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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 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?

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Question 1

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 2

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

  1. (x+2)2=−2(y+2)(x+2)^2=-2(y+2) (correct answer)
  2. (x+2)2=2(y+2)(x+2)^2=2(y+2)
  3. (x−2)2=−2(y+2)(x-2)^2=-2(y+2)
  4. (x+2)2=−2(y−2)(x+2)^2=-2(y-2)

Explanation: To derive this parabola's equation, we use the focus-directrix definition. A parabola consists of all points P(x,y) equidistant from focus F(-2,-3) and directrix y=-1. The distance from P to F is √((x+2)²+(y+3)²), while the distance from P to the horizontal line y=-1 is |y-(-1)| = |y+1|. Setting these equal: √((x+2)²+(y+3)²) = |y+1|. Since the focus (y=-3) is below the directrix (y=-1), the parabola opens downward, and y < -1 for points on the parabola, so |y+1| = -1-y. Squaring both sides: (x+2)²+(y+3)² = (-1-y)², which simplifies to (x+2)² = -2(y+2). Students often forget to determine the parabola's orientation before removing absolute values.

Question 3

On the coordinate plane, the focus is F(−3,1)F(-3,1) and the directrix is the vertical line x=1x=1. The parabola opens to the left.

Which equation follows from the focus-directrix definition?

  1. (y−1)2=−8(x+1)(y-1)^2=-8(x+1) (correct answer)
  2. (y−1)2=8(x+1)(y-1)^2=8(x+1)
  3. (y+1)2=−8(x+1)(y+1)^2=-8(x+1)
  4. (y−1)2=−8(x−1)(y-1)^2=-8(x-1)

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 (-3,1) and the directrix x = 1. 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-1)²) = |x - 1|. Setting up this equality and squaring both sides eliminates the square root, leading to (x+3)² + (y-1)² = (x-1)², which simplifies through expansion and cancellation to (y-1)² = -8(x+1). This final form is justified as it places the vertex at (-1,1), midway between the focus and directrix, with the coefficient -8 corresponding to 4p where p=-2 for leftward opening. A common distractor misconception is using a positive coefficient, implying incorrect opening direction. To transfer this strategy, always start from the distance definition and carefully simplify when deriving parabola equations.

Question 4

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.

Question 5

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

  1. (x+2)2=−2(y+2)(x+2)^2=-2(y+2) (correct answer)
  2. (x+2)2=2(y+2)(x+2)^2=2(y+2)
  3. (x−2)2=−2(y+2)(x-2)^2=-2(y+2)
  4. (x+2)2=−2(y−2)(x+2)^2=-2(y-2)

Explanation: To derive this parabola's equation, we use the focus-directrix definition. A parabola consists of all points P(x,y) equidistant from focus F(-2,-3) and directrix y=-1. The distance from P to F is √((x+2)²+(y+3)²), while the distance from P to the horizontal line y=-1 is |y-(-1)| = |y+1|. Setting these equal: √((x+2)²+(y+3)²) = |y+1|. Since the focus (y=-3) is below the directrix (y=-1), the parabola opens downward, and y < -1 for points on the parabola, so |y+1| = -1-y. Squaring both sides: (x+2)²+(y+3)² = (-1-y)², which simplifies to (x+2)² = -2(y+2). Students often forget to determine the parabola's orientation before removing absolute values.

Question 6

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 7

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 8

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 9

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 10

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.