Geometry Quiz: Derive The Equation Of A Parabola
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Derive The Equation Of A ParabolaQuestion 1 of 20

A satellite dish is designed so that its cross-section forms a parabola. The receiver is placed at the focus, which is 6 inches from the vertex. If the vertex is at the origin and the dish opens upward, what is the equation of the parabolic cross-section?

x2=6yx^2 = 6y
x2=12yx^2 = 12y
x2=24yx^2 = 24y
y2=24xy^2 = 24x
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Geometry Quiz: Derive The Equation Of A Parabola

Practice Derive The Equation Of A Parabola 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 The Equation Of A Parabola, giving you a quick way to practice the rules, question types, and explanations that matter most for Geometry.

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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.

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Question 1

A satellite dish is designed so that its cross-section forms a parabola. The receiver is placed at the focus, which is 6 inches from the vertex. If the vertex is at the origin and the dish opens upward, what is the equation of the parabolic cross-section?

  1. x2=6yx^2 = 6y
  2. x2=12yx^2 = 12y
  3. x2=24yx^2 = 24y (correct answer)
  4. y2=24xy^2 = 24x
Explanation: With vertex at origin (0, 0) and opening upward, the standard form is x2=4pyx^2 = 4py where p is the distance from vertex to focus. Given that the focus is 6 inches from the vertex, p=6p = 6. Therefore, the equation is x2=4(6)y=24yx^2 = 4(6)y = 24y. Choice A uses p directly instead of 4p. Choice B uses 4p/2 = 12 instead of 4p = 24. Choice D incorrectly assumes the parabola opens horizontally instead of vertically, and uses the wrong variable arrangement.

Question 2

On the coordinate plane, the focus is F(5,1)F(5,1) and the directrix is the vertical line x=1x=1 (shown). The parabola opens to the right. Which statement identifies the vertex?​

  1. The vertex is at (3,1)(3,1). (correct answer)
  2. The vertex is at (1,3)(1,3).
  3. The vertex is at (5,3)(5,3).
  4. The vertex is at (3,5)(3,5).
Explanation: The skill involves deriving the equation of a parabola given its focus and directrix. A parabola is defined as the set of all points equidistant from a fixed point called the focus and a fixed line called the directrix. For any point on the parabola, the distance to the focus F(5,1) equals the distance to the directrix x=1. The vertex is the midpoint between the focus and directrix, calculated as x=(5+1)/2=3 and y=1. This identifies the vertex at (3,1), consistent with the parabola opening to the right. A distractor misconception is miscalculating the midpoint, such as averaging y-coordinates incorrectly leading to (5,3). To derive equations for other parabolas, always start from the distance definition and simplify step by step.

Question 3

A parabola is defined as the set of points equidistant from the focus F(3,−2)F(3,-2) and the directrix y=2y=2. The parabola opens downward.

Which equation represents the parabola?

  1. (x−3)2=−8(y+2)(x-3)^2=-8(y+2)
  2. (x−3)2=−8y(x-3)^2=-8y (correct answer)
  3. (x+3)2=−8(y+2)(x+3)^2=-8(y+2)
  4. (x−3)2=8(y+2)(x-3)^2=8(y+2)
Explanation: The skill is deriving the equation of a parabola from its focus and directrix. A parabola is geometrically defined as the set of points equidistant from the focus at (3,-2) and the directrix y = 2. For any point (x,y) on the parabola, the distance to the focus equals the distance to the directrix, expressed as √((x-3)² + (y+2)²) = |y - 2|. Setting up this equality and squaring both sides eliminates the square root, leading to (x-3)² + (y+2)² = (y-2)², which simplifies through expansion and cancellation to (x-3)² = -8y. This final form is justified as it places the vertex at (3,0), midway between the focus and directrix, with the coefficient -8 corresponding to 4p where p=-2 for downward opening. A common distractor misconception is including an unnecessary y-shift like (y+2), altering the vertex position. To transfer this strategy, always start from the distance definition and carefully simplify when deriving parabola equations.

