Algebra 3 Quiz: Ellipses And Hyperbolas

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Ellipses And HyperbolasQuestion 1 of 20

Consider the ellipse $9x^2+25y^2-36x+50y-164=0$ . What are the endpoints of its major axis?

(7,-1)

and

(-3,-1)

(2,2)

and

(2,-4)

(6,-1)

and

(-2,-1)

(-7,1)

and

(3,1)

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Algebra 3 Quiz

Algebra 3 Quiz: Ellipses And Hyperbolas

Practice Ellipses And Hyperbolas in Algebra 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Ellipses And Hyperbolas, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 3.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Consider the ellipse $9x^2+25y^2-36x+50y-164=0$ . What are the endpoints of its major axis?

$(7,-1)$ and $(-3,-1)$ (correct answer)
$(2,2)$ and $(2,-4)$
$(6,-1)$ and $(-2,-1)$
$(-7,1)$ and $(3,1)$

Explanation: Whenever you see a general quadratic like this, your first move should be completing the square to put the ellipse in standard form. That reveals the center and which axis is major. Group the

x

and

y

terms:

9(x^2-4x)+25(y^2+2y)=164.

Complete the square:

9[(x-2)^2-4]+25[(y+1)^2-1]=164

9(x-2)^2+25(y+1)^2=225.

Divide by 225:

\frac{(x-2)^2}{25}+\frac{(y+1)^2}{9}=1.

The center is

(2,-1)

. Since

25>9

, the major axis is horizontal, with

a=5

. So the endpoints are

(2\pm5,-1)

, which are

(7,-1)

and

(-3,-1)

. The choices

(2,2)

and

(2,-4)

are the endpoints of the minor axis, because they use

b=3

vertically — a classic swap-the-axes error. The choices

(6,-1)

and

(-2,-1)

are actually the foci, since

c=\sqrt{25-9}=4

, not the vertices. Finally,

(-7,1)

and

(3,1)

reverse both the center and the signs: they come from treating the center as

(-2,1)

instead of

(2,-1)

. Your study tip: always write the standard form before finding endpoints. The larger denominator tells you the major axis direction;

a^2

is that denominator, not the distance from the center to the focus. Check that your endpoints are exactly

a

units from the center along that axis.

Question 2

An ellipse is centered at the origin. Its vertices are $(0,6)$ and $(0,-6)$ , and one focus is $(0,\sqrt{11})$ . Which equation represents the ellipse?

$\frac{x^2}{25}+\frac{y^2}{36}=1$ (correct answer)
$\frac{x^2}{36}+\frac{y^2}{25}=1$
$\frac{x^2}{11}+\frac{y^2}{36}=1$
$\frac{x^2}{25}+\frac{y^2}{11}=1$

Explanation: Whenever you see an ellipse centered at the origin, first locate the vertices: they tell you the major axis and the value of

a

. Since the vertices are

(0,6)

and

(0,-6)

, the major axis is vertical, so

a=6

, and

a^2=36

. The focus

(0,\sqrt{11})

gives

c=\sqrt{11}

, so

c^2=11

. For an ellipse,

b^2=a^2-c^2=36-11=25

. Because the major axis is vertical, the larger denominator belongs under

y^2

\frac{x^2}{25}+\frac{y^2}{36}=1

. The choice

\frac{x^2}{36}+\frac{y^2}{25}=1

swaps the denominators, treating the major axis as horizontal. The choice

\frac{x^2}{11}+\frac{y^2}{36}=1

puts

c^2=11

under

x^2

, confusing the focal distance with the semi-minor axis. The choice

\frac{x^2}{25}+\frac{y^2}{11}=1

puts

c^2=11

under

y^2

, losing the fact that the

y

-direction reaches to the vertices. Only the equation with

36

under

y^2

and

25

under

x^2

matches the vertical ellipse. Study tip: always identify whether the major axis is vertical or horizontal first, then use

b^2=a^2-c^2

. That prevents the most common mistake on this exam: swapping the denominators or using

c^2

b^2

Question 3

A hyperbola has equation $\frac{y^2}{25}-\frac{x^2}{16}=1$ . Which statement about its asymptotes is correct?

