Geometry Quiz: Derive Equations Of Ellipses And Parabolas

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Derive Equations Of Ellipses And ParabolasQuestion 1 of 20

The equation of an ellipse is $\frac{(x-2)^2}{36} + \frac{(y+1)^2}{20} = 1$ . What is the distance between the two foci of this ellipse?

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Geometry Quiz

Geometry Quiz: Derive Equations Of Ellipses And Parabolas

Practice Derive Equations Of Ellipses And Parabolas in Geometry 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 Derive Equations Of Ellipses And Parabolas, giving you a quick way to practice the rules, question types, and explanations that matter most for Geometry.

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

The equation of an ellipse is $\frac{(x-2)^2}{36} + \frac{(y+1)^2}{20} = 1$ . What is the distance between the two foci of this ellipse?

4
8 (correct answer)
16
32

Explanation: From the equation, a² = 36 and b² = 20, so a = 6 and b = 2√5. Since a > b, the major axis is horizontal. Using c² = a² - b² = 36 - 20 = 16, we get c = 4. The distance between foci is 2c = 2(4) = 8. Choice A gives c instead of 2c. Choice C uses a² - b² = 16 as the distance. Choice D uses 2(a² - b²) incorrectly.

Question 2

An ellipse has foci at $F_1(0,-2)$ and $F_2(0,2)$ . The sum of distances from any point $P(x,y)$ on the ellipse to the foci is $10$ . Which equation represents the ellipse? (No vertices are given.)

$\dfrac{x^2}{25}+\dfrac{y^2}{21}=1$
$\dfrac{x^2}{21}+\dfrac{y^2}{25}=1$ (correct answer)
$\dfrac{x^2}{25}-\dfrac{y^2}{21}=1$
$\dfrac{x^2}{21}+\dfrac{(y-2)^2}{25}=1$

Explanation: This problem asks for a vertical ellipse equation from the focus definition. An ellipse consists of points P(x,y) where the sum of distances to two foci is constant. The foci F₁(0,-2) and F₂(0,2) are vertical with center at (0,0), and c = 2. The condition PF₁ + PF₂ = 10 gives 2a = 10, so a = 5. Using a² = b² + c² for ellipses, we get 25 = b² + 4, so b² = 21. Since the foci are vertical, the major axis is vertical, making the equation x²/b² + y²/a² = 1, which gives x²/21 + y²/25 = 1. A common error is always putting a² under x² regardless of orientation - remember that a² goes with the direction of the major axis (foci direction). To derive equations correctly, identify the orientation from the foci first.

Question 3

A parabola opening rightward has vertex at $(-2, 4)$ and focus at $(1, 4)$ . What is the equation of this parabola?

$(x + 2)^2 = 12(y - 4)$
$(y - 4)^2 = 6(x + 2)$
$(y - 4)^2 = 12(x + 2)$ (correct answer)
$(y - 4)^2 = 3(x + 2)$

Explanation: When you encounter a parabola problem with vertex and focus information, you need to identify the parabola's orientation and use the appropriate standard form. Since this parabola opens rightward (horizontal orientation), you'll use the form

(y - k)^2 = 4p(x - h)

where

(h,k)

is the vertex and

p

is the distance from vertex to focus. The vertex is at

(-2, 4)

, so

h = -2

and

k = 4

. The focus is at

(1, 4)

. Since both points have the same y-coordinate and the focus is to the right of the vertex, this confirms rightward opening. The distance

p

from vertex to focus is

1 - (-2) = 3

. Substituting into the standard form:

(y - 4)^2 = 4(3)(x - (-2))

, which simplifies to

(y - 4)^2 = 12(x + 2)

. Option A uses the wrong standard form

(x + 2)^2 = 12(y - 4)

, which describes a parabola opening upward, not rightward. Option B has

(y - 4)^2 = 6(x + 2)

, using the correct form but with

4p = 6

instead of

4p = 12

. This represents a parabola with

p = 1.5

, placing the focus at

(0.5, 4)

instead of

(1, 4)

. Option D also uses the correct orientation but has

4p = 3

, giving

p = 0.75

and placing the focus at

(-1.25, 4)

. Remember: for horizontal parabolas, use

(y - k)^2 = 4p(x - h)

, and always calculate

p

as the actual distance between vertex and focus to get

4p

correct.