Question 4

On the coordinate plane, the focus is F(5,1)F(5,1) and the directrix is the vertical line x=1x=1 (shown). The parabola opens to the right. Which statement identifies the vertex?

  1. The vertex is at (3,1)(3,1). (correct answer)
  2. The vertex is at (1,3)(1,3).
  3. The vertex is at (5,3)(5,3).
  4. The vertex is at (3,5)(3,5).
Explanation: The skill involves deriving the equation of a parabola given its focus and directrix. A parabola is defined as the set of all points equidistant from a fixed point called the focus and a fixed line called the directrix. For any point on the parabola, the distance to the focus F(5,1) equals the distance to the directrix x=1. The vertex is the midpoint between the focus and directrix, calculated as x=(5+1)/2=3 and y=1. This identifies the vertex at (3,1), consistent with the parabola opening to the right. A distractor misconception is miscalculating the midpoint, such as averaging y-coordinates incorrectly leading to (5,3). To derive equations for other parabolas, always start from the distance definition and simplify step by step.

Question 5

A parabola is defined as the set of points equidistant from the focus F(0,1)F(0,1) and the directrix line y=−3y=-3. The parabola opens upward. Which expression represents all points (x,y)(x,y) equidistant from the focus and directrix?

  1. x2+(y−1)2=∣y+3∣\sqrt{x^2+(y-1)^2}=|y+3| (correct answer)
  2. x2+(y+3)2=∣y−1∣\sqrt{x^2+(y+3)^2}=|y-1|
  3. (x−1)2+y2=∣y+3∣\sqrt{(x-1)^2+y^2}=|y+3|
  4. x2+(y−1)2=∣x+3∣\sqrt{x^2+(y-1)^2}=|x+3|
Explanation: The skill here is deriving the equation of a parabola from its focus and directrix. A parabola is defined as the set of all points equidistant from a fixed point called the focus and a fixed line called the directrix. For any point (x,y) on this parabola, the distance to the focus F(0,1) equals the distance to the directrix y=-3. This equality is directly represented by the expression √(x² + (y-1)²) = |y+3|, without yet squaring. This form is justified as it captures the raw geometric definition before algebraic simplification to the parabola equation. A distractor like choice B swaps the focus and directrix distances, which would not satisfy the definition. To derive equations for other parabolas, always start by equating the distance to the focus and to the directrix, then square and simplify if needed.

Question 6

The focus of a parabola is F(−2,−3)F(-2,-3) and the directrix is the horizontal line y=1y=1 (shown on the coordinate plane). The parabola opens downward. Which expression represents all points (x,y)(x,y) equidistant from the focus and the directrix?

  1. (x+2)2+(y+3)2=∣y+1∣\sqrt{(x+2)^2+(y+3)^2}=|y+1|
  2. (x+2)2+(y+3)2=∣y−1∣\sqrt{(x+2)^2+(y+3)^2}=|y-1| (correct answer)
  3. (x−2)2+(y−3)2=∣y−1∣\sqrt{(x-2)^2+(y-3)^2}=|y-1|
  4. (x+2)2+(y−3)2=∣y−1∣\sqrt{(x+2)^2+(y-3)^2}=|y-1|
Explanation: The skill involves deriving the equation of a parabola given its focus and directrix. A parabola is defined as the set of all points equidistant from a fixed point called the focus and a fixed line called the directrix. For any point (x,y) on the parabola, the distance to the focus F(-2,-3) equals the distance to the directrix y=1. This equality is expressed as sqrt((x+2)^2 + (y+3)^2) = |y-1|. This form correctly captures the downward opening with the focus below the directrix. A distractor misconception is using |y+1| instead of |y-1|, confusing the directrix position. To derive equations for other parabolas, always start from the distance definition and simplify step by step.

Question 7

A parabola opens upward with its focus at (2,7)(2, 7) and vertex at (2,4)(2, 4). If the parabola passes through point (6,y)(6, y), what is the value of yy?

