directrix of the parabola

[rachael724]

The equation of the directrix of the parabola 2y = x^2 is

Can someone help me.

I dont think this is the right answer, but would it be y = -1/2

if not can you tell me what I did wrong?

 

[soroban]

Hello, rachael724!

The equation of the directrix of the parabola $\displaystyle 2y\,=\,x^2$ is __.

Would it be $\displaystyle y\,=\,-\frac{1}{2}$? . . . . yes!

The general form is: .$\displaystyle x^2\,=\,4py$, where $\displaystyle p$ is the "focal distance"
. . (the directed distance from the vertex to the focus).

You have: .$\displaystyle x^2\,=\,2y\;\;\Rightarrow\;\;4p\,=\,2\;\;\Rightarrow\;\;p\,=\,\frac{1}{2}$

The vertex is at (0,0) and we go <u>up</u> distance $\displaystyle p$ for the focus: $\displaystyle (0,\frac{1}{2})$.

The directrix is always in the opposite direction.
. . Hence: .$\displaystyle y\,=\,-\frac{1}{2}$.