Question 4

A parabola has focus at $(3, -2)$ and vertex at $(3, 1)$ . Which point lies on the directrix of this parabola?

$(3, -5)$
$(3, 1)$
$(3, -2)$
$(0, 4)$ (correct answer)

Explanation: When you encounter a parabola problem involving focus and vertex, remember that the directrix is always positioned so that it's equidistant from the vertex as the focus, but on the opposite side. First, let's find the directrix. The vertex is at

(3, 1)

and the focus is at

(3, -2)

. Since both points have the same x-coordinate, this is a vertical parabola. The distance from vertex to focus is

|1 - (-2)| = 3

units downward from the vertex. The directrix must be 3 units upward from the vertex, placing it at

y = 1 + 3 = 4

. So the directrix is the horizontal line

y = 4

. Now let's check which point lies on this line. Point D

(0, 4)

has a y-coordinate of 4, so it lies on the directrix

y = 4

. Why the other answers are wrong: Point A

(3, -5)

lies on the vertical line through the vertex and focus, but below the focus. Point B

(3, 1)

is the vertex itself, not on the directrix. Point C

(3, -2)

is the focus, which by definition cannot be on the directrix. Study tip: For any parabola, the vertex is always the midpoint between the focus and directrix. Once you know the focus and vertex, you can find the directrix by reflecting the focus across the vertex. Remember that the directrix is always a line perpendicular to the axis of symmetry.

Question 5

A hyperbola has foci $F_1(0,-10)$ and $F_2(0,10)$ , and the distance condition is $|PF_2-PF_1|=12$ for any point $P(x,y)$ on the hyperbola. Which equation represents the hyperbola? (No asymptotes are given.)

$\dfrac{y^2}{36}-\dfrac{x^2}{64}=1$ (correct answer)
$\dfrac{x^2}{36}-\dfrac{y^2}{64}=1$
$\dfrac{y^2}{64}-\dfrac{x^2}{36}=1$
$\dfrac{y^2}{36}+\dfrac{x^2}{64}=1$

Explanation: This problem asks for a vertical hyperbola equation from the focus definition. A hyperbola consists of points P(x,y) where the absolute difference of distances to foci is constant. The foci F₁(0,-10) and F₂(0,10) are vertical with center at (0,0), and c = 10. The condition |PF₂ - PF₁| = 12 gives 2a = 12, so a = 6. Using c² = a² + b² for hyperbolas, we get 100 = 36 + b², so b² = 64. Since the foci are vertical, the transverse axis is vertical, making the equation y²/a² - x²/b² = 1, which gives y²/36 - x²/64 = 1. A common error is using the wrong orientation - vertical foci mean the y² term is positive and comes first. To derive hyperbola equations systematically, always identify the transverse axis direction from the foci arrangement.

Question 6

A hyperbola is the set of all points $P(x,y)$ such that the absolute value of the difference of distances from $P$ to two foci is constant. The foci are $F_1(0,-5)$ and $F_2(0,5)$ , and for every point on the hyperbola, $|PF_2-PF_1|=6$ . Which equation represents the hyperbola?

$\dfrac{y^2}{9}-\dfrac{x^2}{16}=1$ (correct answer)
$\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$
$\dfrac{y^2}{25}+\dfrac{x^2}{16}=1$
$\dfrac{(y-5)^2}{9}-\dfrac{x^2}{16}=1$

Explanation: This problem requires deriving a hyperbola equation from its geometric definition. A hyperbola is the set of all points where the absolute difference of distances to two foci is constant. The foci F₁(0,-5) and F₂(0,5) lie on the y-axis with center at origin and c=5, and |PF₂-PF₁|=6 means 2a=6, so a=3. For a hyperbola, c²=a²+b², giving 25=9+b², so b²=16. Since the foci are on the y-axis, this is a vertical hyperbola with equation y²/a²-x²/b²=1, yielding y²/9-x²/16=1. A common mistake is using the horizontal form x²/a²-y²/b²=1, but the foci location determines orientation. Always start by identifying whether foci align vertically or horizontally to choose the correct standard form.