They intersect at $(5,4)$ and have slopes $\pm\frac{5}{4}$ .
They intersect at $(0,0)$ and have slopes $\pm\frac{4}{5}$ .
They intersect at $(0,0)$ and have slopes $\pm\frac{5}{4}$ . (correct answer)
They intersect at $(0,0)$ and have slopes $\pm\frac{25}{16}$ .

Explanation: Whenever you see a hyperbola in standard form, first identify which squared term is positive. Here

\frac{y^2}{25}-\frac{x^2}{16}=1

is a vertical hyperbola because

y^2

comes first, so its center is

(0,0)

and its asymptotes pass through the center with slopes

\pm \frac{a}{b}

, where

a^2=25

and

b^2=16

. Thus

a=5

b=4

, giving slopes

\pm \frac{5}{4}

. The correct statement is the one saying they intersect at

(0,0)

and have slopes

\pm\frac{5}{4}

. The choice claiming they intersect at

(5,4)

confuses the center with a point on or related to the rectangle; the asymptotes always intersect at the hyperbola's center, not at

(a,b)

. The choice with slopes

\pm\frac{4}{5}

uses

\frac{b}{a}

, which is the slope formula for a horizontal hyperbola like

\frac{x^2}{25}-\frac{y^2}{16}=1

, not this one. The choice with slopes

\pm\frac{25}{16}

incorrectly squares the denominators instead of taking square roots to get

a

and

b

. A reliable strategy: for hyperbolas, find the center from the numerators, then read

a^2

and

b^2

from the denominators. If

y^2

is positive, slopes are

\pm\frac{a}{b}

; if

x^2

is positive, slopes are

\pm\frac{b}{a}

. Also remember the center is always the asymptotes' intersection point.

Question 4

A hyperbola centered at $(-2,4)$ has a vertex at $(-2,7)$ and a focus at $(-2,9)$ . Which equation represents the hyperbola?

$\frac{(x+2)^2}{16}-\frac{(y-4)^2}{9}=1$
$\frac{(y-4)^2}{9}-\frac{(x-2)^2}{16}=1$
$\frac{(y-4)^2}{9}-\frac{(x+2)^2}{16}=1$ (correct answer)
$\frac{(y+4)^2}{9}-\frac{(x-2)^2}{16}=1$

Explanation: Whenever you see a hyperbola question, start by locating the center and deciding whether the transverse axis is horizontal or vertical. Here the center is

(-2,4)

, and both the vertex

(-2,7)

and focus

(-2,9)

lie directly above it, so the hyperbola opens vertically — the

y

-term comes first. The distance from the center to the vertex is

a=3

. The distance from the center to the focus is

c=5

. For a hyperbola,

c^2=a^2+b^2

, so

25=9+b^2

, giving

b^2=16

. The vertical standard form is

\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1

. Substituting

h=-2

k=4

gives

\frac{(y-4)^2}{9}-\frac{(x+2)^2}{16}=1

, the correct equation. The choice

\frac{(x+2)^2}{16}-\frac{(y-4)^2}{9}=1

would open horizontally, but your vertices and focus are above/below the center, not left/right. It also uses the denominators in the wrong positions. The option

\frac{(y-4)^2}{9}-\frac{(x-2)^2}{16}=1

has the right vertical orientation butn

x-2

shifts the center to

(2,4)

, when you need

x+2

for center

x=-2

. The choice

\frac{(y+4)^2}{9}-\frac{(x-2)^2}{16}=1

shifts both coordinates, placing the center at

(2,-4)

, which is incorrect. A quick check: from center, the vertex and focus both move in the

y

-direction, so vertical form. Then the signs in parentheses must match the center:

x+2

and

y-4

. Memorize this pattern — orientation decides which term is positive, and the parentheses reveal the center.

Question 5

A hyperbola is centered at $(-1,3)$ , has vertices $(-4,3)$ and $(2,3)$ , and its asymptotes have slopes $\pm\frac{4}{3}$ . Which equation represents the hyperbola?

