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: .\(\displaystyle y\,=\,-\frac{1}{2}\).