What is the equation in standard form of the parabola with a focus at (-14,-1) and a directrix of y= -32?

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What is the equation in standard form of the parabola with a focus at (-14,-1) and a directrix of y= -32?

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Answer 1 · 2015-12-30T14:30:32

$62 y = x^{2} + 28 x - 860$ $62y = x^2 +28x - 860$

Explanation:

The equation of the parabola can be written as $4 p (y - k) = {(x - h)}^{2}$ $4p(y- k)= (x-h)^2$ from which the vertex is $(h, k)$ $(h,k)$ and $p$ $p$ is the distance from the vertex to the focus (or the directrix since the directrix is an equal distance the other side of the vertex).

The focus and directrix are $(- 1 - (- 32))$ $(-1 - (-32))$ units apart and p is therefore $\frac{1}{2} (31) = \frac{31}{2}$ $1/2(31) = 31/2$
The vertex is at $(- 14, - 1 - \frac{31}{2}) = (- 14, - \frac{33}{2})$ $(-14,-1-31/2) = (-14,-33/2)$
The equation is therefore $4 \cdot \frac{31}{2} (y + \frac{33}{2}) = {(x + 14)}^{2}$ $4*31/2 (y+33/2) = (x+14)^2$
$62 y + 32 \cdot 33 = x^{2} + 28 x + 14^{2}$ $62y +32*33 = x^2 +28x +14^2$
$62 y = x^{2} + 28 x + 196 - 1056$ $62y = x^2 +28x +196 - 1056$
$62 y = x^{2} + 28 x - 860$ $62y = x^2 +28x - 860$

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What is the equation in standard form of the parabola with a focus at (-14,-1) and a directrix of y= -32?

1 Answer

Explanation:

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