$\frac{(y-3)^2}{9}-\frac{(x+1)^2}{16}=1$
$\frac{(x+1)^2}{9}-\frac{(y-3)^2}{16}=1$ (correct answer)
$\frac{(x+1)^2}{16}-\frac{(y-3)^2}{9}=1$
$\frac{(x-1)^2}{9}-\frac{(y+3)^2}{16}=1$

Explanation: Whenever you see a hyperbola, start by translating the center into parentheses. Since the center is

(-1,3)

, the equation must contain

(x+1)

and

(y-3)

. The vertices

(-4,3)

and

(2,3)

lie horizontally from the center, so the hyperbola opens left/right: the

x

-term must be positive, and the distance from center to a vertex is

3

, so

a^2=9

. For a horizontal hyperbola, asymptote slopes are

\pm \frac{b}{a}

. Given

\pm\frac{4}{3}

and

a=3

, you get

b=4

, so

b^2=16

. This yields

\frac{(x+1)^2}{9}-\frac{(y-3)^2}{16}=1

. The equation with

\frac{(y-3)^2}{9}-\frac{(x+1)^2}{16}=1

reverses the transverse axis: it would open vertically, with vertices at

y=3\pm3

, and its asymptote slopes would be

\pm\frac{3}{4}

. The choice

\frac{(x+1)^2}{16}-\frac{(y-3)^2}{9}=1

uses

a=4

and

b=3

, giving vertices

4

units from center and slopes

\pm\frac{3}{4}

. The choice

\frac{(x-1)^2}{9}-\frac{(y+3)^2}{16}=1

has center

(1,-3)

, not

(-1,3)

, because the signs inside the parentheses are reversed. Strategy: find the center, check whether the vertices move horizontally or vertically, then use the asymptote slope

\frac{b}{a}

to find

b

. This three-step check prevents the most common sign and axis mistakes.

Question 6

The conic $4x^2-9y^2+8x+36y-16=0$ is a hyperbola. What are its vertices?

The vertices are $\left(-1,2+\frac{2\sqrt{13}}{3}\right)$ and $\left(-1,2-\frac{2\sqrt{13}}{3}\right)$ .
The vertices are $(-1,4)$ and $(-1,0)$ .
The vertices are $\left(1,\frac{10}{3}\right)$ and $\left(1,\frac{2}{3}\right)$ .
The vertices are $\left(-1,\frac{10}{3}\right)$ and $\left(-1,\frac{2}{3}\right)$ . (correct answer)

Explanation: When you see a general second-degree conic like this, completing the square reveals both its center and orientation. Group terms:

4(x^2+2x)-9(y^2-4y)-16=0.

Complete the squares:

4[(x+1)^2-1]-9[(y-2)^2-4]-16=0,

which simplifies to

9(y-2)^2-4(x+1)^2=16.

Divide by

16

\frac{(y-2)^2}{16/9}-\frac{(x+1)^2}{4}=1.

This is a vertical hyperbola centered at

(-1,2)

, with

a^2=\frac{16}{9}

, so

a=\frac43

. Therefore the vertices are

\frac43

units above and below the center, along

x=-1

\left(-1,2+\frac43\right)=\left(-1,\frac{10}{3}\right),\qquad \left(-1,2-\frac43\right)=\left(-1,\frac23\right).

The choices with

x=1

, such as

\left(1,\frac{10}{3}\right)

and

\left(1,\frac23\right)

, use the wrong center sign from

(x+1)^2

. The choice

\left(-1,2\pm\frac{2\sqrt{13}}3\right)

gives the foci, since

c=\sqrt{a^2+b^2}=\sqrt{\frac{16}{9}+4}=\frac{2\sqrt{13}}3

, not the vertices. The choice

(-1,4)

and

(-1,0)

uses distance

2

from the center, which is

b

, not

a

; the relevant denominator is

\frac{16}{9}

, not

4

. Study tip: after writing a hyperbola in standard form, look at which squared term is positive — that term gives the transverse axis and

a^2

. Vertices are center

\pm a

along that axis; foci come from

c^2=a^2+b^2

Question 7

Which of the following are the foci of the ellipse $\frac{(x+1)^2}{20}+\frac{(y-3)^2}{36}=1$ ?