  1. y=7y = 7
  2. y=193y = \frac{19}{3}
  3. y=203y = \frac{20}{3} (correct answer)
  4. y=8y = 8
Explanation: First, find the parabola equation. The vertex is (2, 4) and focus is (2, 7), so p = 7 - 4 = 3. The equation is (x−2)2=4p(y−4)=12(y−4)(x - 2)^2 = 4p(y - 4) = 12(y - 4). For point (6, y): (6−2)2=12(y−4)(6 - 2)^2 = 12(y - 4), so 16=12(y−4)16 = 12(y - 4). Solving: 1612=y−4\frac{16}{12} = y - 4, so 43=y−4\frac{4}{3} = y - 4, giving y=4+43=12+43=203y = 4 + \frac{4}{3} = \frac{12 + 4}{3} = \frac{20}{3}. Choice A incorrectly assumes y equals the focus y-coordinate. Choice B results from arithmetic error: 1612+4=43+123=163\frac{16}{12} + 4 = \frac{4}{3} + \frac{12}{3} = \frac{16}{3} but written as 193\frac{19}{3}. Choice D uses incorrect formula or calculation.

Question 8

A parabola has focus F(−6,0)F(-6,0) and directrix x=−2x=-2. Which statement identifies the vertex?​

  1. The vertex is at (−2,0)(-2,0).
  2. The vertex is at (−4,0)(-4,0). (correct answer)
  3. The vertex is at (−6,0)(-6,0).
  4. The vertex is at (2,0)(2,0).
Explanation: To find the vertex of this parabola, we use the focus-directrix definition. A parabola consists of points equidistant from focus F(-6,0) and directrix x=-2. The vertex is the point on the parabola that lies on the axis of symmetry, exactly halfway between the focus and directrix. For a horizontal parabola, the vertex's x-coordinate is the average of the focus's x-coordinate and the directrix's x-value: (-6+(-2))/2 = -4. The vertex shares the same y-coordinate as the focus, so the vertex is at (-4,0). Students often miscalculate by subtracting instead of averaging, or they confuse which coordinate changes. Remember: for horizontal parabolas, average the x-values; for vertical parabolas, average the y-values.

Question 9

A parabola is defined by focus F(2,−1)F(2,-1) and directrix y=3y=3. Which equation represents the parabola?

  1. (x−2)2=−8(y−1)(x-2)^2=-8(y-1) (correct answer)
  2. (x−2)2=8(y−1)(x-2)^2=8(y-1)
  3. (x+2)2=−8(y−1)(x+2)^2=-8(y-1)
  4. (x−2)2=−8(y+1)(x-2)^2=-8(y+1)
Explanation: This problem requires deriving a parabola's equation from its focus-directrix definition. A parabola is the locus of points equidistant from focus F(2,-1) and directrix y=3. For any point P(x,y) on the parabola, we have: distance from P to F equals distance from P to directrix. This gives us √((x-2)²+(y+1)²) = |y-3|. Since the focus is below the directrix, the parabola opens downward, and y < 3 for points on the parabola, so |y-3| = 3-y. Squaring both sides: (x-2)²+(y+1)² = (3-y)², which simplifies to (x-2)² = -8(y-1). A common mistake is assuming the parabola always opens upward or rightward without checking the focus-directrix positions.

Question 10

The focus of a parabola is F(−2,−3)F(-2,-3) and the directrix is the horizontal line y=1y=1 (shown on the coordinate plane). The parabola opens downward. Which expression represents all points (x,y)(x,y) equidistant from the focus and the directrix?​

  1. (x+2)2+(y+3)2=∣y+1∣\sqrt{(x+2)^2+(y+3)^2}=|y+1|
  2. (x+2)2+(y+3)2=∣y−1∣\sqrt{(x+2)^2+(y+3)^2}=|y-1| (correct answer)
  3. (x−2)2+(y−3)2=∣y−1∣\sqrt{(x-2)^2+(y-3)^2}=|y-1|
  4. (x+2)2+(y−3)2=∣y−1∣\sqrt{(x+2)^2+(y-3)^2}=|y-1|
Explanation: The skill involves deriving the equation of a parabola given its focus and directrix. A parabola is defined as the set of all points equidistant from a fixed point called the focus and a fixed line called the directrix. For any point (x,y) on the parabola, the distance to the focus F(-2,-3) equals the distance to the directrix y=1. This equality is expressed as sqrt((x+2)^2 + (y+3)^2) = |y-1|. This form correctly captures the downward opening with the focus below the directrix. A distractor misconception is using |y+1| instead of |y-1|, confusing the directrix position. To derive equations for other parabolas, always start from the distance definition and simplify step by step.

