What is the standard form equation of the parabola with a vertex at (0,0) and directrix at x= -2?

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Answer 1 · 2017-05-30T16:54:04

$x = 1/8y^2$

Please observe that the directrix is a vertical line, therefore, the vertex form is of the equation is:

$x = a(y-k)^2+h" [1]"$

where $(h,k)$ is the vertex and the equation of the directrix is $x = k - 1/(4a)" [2]"$ .

Substitute the vertex, $(0,0)$ , into equation [1]:

$x = a(y-0)^2 + 0$

Simplify:

$x = ay^2" [3]"$

Solve equation [2] for "a" given that $k = 0$ and $x = -2$ :

$-2 = 0 - 1/(4a)$

$4a = 1/2$

$a = 1/8$

Substitute for "a" into equation [3]:

$x = 1/8y^2 larr$ answer

Here is a graph of the parabola with the vertex and the directrix:

![Desmos.com]( useruploads.socratic.org )