Question

what's the directrix in this:
https://media.glynlyon.com/a_matalg02_2011/7/26.gif
&
https://media.glynlyon.com/a_matalg02_2011/7/24.gif

Answer 1

So in this question, I guess y would be 'x' in normal problems.

Answer 2

yeah.

Answer 3

And we know that the standard form of a parabola is \(y = a(x-h)^2+k\)
Let's plug our things in
\(y = -\frac{1}{8}x^2\)
Move away the -1/8
\(y \div -\frac{1}{8}= x^2\)
\(-8y = x^2\)

Factor out 4 out of -8
\(4(-2)y = x^2\)

So our focus is at (0, -2) h is 0 so 0.

Answer 4

but you would switch it, so it would be -2, 0
and the directrix would be 2?

Answer 5

Found your focus!

We know that the directrix is at the same distance as the distance to the focus.

Focus (0, -2)

We need our vertex, which is (0,0)