Question 11

On the coordinate plane, the focus is the point F(2,3)F(2,3) and the directrix is the vertical line x=−2x=-2 (shown). The parabola opens to the right. Which equation follows from the focus–directrix definition?

  1. (y−3)2=−8(x−0)(y-3)^2=-8(x-0)
  2. (x−3)2=8(y−2)(x-3)^2=8(y-2)
  3. (y−3)2=8(x−0)(y-3)^2=8(x-0) (correct answer)
  4. (y−2)2=8(x−3)(y-2)^2=8(x-3)
Explanation: The skill involves deriving the equation of a parabola given its focus and directrix. A parabola is defined as the set of all points equidistant from a fixed point called the focus and a fixed line called the directrix. For any point (x,y) on the parabola, the distance to the focus F(2,3) equals the distance to the directrix x=-2. This equality leads to the equation sqrt((x-2)^2 + (y-3)^2) = |x + 2|, which simplifies after squaring to (y-3)^2 = 8x. The final form (y-3)^2 = 8(x-0) justifies the vertex at (0,3) and opening to the right with parameter 4p=8 based on the distance from vertex to focus being 2. A distractor misconception is swapping x and y terms, like in choice B, which assumes vertical opening instead of horizontal. To derive equations for other parabolas, always start from the distance definition and simplify step by step.

Question 12

A parabola is defined as the set of all points equidistant from a focus and a directrix. The focus is F(0,2)F(0,2) and the directrix is the line y=−2y=-2. Which equation follows from the focus–directrix definition?

  1. x2=8(y−2)x^2=8(y-2)
  2. x2=8yx^2=8y (correct answer)
  3. x2=−8yx^2=-8y
  4. x2=8(y+2)x^2=8(y+2)
Explanation: This problem asks us to derive the equation of a parabola from its focus-directrix definition. A parabola is the set of all points that are equidistant from a fixed point (the focus) and a fixed line (the directrix). For any point P(x,y) on the parabola, the distance from P to focus F(0,2) equals the distance from P to the directrix y=-2. Setting these distances equal: √(x²+(y-2)²) = |y-(-2)| = |y+2|. Since y > -2 for points above the directrix, we have |y+2| = y+2, so √(x²+(y-2)²) = y+2. Squaring both sides and simplifying: x²+(y-2)² = (y+2)², which gives x² = 8y. A common error is forgetting to square both sides properly or misidentifying which direction the parabola opens.

Question 13

A parabola is the set of points equidistant from the focus F(−2,0)F(-2,0) and the directrix x=2x=2. The parabola opens to the left.

Which expression represents all points equidistant from the focus and directrix?

  1. (x+2)2+y2=∣x−2∣\sqrt{(x+2)^2+y^2}=|x-2| (correct answer)
  2. (x−2)2+y2=∣x+2∣\sqrt{(x-2)^2+y^2}=|x+2|
  3. (x+2)2+y2=∣x−2∣(x+2)^2+y^2=|x-2|
  4. (x+2)2+y2=∣y−2∣\sqrt{(x+2)^2+y^2}=|y-2|
Explanation: The skill is deriving the equation of a parabola from its focus and directrix. A parabola is geometrically defined as the set of points equidistant from the focus at (-2,0) and the directrix x = 2. For any point (x,y) on the parabola, the distance to the focus equals the distance to the directrix, expressed directly as √((x+2)² + y²) = |x - 2| before squaring. This expression arises by applying the distance formulas without simplification, capturing the equality condition. The form is justified as it correctly uses absolute value for the directrix distance and square root for the point-to-focus distance. A common distractor misconception is omitting the square root or using incorrect coordinates like |y-2|. To transfer this strategy, always start from the distance definition and set up the unsimplified equality when deriving parabola equations.