$(1,7)$ and $(1,-1)$
$(3,3)$ and $(-5,3)$
$(-1,3+2\sqrt{5})$ and $(-1,3-2\sqrt{5})$
$(-1,7)$ and $(-1,-1)$ (correct answer)

Explanation: Whenever you see an ellipse question, first find the center and decide whether the major axis is horizontal or vertical. The center comes from the numerators: here

(x+1)^2

and

(y-3)^2

mean center

(-1,3)

. Since the larger denominator

36

is under the

y

-term, the ellipse is vertical. For a vertical ellipse, the foci are

(h, k \pm c)

, where

c^2 = a^2 - b^2

. Here

a^2=36

, so

a=6

, and

b^2=20

, so

c^2=36-20=16

, giving

c=4

. Thus the foci are

(-1, 3+4)=(-1,7)

and

(-1,3-4)=(-1,-1)

. The choice

(1,7)

and

(1,-1)

uses the correct vertical shifts but moves the center's

x

-coordinate incorrectly; the center is

-1

, not

1

. The choice

(3,3)

and

(-5,3)

treats the major axis as horizontal; those would be the foci if the larger denominator were under

x

, not

y

. The choice

(-1,3+2\sqrt{5})

and

(-1,3-2\sqrt{5})

uses

b=\sqrt{20}=2\sqrt{5}

as the focal distance, but the focus distance must be

c=\sqrt{a^2-b^2}

, not

b

itself. Your key move: always compare

a^2

and

b^2

to determine orientation, then compute

c=\sqrt{a^2-b^2}

. The foci lie along the major axis,

c

units from the center.

Question 8

An elliptical archway is modeled by the upper half of an ellipse centered at the origin. The archway is 20 feet wide at its base and 6 feet high at the center. Which equation describes the full ellipse?

$\frac{x^2}{36}+\frac{y^2}{100}=1$
$\frac{x^2}{100}+\frac{y^2}{36}=1$ (correct answer)
$\frac{x^2}{100}+\frac{y^2}{12}=1$
$\frac{x^2}{400}+\frac{y^2}{36}=1$

Explanation: Whenever you see an ellipse problem, start by recalling the standard form:

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

where

a

is the horizontal semi-axis and

b

is the vertical semi-axis. Here, the archway is the upper half of an ellipse centered at the origin, so its base lies on the

x

-axis. A width of 20 feet means the full ellipse crosses from

-10

10

on the

x

-axis, so

a=10

and

a^2=100

. The height of 6 feet at the center means the top of the arch reaches

y=6

, so

b=6

and

b^2=36

. Therefore the full ellipse is

\frac{x^2}{100}+\frac{y^2}{36}=1.

The choice

\frac{x^2}{36}+\frac{y^2}{100}=1

swaps the axes, giving a 12-foot-wide, 20-foot-tall ellipse. The choice

\frac{x^2}{100}+\frac{y^2}{12}=1

uses 12 instead of

6^2

, confusing the height with its square. The choice

\frac{x^2}{400}+\frac{y^2}{36}=1

squares the full width 20 instead of the semi-width 10, which would make the ellipse 40 feet wide. On exam day, remember: the denominators in the ellipse equation are the squares of the semi-axes, not the full widths or heights. Always halve the total width before squaring it.

Question 9

Consider the ellipse $x^2+4y^2=36$ . Which points are the endpoints of its minor axis?

$(6,0)$ and $(-6,0)$
$(0,3)$ and $(0,-3)$ (correct answer)
$(3\sqrt{3},0)$ and $(-3\sqrt{3},0)$
$(0,6)$ and $(0,-6)$

Explanation: Whenever you're asked about the axes of an ellipse, start by rewriting the equation in standard form. Dividing

x^2+4y^2=36

by 36 gives

\frac{x^2}{36}+\frac{y^2}{9}=1

. Since the denominator under

x^2

is larger, the major axis is horizontal, with semi-axis length

a=\sqrt{36}=6

, so its endpoints are

(6,0)

and

(-6,0)

. The minor axis is vertical, with semi-axis length

b=\sqrt{9}=3

, so its endpoints are

(0,3)

and

(0,-3)

. The pair

(6,0)

and

(-6,0)

are actually the major-axis endpoints, not the minor-axis endpoints. The points

(3\sqrt{3},0)

and

(-3\sqrt{3},0)

are the foci, since

c=\sqrt{a^2-b^2}=\sqrt{36-9}=3\sqrt{3}

; foci always lie on the major axis, so they never belong to the minor axis. The pair

(0,6)

and

(0,-6)

reverses the roles of 6 and 3, treating the vertical direction as the longer axis. A reliable strategy: after putting the ellipse in standard form, the smaller denominator tells you which axis is the minor axis. On this exam, the most common trap is confusing the foci with the minor-axis endpoints — remember that the foci are on the major axis, not on the minor axis.

