How do you find the equation of the parabola whose focus is at (3, 3) and whose directrix is at x = 7?

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Answer 1 · 2018-03-13T23:06:37

$-8(x-5)=(y-3)^2$

A quadratic relation (or function) can be in the form:

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

Since our diretrix is a vertical line, our parabola is horizontal.

We, therefore, use the form $4p(x-h)=(y-k)^2$

In this form, the focus is at $(h+p,k)$ and the diretrix is at $x=h-p$

Using our information, the focus is at $(3,3)$ and the diretrix is at $x=7$ .

Therefore, $k=3$ , $h+p=3$ , and $h-p=7$

We add the second and the third equations to get:

$2h=10$

$=>h=5$ You could use this to find that $p=-2$

Let's plug in what we know:

$4*-2(x-5)=(y-3)^2$

$=>-8(x-5)=(y-3)^2$