Question 7

An ellipse is defined as the set of points $P(x,y)$ such that $PF_1+PF_2$ is constant. The foci are $F_1(0,-4)$ and $F_2(0,4)$ , and $PF_1+PF_2=12$ . Which equation represents the ellipse?

$\dfrac{y^2}{36}+\dfrac{x^2}{20}=1$ (correct answer)
$\dfrac{x^2}{36}+\dfrac{y^2}{20}=1$
$\dfrac{y^2}{16}+\dfrac{x^2}{20}=1$
$\dfrac{y^2}{36}-\dfrac{x^2}{20}=1$

Explanation: To derive this ellipse equation, we apply the definition: points where the sum of distances to two foci equals a constant. The foci F₁(0,-4) and F₂(0,4) lie on the y-axis with center at origin, and c=4. Given PF₁+PF₂=12, we have 2a=12, so a=6. Using the ellipse relationship c²=a²-b², we get 16=36-b², so b²=20. Since foci are vertical, this is a vertical ellipse with form y²/a²+x²/b²=1, yielding y²/36+x²/20=1. Students often write x²/36+y²/20=1 by habit, but the larger denominator (a²=36) must go under the variable matching the foci direction (y here). To verify, check that any point satisfying the equation maintains the sum of distances equal to 12.

Question 8

An ellipse is defined as the set of all points $P(x,y)$ such that the sum of its distances to the foci $F_1(-3,0)$ and $F_2(3,0)$ is constant. In this case, $PF_1+PF_2=10$ . Which equation represents the ellipse? (No directrix is given.)

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

Explanation: This problem asks us to derive the equation of an ellipse from its focus definition. An ellipse is the set of all points P(x,y) where the sum of distances to two fixed points (foci) is constant. The foci are F₁(-3,0) and F₂(3,0), centered at the origin with c = 3. The distance condition PF₁ + PF₂ = 10 means 2a = 10, so a = 5. For an ellipse, b² = a² - c² = 25 - 9 = 16, giving the equation x²/25 + y²/16 = 1. A common error is reversing a² and b², placing 16 under x² instead of y². To derive equations correctly, always start from the focus definition and identify which axis contains the foci.

Question 9

A hyperbola is defined as the set of all points $P(x,y)$ such that the absolute difference of its distances to the foci $F_1(0,-5)$ and $F_2(0,5)$ is constant. Here, $|PF_2-PF_1|=6$ . Which equation represents the hyperbola? (No asymptotes are given.)

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

Explanation: This problem requires deriving a hyperbola equation from its focus definition. A hyperbola is the set of all points P(x,y) where the absolute difference of distances to two foci is constant. The foci F₁(0,-5) and F₂(0,5) lie on the y-axis, making this a vertical hyperbola centered at the origin with c = 5. The condition |PF₂ - PF₁| = 6 means 2a = 6, so a = 3. For a hyperbola, c² = a² + b², so b² = c² - a² = 25 - 9 = 16, giving y²/9 - x²/16 = 1. A common mistake is using the ellipse relationship b² = a² - c² instead of the hyperbola relationship. Always verify that c > a for hyperbolas and use the correct formula.

Question 10

A hyperbola is defined as the set of all points $P(x,y)$ such that the absolute difference of distances to the foci $F_1(-2,0)$ and $F_2(6,0)$ is constant. Here, $|PF_2-PF_1|=2$ . Which equation follows from this focus definition? (No asymptotes are given.)

$\dfrac{(x-2)^2}{1}-\dfrac{y^2}{15}=1$ (correct answer)
$\dfrac{(x-2)^2}{4}-\dfrac{y^2}{12}=1$
$\dfrac{(x-2)^2}{1}+\dfrac{y^2}{15}=1$
$\dfrac{y^2}{1}-\dfrac{(x-2)^2}{15}=1$

Explanation: This problem requires deriving a hyperbola equation from its focus definition. A hyperbola consists of all points P(x,y) where |PF₂ - PF₁| is constant. The foci F₁(-2,0) and F₂(6,0) lie on the x-axis, creating a horizontal hyperbola centered at ((-2+6)/2, 0) = (2,0). Given |PF₂ - PF₁| = 2, we have 2a = 2, so a = 1 and a² = 1. The focal distance is 2c = 8, so c = 4, and b² = c² - a² = 16 - 1 = 15, giving (x-2)²/1 - y²/15 = 1. Students might incorrectly use b² = 12 by confusing formulas; always use b² = c² - a² for hyperbolas. Start from the geometric definition to avoid formula confusion.