Question 10

A hyperbola centered at the origin has vertices $(\pm3,0)$ and asymptotes $y=\pm2x$ . Which equation represents the hyperbola?

$\frac{x^2}{36}-\frac{y^2}{9}=1$
$\frac{x^2}{9}-\frac{y^2}{18}=1$
$\frac{y^2}{9}-\frac{x^2}{36}=1$
$\frac{x^2}{9}-\frac{y^2}{36}=1$ (correct answer)

Explanation: When you see a hyperbola centered at the origin with given vertices and asymptotes, start by locating the transverse axis. The vertices

(\pm3,0)

tell you the hyperbola opens left and right, so its equation has the form

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1

, with

a=3

, hence

a^2=9

. The asymptotes

y=\pm2x

tell you the slope ratio

\frac{b}{a}=2

. Since

a=3

, that gives

b=6

, so

b^2=36

. Therefore the equation must be

\frac{x^2}{9}-\frac{y^2}{36}=1

. Now check the traps. The equation

\frac{x^2}{36}-\frac{y^2}{9}=1

reverses

a^2

and

b^2

; it would have vertices

(\pm6,0)

and asymptote slopes

\pm\frac{1}{2}

, not

\pm2

. The equation

\frac{x^2}{9}-\frac{y^2}{18}=1

has the correct

a^2=9

, but

b^2=18

makes

b=\sqrt{18}

, so the asymptote slope is

\frac{\sqrt{18}}{3}\approx1.41

, not

2

. The equation

\frac{y^2}{9}-\frac{x^2}{36}=1

represents a vertical hyperbola with vertices

(0,\pm3)

, contradicting the given

(\pm3,0)

; its asymptotes would be

y=\pm\frac{1}{2}x

. Your key move: from the vertices get

a^2

, then use the asymptote slope to solve for

b^2

. Always confirm the transverse axis direction first — that single check eliminates several wrong choices immediately.

Question 11

An ellipse centered at the origin has foci $(\pm4,0)$ and vertices $(\pm5,0)$ . Which equation represents the ellipse?

$\frac{x^2}{25}+\frac{y^2}{16}=1$
$\frac{x^2}{9}+\frac{y^2}{25}=1$
$\frac{x^2}{25}+\frac{y^2}{9}=1$ (correct answer)
$\frac{x^2}{16}+\frac{y^2}{9}=1$

Explanation: Whenever you see an ellipse centered at the origin, start by locating the major axis. Here, both foci (±4,0) and vertices (±5,0) lie on the x-axis, so the ellipse is horizontal: the larger denominator belongs under

x^2

. The vertices give

a=5

, so

a^2=25

. The foci give

c=4

, so

c^2=16

. For an ellipse,

c^2=a^2-b^2

, so

16=25-b^2

, meaning

b^2=9

. Thus the equation is

\frac{x^2}{25}+\frac{y^2}{9}=1

. The choice

\frac{x^2}{25}+\frac{y^2}{16}=1

has the correct vertices but treats the focal distance 4 as

b^2

, giving

c^2=25-16=9

, so the foci would be (±3,0), not (±4,0). The choice

\frac{x^2}{9}+\frac{y^2}{25}=1

swaps the denominators, making the major axis vertical with vertices (0,±5) and foci (0,±4) — the wrong orientation. The choice

\frac{x^2}{16}+\frac{y^2}{9}=1

uses

a^2=16

, so the vertices would be (±4,0), which are actually the foci given; it confuses the roles of

a

and

c

. A quick study tip: in ellipse problems, the vertices always determine

a^2

, and the foci determine

c^2

. Then find

b^2

using

c^2=a^2-b^2

. Also, check which variable has the larger denominator — that tells you whether the ellipse is horizontal or vertical.

Question 12

Which of the following is an equation of one asymptote of the hyperbola $\frac{(x-3)^2}{16}-\frac{(y+2)^2}{9}=1$ ?