Question 14

A parabola consists of all points P(x,y)P(x,y) such that the distance from PP to the focus equals the distance from PP to the directrix. The focus is F(−1,3)F(-1,3) and the directrix is the vertical line x=−5x=-5. Which equation follows from the focus–directrix definition?​

  1. (y−3)2=8(x+3)(y-3)^2=8(x+3) (correct answer)
  2. (y−3)2=4(x+3)(y-3)^2=4(x+3)
  3. (y−3)2=8(x−3)(y-3)^2=8(x-3)
  4. (y+3)2=8(x+3)(y+3)^2=8(x+3)
Explanation: To derive a parabola's equation, we use the focus-directrix definition. A parabola consists of all points P(x,y) equidistant from the focus F(-1,3) and the directrix x=-5. The distance from P to F is √((x+1)²+(y-3)²), and the distance from P to the vertical line x=-5 is |x-(-5)| = |x+5|. Setting these equal: √((x+1)²+(y-3)²) = |x+5|. Since the focus is to the right of the directrix, x > -5 for points on the parabola, so |x+5| = x+5. Squaring both sides: (x+1)²+(y-3)² = (x+5)², which simplifies to (y-3)² = 8(x+3). Students often confuse horizontal and vertical parabolas or miscalculate the coefficient 4p = 8.

Question 15

A parabola is defined as the set of all points equidistant from a focus and a directrix. The focus is F(0,2)F(0,2) and the directrix is the line y=−2y=-2. Which equation follows from the focus–directrix definition?​

  1. x2=8(y−2)x^2=8(y-2)
  2. x2=8yx^2=8y (correct answer)
  3. x2=−8yx^2=-8y
  4. x2=8(y+2)x^2=8(y+2)
Explanation: This problem asks us to derive the equation of a parabola from its focus-directrix definition. A parabola is the set of all points that are equidistant from a fixed point (the focus) and a fixed line (the directrix). For any point P(x,y) on the parabola, the distance from P to focus F(0,2) equals the distance from P to the directrix y=-2. Setting these distances equal: √(x²+(y-2)²) = |y-(-2)| = |y+2|. Since y > -2 for points above the directrix, we have |y+2| = y+2, so √(x²+(y-2)²) = y+2. Squaring both sides and simplifying: x²+(y-2)² = (y+2)², which gives x² = 8y. A common error is forgetting to square both sides properly or misidentifying which direction the parabola opens.

Question 16

A parabola consists of all points P(x,y)P(x,y) such that the distance from PP to the focus equals the distance from PP to the directrix. The focus is F(−1,3)F(-1,3) and the directrix is the vertical line x=−5x=-5. Which equation follows from the focus–directrix definition?

  1. (y−3)2=8(x+3)(y-3)^2=8(x+3) (correct answer)
  2. (y−3)2=4(x+3)(y-3)^2=4(x+3)
  3. (y−3)2=8(x−3)(y-3)^2=8(x-3)
  4. (y+3)2=8(x+3)(y+3)^2=8(x+3)
Explanation: To derive a parabola's equation, we use the focus-directrix definition. A parabola consists of all points P(x,y) equidistant from the focus F(-1,3) and the directrix x=-5. The distance from P to F is √((x+1)²+(y-3)²), and the distance from P to the vertical line x=-5 is |x-(-5)| = |x+5|. Setting these equal: √((x+1)²+(y-3)²) = |x+5|. Since the focus is to the right of the directrix, x > -5 for points on the parabola, so |x+5| = x+5. Squaring both sides: (x+1)²+(y-3)² = (x+5)², which simplifies to (y-3)² = 8(x+3). Students often confuse horizontal and vertical parabolas or miscalculate the coefficient 4p = 8.