Question 11

A hyperbola has foci at $F_1(0,-6)$ and $F_2(0,6)$ . For any point $P(x,y)$ on the hyperbola, $|PF_2-PF_1|=8$ . Which equation represents the hyperbola? (No asymptotes are given.)

$\dfrac{y^2}{16}-\dfrac{x^2}{20}=1$ (correct answer)
$\dfrac{x^2}{16}-\dfrac{y^2}{20}=1$
$\dfrac{y^2}{20}-\dfrac{x^2}{16}=1$
$\dfrac{(y-6)^2}{16}-\dfrac{x^2}{20}=1$

Explanation: This problem asks for a vertical hyperbola equation from the focus definition. A hyperbola consists of points P(x,y) where the absolute difference of distances to two foci is constant. The foci F₁(0,-6) and F₂(0,6) are vertical with center at (0,0), and c = 6. The condition |PF₂ - PF₁| = 8 means 2a = 8, so a = 4. For hyperbolas, c² = a² + b², giving 36 = 16 + b², so b² = 20. Since the foci are vertical, the transverse axis is vertical, and the equation is y²/a² - x²/b² = 1, yielding y²/16 - x²/20 = 1. A common mistake is using the horizontal form x²/a² - y²/b² = 1 when foci are vertical. To avoid errors, identify the direction of the transverse axis from the foci positions before writing the equation.

Question 12

An ellipse is defined as the set of all points $P(x,y)$ such that the sum of the distances from $P$ to the two foci is constant. The foci are $F_1(-2,0)$ and $F_2(2,0)$ , and the constant sum of distances is $10$ . Which equation represents the ellipse? (No directrix is given.)

$\dfrac{x^2}{25}+\dfrac{y^2}{21}=1$ (correct answer)
$\dfrac{x^2}{21}+\dfrac{y^2}{25}=1$
$\dfrac{x^2}{25}-\dfrac{y^2}{21}=1$
$\dfrac{(x-2)^2}{25}+\dfrac{y^2}{21}=1$

Explanation: This problem asks us to derive the equation of an ellipse from its geometric definition. An ellipse is the set of all points P(x,y) where the sum of distances to two fixed points (foci) is constant. The foci are F₁(-2,0) and F₂(2,0), and the center is at their midpoint (0,0). The distance condition PF₁ + PF₂ = 10 means 2a = 10, so a = 5. Since c = 2 (distance from center to focus) and a² = b² + c², we get 25 = b² + 4, so b² = 21. The standard form for a horizontal ellipse centered at origin is x²/a² + y²/b² = 1, giving us x²/25 + y²/21 = 1. A common error is confusing which denominator goes with which variable - remember that a corresponds to the direction of the foci. To derive equations systematically, always start with the distance condition and work through the algebra to eliminate radicals.

Question 13

A parabola has vertex at $(-1, 3)$ and passes through the point $(3, 7)$ . If the parabola opens upward, what is the equation of its directrix?

$y = 2$ (correct answer)
$y = 1$
$y = 4$
$y = 5$

Explanation: Since the parabola opens upward with vertex (-1,3), the equation has form (x+1)² = 4p(y-3). Using point (3,7): (3+1)² = 4p(7-3), so 16 = 16p, giving p = 1. The directrix is p units below the vertex, so it's at y = 3 - 1 = 2. Choice B places the directrix too far below the vertex. Choice C places it above the vertex. Choice D places it even further above the vertex.

Question 14

An ellipse has foci at $(-2, 1)$ and $(4, 1)$ and passes through the point $(1, 5)$ . What is the value of $a^2$ in the standard form equation of this ellipse?