$y+2=\frac{3}{4}(x-3)$ (correct answer)
$y-2=\frac{3}{4}(x+3)$
$y+2=\frac{4}{3}(x-3)$
$y-2=\frac{4}{3}(x+3)$

Explanation: When you see a hyperbola in standard form, your first move is to locate its center and decide whether it opens left/right or up/down. Here the equation is

\frac{(x-3)^2}{16}-\frac{(y+2)^2}{9}=1

, so the center is

(3,-2)

. Because the

x

-term is positive, the transverse axis is horizontal; thus the asymptote slopes are

\pm \frac{b}{a}

, where

a^2=16

gives

a=4

and

b^2=9

gives

b=3

. Asymptotes pass through the center, so they have the form

y - (-2)=\pm \frac{3}{4}(x-3)

, or

y+2=\pm \frac{3}{4}(x-3)

. That makes

y+2=\frac{3}{4}(x-3)

the correct asymptote. The choice

y-2=\frac{3}{4}(x+3)

uses the wrong center: it treats the center as

(-3,2)

instead of

(3,-2)

. The choice

y+2=\frac{4}{3}(x-3)

has the right center but uses the reciprocal slope

\frac{a}{b}=\frac{4}{3}

, which would be the slope if the hyperbola opened vertically. The remaining choice

y-2=\frac{4}{3}(x+3)

combines both errors: wrong center and reversed slope ratio. A reliable study tip: for any hyperbola, write the center from the shifted coordinates, identify which squared term is positive to know the direction, and then form asymptotes using

\frac{b}{a}

for horizontal hyperbolas but

\frac{a}{b}

for vertical ones. Always keep the slope paired with the actual center of the hyperbola.

Question 13

Consider the conic $9x^2 - 4y^2 - 54x - 16y - 79 = 0$ . Which statement about its graph is true?

The vertices are $(-1,-2)$ and $(7,-2)$ . (correct answer)
The center is $(3,2)$ , and the vertices are $(-1,2)$ and $(7,2)$ .
The foci are $(3,-2 \pm 2\sqrt{13})$ .
The asymptotes are $y+2 = \pm \frac{2}{3}(x-3)$ .

Explanation: Completing the square gives

9(x-3)^2 - 4(y+2)^2 = 144

, so

\frac{(x-3)^2}{16} - \frac{(y+2)^2}{36}=1

. The center is

(3,-2)

a=4

, and

b=6

. The transverse axis is horizontal, so the vertices are

(3 \pm 4, -2)

, namely

(-1,-2)

and

(7,-2)

. The focal distance is

c=\sqrt{52}=2\sqrt{13}

, so the foci are

(3 \pm 2\sqrt{13}, -2)

, not vertically displaced. The asymptote slope is

6/4=3/2

, not

2/3

, and the center is

(3,-2)

, not

(3,2)

Question 14

A hyperbola with center $(-2,3)$ has vertices $(-6,3)$ and $(2,3)$ . One of its asymptotes is $y-3=\frac{3}{4}(x+2)$ . What are the foci?

$(-6,3)$ and $(2,3)$
$(-5,3)$ and $(1,3)$
$(-7,3)$ and $(3,3)$ (correct answer)
$(-2,-2)$ and $(-2,8)$

Explanation: Each vertex is 4 units from the center, so

a=4

. For a horizontal hyperbola the asymptote slope is

b/a=3/4

, so

b=3

. Then

c^2=a^2+b^2=16+9=25

, so

c=5

. The foci are

(-2 \pm 5, 3)

, giving

(-7,3)

and

(3,3)

. The pair

(-6,3)

(2,3)

are the vertices, and the vertical pair would come from treating the transverse axis as vertical.

Question 15

An ellipse has center $(2,-3)$ , one vertex at $(7,-3)$ , and one focus at $(5,-3)$ . Which equation represents the ellipse?

$\frac{(x-2)^2}{25}+\frac{(y+3)^2}{16}=1$ (correct answer)
$\frac{(x-2)^2}{16}+\frac{(y+3)^2}{25}=1$
$\frac{(x+2)^2}{25}+\frac{(y-3)^2}{16}=1$
$\frac{(x-2)^2}{25}-\frac{(y+3)^2}{16}=1$

Explanation: The vertex is 5 units from the center, so

a=5

. The focus is 3 units from the center, so

c=3

. Thus

b^2=25-9=16

. Since both the vertex and focus lie on the horizontal line

y=-3

, the major axis is horizontal, giving

\frac{(x-2)^2}{25}+\frac{(y+3)^2}{16}=1

Question 16

A conic has vertices $(\pm 2,0)$ and foci $(\pm \sqrt{13},0)$ . Which equation represents it?

