QuestionApril 19, 2025

Derive the equation of a parabola with a focus at (6,-2) and a directrix at y=0 (1 point) y=(1)/(4)x^2+3x+8 y=(1)/(4)x^2+x+4 y=-(1)/(4)x^2+3x-10 y=-(1)/(4)x^2+2x-6

Derive the equation of a parabola with a focus at (6,-2) and a directrix at y=0 (1 point) y=(1)/(4)x^2+3x+8 y=(1)/(4)x^2+x+4 y=-(1)/(4)x^2+3x-10 y=-(1)/(4)x^2+2x-6
Derive the equation of a parabola with a focus at (6,-2) and a directrix at y=0 (1 point) y=(1)/(4)x^2+3x+8 y=(1)/(4)x^2+x+4 y=-(1)/(4)x^2+3x-10 y=-(1)/(4)x^2+2x-6

Solution
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Answer

y = \frac{1}{4}x^2 - 3x + 8 Explanation 1. Identify the vertex The vertex is midway between the focus (6, -2) and the directrix y = 0. Midpoint is (6, -1). 2. Determine the orientation Since the directrix is horizontal (y=0), the parabola opens vertically. 3. Calculate distance from vertex to focus Distance p = -1 - (-2) = 1. 4. Write the standard form of the parabola For a vertical parabola with vertex (h, k): y = \frac{1}{4p}(x-h)^2 + k. 5. Substitute values into the formula y = \frac{1}{4 \times 1}(x-6)^2 - 1 = \frac{1}{4}(x-6)^2 - 1. 6. Expand and simplify y = \frac{1}{4}(x^2 - 12x + 36) - 1 = \frac{1}{4}x^2 - 3x + 9 - 1 = \frac{1}{4}x^2 - 3x + 8.

Explanation

1. Identify the vertex<br /> The vertex is midway between the focus $(6, -2)$ and the directrix $y = 0$. Midpoint is $(6, -1)$.<br />2. Determine the orientation<br /> Since the directrix is horizontal ($y=0$), the parabola opens vertically.<br />3. Calculate distance from vertex to focus<br /> Distance $p = -1 - (-2) = 1$.<br />4. Write the standard form of the parabola<br /> For a vertical parabola with vertex $(h, k)$: $y = \frac{1}{4p}(x-h)^2 + k$.<br />5. Substitute values into the formula<br /> $y = \frac{1}{4 \times 1}(x-6)^2 - 1 = \frac{1}{4}(x-6)^2 - 1$.<br />6. Expand and simplify<br /> $y = \frac{1}{4}(x^2 - 12x + 36) - 1 = \frac{1}{4}x^2 - 3x + 9 - 1 = \frac{1}{4}x^2 - 3x + 8$.
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