18
25 (correct answer)
36
45

Explanation: The sum of distances from point (1,5) to both foci equals 2a. Distance from (1,5) to (-2,1) is √[(1-(-2))² + (5-1)²] = √(9+16) = 5. Distance from (1,5) to (4,1) is √[(1-4)² + (5-1)²] = √(9+16) = 5. Therefore 2a = 5 + 5 = 10, so a = 5 and a² = 25. Choice A gives a² = 18 (using incorrect calculation). Choice C uses a = 6 instead of a = 5. Choice D incorrectly uses the sum of squared distances instead of sum of distances.

Question 15

An ellipse has center at the origin and passes through points $(5, 0)$ and $(0, 3)$ . If the foci lie on the x-axis, what are the coordinates of the foci?

$(\pm 2, 0)$
$(\pm \sqrt{16}, 0)$
$(\pm \sqrt{34}, 0)$
$(\pm 4, 0)$ (correct answer)

Explanation: When you encounter an ellipse problem with given points and foci locations, you need to identify the semi-major axis

a

, semi-minor axis

b

, and then use the relationship

c^2 = a^2 - b^2

to find the focal distance

c

. Since the ellipse is centered at the origin with foci on the x-axis, it has the standard form

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

, where

a > b

. The point

(5, 0)

tells us the ellipse extends 5 units along the x-axis, so

a = 5

. The point

(0, 3)

tells us it extends 3 units along the y-axis, so

b = 3

. Now you can find the focal distance:

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

, so

c = 4

. The foci are located at

(\pm c, 0) = (\pm 4, 0)

, which is answer D. Looking at the wrong answers: A gives

(\pm 2, 0)

, which would result from incorrectly calculating

c^2 = 9 - 25 = -16

and taking

c = 2

. B shows

(\pm \sqrt{16}, 0)

, which equals

(\pm 4, 0)

but leaves the square root unsimplified—this might tempt you if you forgot to simplify

\sqrt{16} = 4

. C gives

(\pm \sqrt{34}, 0)

, which comes from incorrectly adding

a^2 + b^2 = 25 + 9 = 34

instead of subtracting. Remember: for ellipses, always verify that

a > b

when foci lie on the x-axis, and use

c^2 = a^2 - b^2

, not addition.

Question 16

An ellipse has foci at $F_1(-3, 0)$ and $F_2(3, 0)$ . If the point $(0, 4)$ lies on the ellipse, what is the equation of the ellipse in standard form?

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

Explanation: For an ellipse, the sum of distances from any point to the two foci is constant. Using point (0,4): distance from (0,4) to (-3,0) is √(9+16) = 5, and distance from (0,4) to (3,0) is √(9+16) = 5. So 2a = 10, thus a = 5. Since c = 3 (distance from center to focus), we have b² = a² - c² = 25 - 9 = 16, so b = 4. The equation is x²/25 + y²/16 = 1. Choice B swaps a and b values. Choice C uses b² = 9 instead of 16. Choice D incorrectly uses a = 3 and b = 5.

Question 17

An ellipse has foci at $F_1(-1,4)$ and $F_2(7,4)$ . For any point $P(x,y)$ on the ellipse, $PF_1+PF_2=18$ . Which equation follows from the focus definition? (No directrix is given.)

$\dfrac{(x-3)^2}{81}+\dfrac{(y-4)^2}{65}=1$ (correct answer)
$\dfrac{(x-3)^2}{65}+\dfrac{(y-4)^2}{81}=1$
$\dfrac{(x-3)^2}{81}-\dfrac{(y-4)^2}{65}=1$
$\dfrac{(x+1)^2}{81}+\dfrac{(y-4)^2}{65}=1$

Explanation: This problem involves deriving an ellipse equation with both horizontal and vertical shifts. An ellipse is defined by points P(x,y) where PF₁ + PF₂ equals a constant. The foci F₁(-1,4) and F₂(7,4) are horizontal with center at (3,4), and the distance between foci is 8, so c = 4. The condition PF₁ + PF₂ = 18 gives 2a = 18, so a = 9. Using a² = b² + c² for ellipses, we get 81 = b² + 16, so b² = 65. For a horizontal ellipse centered at (h,k), the equation is (x-h)²/a² + (y-k)²/b² = 1, yielding (x-3)²/81 + (y-4)²/65 = 1. A common mistake is incorrectly calculating the center - it must be the midpoint of the foci. Always verify your center coordinates before writing the shifted equation form.