$\frac{x^2}{4}+\frac{y^2}{9}=1$
$\frac{x^2}{4}-\frac{y^2}{9}=1$ (correct answer)
$\frac{x^2}{9}-\frac{y^2}{4}=1$
$\frac{x^2}{9}+\frac{y^2}{4}=1$

Explanation: Since the foci are farther from the origin than the vertices, the conic is a hyperbola. For a horizontal hyperbola,

a=2

and

c=\sqrt{13}

, so

b^2=c^2-a^2=13-4=9

. The equation is

\frac{x^2}{4}-\frac{y^2}{9}=1

. The choice

\frac{x^2}{9}-\frac{y^2}{4}=1

has the same focal distance but vertices at

(\pm 3,0)

, not

(\pm 2,0)

Question 17

A student converts $4x^2 - 9y^2 + 16x + 54y - 29 = 0$ into $\frac{(x+2)^2}{9}-\frac{(y-3)^2}{4}=1$ and concludes the vertices are $(-5,3)$ and $(1,3)$ . What are the actual vertices?

$(-5,3)$ and $(1,3)$
$(-2,-1)$ and $(-2,7)$
$(-2,3 \pm \sqrt{13})$
$(-2,1)$ and $(-2,5)$ (correct answer)

Explanation: Completing the square correctly gives

4(x+2)^2 - 9(y-3)^2 = -36

, equivalent to

\frac{(y-3)^2}{4}-\frac{(x+2)^2}{9}=1

. This is a vertical hyperbola centered at

(-2,3)

with

a=2

, so the vertices are

(-2,3 \pm 2)

, namely

(-2,1)

and

(-2,5)

. The student's equation used the wrong sign on the constant and produced a horizontal hyperbola.

Question 18

Which equation represents a hyperbola with center at the origin, focus $(0,\sqrt{34})$ , and asymptote $y=\frac{3}{5}x$ ?

$\frac{x^2}{25}-\frac{y^2}{9}=1$
$\frac{y^2}{25}-\frac{x^2}{9}=1$
$\frac{y^2}{9}-\frac{x^2}{25}=1$ (correct answer)
$\frac{x^2}{9}-\frac{y^2}{25}=1$

Explanation: The focus lies on the

y

-axis, so the vertical form

\frac{y^2}{a^2}-\frac{x^2}{b^2}=1

is required. Its asymptote slope is

a/b

. Setting

a/b=3/5

and

c=\sqrt{34}

, the values

a=3

b=5

work because

3^2+5^2=34

. Hence the equation is

\frac{y^2}{9}-\frac{x^2}{25}=1

Question 19

A conic with center at the origin has vertices $(\pm 4,0)$ and passes through $(2,3)$ . Which point is a focus of this conic?

$(2,0)$ (correct answer)
$(\sqrt{7},0)$
$(0,2)$
$(4,0)$

Explanation: If it were a hyperbola,

\frac{x^2}{16}-\frac{y^2}{b^2}=1

would require

\frac{1}{4}-\frac{9}{b^2}=1

, which is impossible. Thus it is an ellipse

\frac{x^2}{16}+\frac{y^2}{b^2}=1

. Substituting

(2,3)

gives

\frac{1}{4}+\frac{9}{b^2}=1

, so

b^2=12

. Then

c^2=16-12=4

, so the foci are

(\pm 2,0)

;

(2,0)

is one of them.

Question 20

A hyperbola has vertices $(1,3)$ and $(7,3)$ , and one focus is at $(9,3)$ . What are its asymptotes?

$y-3=\pm \frac{3}{4}(x-4)$
$y-3=\pm \frac{5}{3}(x-4)$
$y-3=\pm \frac{4}{3}(x-4)$ (correct answer)
$y-3=\pm \frac{4}{5}(x-4)$

Explanation: The center is

(4,3)

, and

a=3

. The focus is 5 units from the center, so

c=5

. Then

b^2=25-9=16

, so

b=4

. For a horizontal hyperbola, the asymptote slope is

b/a=4/3

, giving

y-3=\pm \frac{4}{3}(x-4)