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Geometry Help: Derive The Equation Of A Parabola

Review real example questions for Derive The Equation Of A Parabola in Geometry.

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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?

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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.