Question 18

An ellipse has foci at $F_1(-1, 2)$ and $F_2(5, 2)$ . If point $Q(2, 6)$ lies on the ellipse, what is the equation of the ellipse in standard form centered at its center?

$\frac{(x-2)^2}{25} + \frac{(y-2)^2}{9} = 1$
$\frac{(x-2)^2}{16} + \frac{(y-2)^2}{25} = 1$
$\frac{(x-2)^2}{25} + \frac{(y-2)^2}{16} = 1$ (correct answer)
$\frac{(x-2)^2}{100} + \frac{(y-2)^2}{91} = 1$

Explanation: When you encounter an ellipse problem with given foci and a point on the ellipse, you need to use the fundamental definition: an ellipse is the set of all points where the sum of distances to two foci is constant. First, find the center by averaging the foci coordinates:

(\frac{-1+5}{2}, \frac{2+2}{2}) = (2, 2)

. Since both foci have the same y-coordinate, this is a horizontal ellipse. The distance between foci is

|5-(-1)| = 6

, so

2c = 6

and

c = 3

. Now calculate the sum of distances from point

Q(2, 6)

to both foci:

Distance to $F_1(-1, 2)$ : $\sqrt{(2-(-1))^2 + (6-2)^2} = \sqrt{9+16} = 5$
Distance to $F_2(5, 2)$ : $\sqrt{(2-5)^2 + (6-2)^2} = \sqrt{9+16} = 5$

The sum is

2a = 10

, so

a = 5

. Using the relationship

c^2 = a^2 - b^2

9 = 25 - b^2

, which gives

b^2 = 16

and

b = 4

. The standard form is

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

, which is answer C. Answer A incorrectly swaps the denominators, placing 9 under the y-term instead of 16. Answer B mistakenly puts the larger denominator (25) under the y-term, suggesting a vertical ellipse when it should be horizontal. Answer D uses incorrect values (100 and 91) that don't follow from the given information. Remember: always verify your ellipse orientation by checking which foci coordinate varies—horizontal variation means horizontal ellipse with

a^2

under the x-term.

Question 19

A hyperbola has foci $F_1(-6,3)$ and $F_2(2,3)$ . For every point $P$ on the hyperbola, $|PF_2-PF_1|=6$ . Which equation represents the hyperbola?

$\dfrac{(x+2)^2}{9}-\dfrac{(y-3)^2}{7}=1$ (correct answer)
$\dfrac{(x+2)^2}{7}-\dfrac{(y-3)^2}{9}=1$
$\dfrac{(y-3)^2}{9}-\dfrac{(x+2)^2}{7}=1$
$\dfrac{(x-2)^2}{9}-\dfrac{(y-3)^2}{7}=1$

Explanation: This hyperbola derivation begins with the geometric definition: points where the absolute difference of distances to foci is constant. The foci F₁(-6,3) and F₂(2,3) share y-coordinate 3, indicating a horizontal hyperbola centered at ((-6+2)/2,3)=(-2,3). The focal distance is 2c=8, so c=4, and |PF₂-PF₁|=6 gives 2a=6, so a=3. Using c²=a²+b² for hyperbolas, we get 16=9+b², so b²=7. The horizontal hyperbola form is (x-h)²/a²-(y-k)²/b²=1, yielding (x+2)²/9-(y-3)²/7=1. A common mistake is using the vertical form or incorrect center, but foci alignment determines orientation. Always verify that the center-to-focus distance equals c and that c>a for valid hyperbola geometry.

Question 20

An ellipse has equation $\frac{x^2}{16} + \frac{y^2}{25} = 1$ . A point $P$ is on this ellipse such that the sum of its distances to the foci is 12. What is the length of the major axis?

8
10 (correct answer)
12
20

Explanation: For any ellipse, the sum of distances from any point on the ellipse to the two foci equals 2a, where a is the semi-major axis length. From the equation, a² = 25 and b² = 16, so a = 5. The information that the sum of distances is 12 is extraneous since this sum is always 2a = 2(5) = 10 for any point on the ellipse. The major axis length is 2a = 10. Choice A gives the semi-major axis length instead of major axis length. Choice C uses the given sum of distances. Choice D uses 2(a+b) incorrectly.