Question

How do you find the vertex, focus and directrix of `4y=x^2+4`?

Answer 1

The vertex is V(0, 1). The size a = 1. The axis is the positive y-axis, x = 0. The focus is S( 0, 0 )/ The directrix is along y = -1.

Explanation:

In the standard form, the equation is `(x-0)^2=4(y-1)`

The vertex is V(0, 1).

The size a = 1.

The axis is the positive y-axis, x = 0.

The focus S( 0, 0 ), is below V, with SV = a = 1.

The directrix is perpendicular to x-axis, at a further depth a = 1, and

so, it is along y = -1.

graph{y=x^2/4+1 [-10, 10, -5, 5]}