Question 17

A parabola is the set of points equidistant from focus F(4,1)F(4,1) and directrix x=0x=0. Which equation follows from the focus–directrix definition?​

  1. (y−1)2=16(x−2)(y-1)^2=16(x-2) (correct answer)
  2. (y+1)2=16(x−2)(y+1)^2=16(x-2)
  3. (y−1)2=8(x−2)(y-1)^2=8(x-2)
  4. (y−1)2=16(x+2)(y-1)^2=16(x+2)
Explanation: To derive this parabola's equation, we apply the focus-directrix definition. A parabola contains all points P(x,y) equidistant from focus F(4,1) and directrix x=0. The distance from P to F is √((x-4)²+(y-1)²), while the distance from P to the vertical line x=0 is |x-0| = |x|. Setting these equal: √((x-4)²+(y-1)²) = |x|. Since the focus is at x=4 (right of the directrix), points on the parabola have x > 0, so |x| = x. Squaring both sides: (x-4)²+(y-1)² = x², which expands and simplifies to (y-1)² = 16(x-2). Students often forget that 4p represents the distance from vertex to focus, leading to incorrect coefficients.

Question 18

The focus of a parabola is at F(−3,1)F(-3, 1) and the directrix is y=−3y = -3. Which point lies exactly halfway between the focus and the directrix?

  1. (−3,−1)(-3, -1) (correct answer)
  2. (−3,2)(-3, 2)
  3. (−3,−2)(-3, -2)
  4. (0,−1)(0, -1)
Explanation: The vertex of a parabola is always exactly halfway between the focus and directrix. The focus is at (-3, 1) and the directrix is y = -3. The x-coordinate of the vertex is the same as the focus: x = -3. The y-coordinate is the midpoint between y = 1 (focus) and y = -3 (directrix): y=1+(−3)2=−22=−1y = \frac{1 + (-3)}{2} = \frac{-2}{2} = -1. Therefore, the vertex is at (-3, -1). Choice B incorrectly calculates the y-coordinate as 2 instead of -1. Choice C uses -2, which is one unit below the correct position. Choice D uses the correct y-coordinate but wrong x-coordinate (0 instead of -3).

Question 19

A parabola has the equation (x−4)2=−8(y+2)(x - 4)^2 = -8(y + 2). What is the distance between the focus and the directrix of this parabola?

  1. 2 units
  2. 4 units (correct answer)
  3. 6 units
  4. 8 units
Explanation: From the equation (x−4)2=−8(y+2)(x - 4)^2 = -8(y + 2), we can identify that this is a downward-opening parabola with vertex (4, -2). Comparing to standard form (x−h)2=4p(y−k)(x - h)^2 = 4p(y - k), we have 4p=−84p = -8, so p=−2p = -2. The absolute value ∣p∣=2|p| = 2 represents the distance from the vertex to the focus (and also from the vertex to the directrix). The total distance between focus and directrix is 2∣p∣=2(2)=42|p| = 2(2) = 4 units. Choice A gives only |p| instead of 2|p|. Choice C incorrectly calculates as 3|p|. Choice D mistakes the coefficient -8 as the distance.

Question 20

Which equation follows from the focus–directrix definition for the parabola with focus F(2,1)F(2,1) and directrix x=−2x=-2 (opening to the right)?

  1. (y−1)2=8(x−0)(y-1)^2=8(x-0) (correct answer)
  2. (x−1)2=8(y−0)(x-1)^2=8(y-0)
  3. (y−1)2=−8(x−0)(y-1)^2=-8(x-0)
  4. (y+1)2=8(x−0)(y+1)^2=8(x-0)
Explanation: This problem asks us to derive the equation of a parabola from its focus-directrix definition. A parabola is the set of all points that are equidistant from a fixed point (the focus) and a fixed line (the directrix). For any point (x,y) on the parabola, the distance to focus F(2,1) equals the distance to directrix x=-2. The distance to F(2,1) is √[(x-2)² + (y-1)²], and the distance to the vertical line x=-2 is |x-(-2)| = |x+2|. Setting these equal and squaring both sides gives (x-2)² + (y-1)² = (x+2)². Expanding and simplifying: x²-4x+4 + (y-1)² = x²+4x+4, which reduces to (y-1)² = 8x. A common error is confusing which variable gets squared based on the directrix orientation. Since we have a vertical directrix and the parabola opens horizontally (to the right), the y-term is squared.