Geometry · Learn by Concept

Geometry Help: Derive Equations Of Ellipses And Parabolas

Review real example questions for Derive Equations Of Ellipses And Parabolas in Geometry.

Question 1 / 10

0 of 10 answered

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?

All questions

Question 1

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 2

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 3

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 4

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 5

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 6

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 7

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 8

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 9

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 10

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.