What is the equation in standard form of the parabola with a focus at (13,0) and a directrix of x= -5?

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What is the equation in standard form of the parabola with a focus at (13,0) and a directrix of x= -5?

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Answer 1 · 2016-04-28T15:07:39

${(y - 0)}^{2} = 36 (x - 4)$ $(y-0)^2=36(x-4)" "$ Vertex Form
or $y^{2} = 36 (x - 4)$ $y^2=36(x-4)$

Explanation:

With the given point $(13, 0)$ $(13, 0)$ and directrix $x = - 5$ $x=-5$ , we can calculate the $p$ $p$ in the equation of the parabola which opens to the right. We know that it opens to the right because of the position of the focus and directrix.

${(y - k)}^{2} = 4 p (x - h)$ $(y-k)^2=4p(x-h)$

From $- 5$ $-5$ to $+ 13$ $+13$ , that is 18 units, and that means the vertex is at $(4, 0)$ $(4, 0)$ . With $p = 9$ $p=9$ which is 1/2 the distance from focus to directrix.

The equation is

${(y - 0)}^{2} = 36 (x - 4)$ $(y-0)^2=36(x-4)" "$ Vertex Form
or $y^{2} = 36 (x - 4)$ $y^2=36(x-4)$

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What is the equation in standard form of the parabola with a focus at (13,0) and a directrix of x= -5?

1 Answer

Explanation:

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