What is the vertex form of the equation of the parabola with a focus at (6,-13) and a directrix of $y=13$ ?

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Answer 1 · 2017-09-12T22:18:18

$y=\frac{1}{-52}(x-6)^2+0$

Given the focus and directrix of a parabola, you can find the parabola's equation with the formula:

$y=\frac{1}{2(b-k)}(x-a)^2+\frac{1}{2}(b+k)$ ,

where:

$k$ is the directrix &

$(a,b)$ is the focus

Plugging in the values of those variables gives us:

$y=\frac{1}{2(-13-13)}(x-6)^2+\frac{1}{2}(-13+13)$

Simplifying gives us:

$y=\frac{1}{-52}(x-6)^2+0$

What is the vertex form of the equation of the parabola with a focus at (6,-13) and a directrix of y=13 y=13 ?