Question

How do you find the focus and directrix of `x^2 = 2y`?

Answer 1

Vertex: V(0, 0). Focus : `S(0, 1/2)`. Directrix DR : `y = -1/2` see the illustrative Socratic graph.

Explanation:

This is in the standard form #x^2=4ax, with

size `a = 1/2`.

The vertex is V(0, 0).

The axis is the y-axis x = 0.

The tangent at the vertex VX is the x-axis y = 0.

The directrix DR is below and parallel to VX ,

at a depth `a = 1/2`, and so,

its equation is `y = -1/2`.

The focus S is on the axis x = 0 and is above V, at a distance a = 1/2,

and so, S is at `(0. 1/2)`.

graph{(x^2-2y)((y-1/2) ^2+x^2-.03)xy(y+1/2)=0 [-5, 5, -2.5, 2.5]}