Question

Y^2=-8x+16 Find the focus?

Answer 1

`Focus (0,0)`

Explanation:

Standard form of verical positive parabola io

`(y-k)^2 = 4 a (x-h)`

`y^2 = 8x + 16`

`(y-0)^2 = 4 * 2 * (x + 2)`

Vertex (-2, 0) & a = 2

Axis of symmetry is y and a positive.

`:.Focus ((-2+2), 0) => (0,0)`

graph{(x^2 - 16) / 8 [-10, 10, -5, 5]}