, k −a). Therefore, the vertices

of the hyperbola are (−1,7) and (−1,−1).

Step 3: Determine the foci of the hyperbola.

The distance from the center to the foci along the transverse axis is equal

to c. We can find cusing the relationship:

c2=a2+b2

In this case, a2= 16 and b2= 9, so a2+b2= 25. Hence, c= 5.

The foci are located at (h, k +c)and (h, k −c). Therefore, the foci of the

hyperbola are (−1,8) and (−1,2).

Question 23

Question

Given the equation of a hyperbola: (y+2)2

25 −(x−1)2

16 = 1, identify the center,

vertices, foci, and asymptotes of the hyperbola.