Question

Parabola help again?
Find the vertex, focus, directrix, and focal width of the parabola.

x = 4y^2

Answer 1

This one is backwards.. I figured the vertex was (0,0), but that's all I got.. @campbell_st

Answer 2

the vertex is (0,0)

Answer 3

ok so this is a parabola that looks like



the standard form is

\[(y - k)^2 = 4a(x - h)\]

(h, k) is the focus

and a is the focal length

so looking at your question the vertex is at the origin

Vertex(0, 0)

so re-write it

\[y^2 = \frac{1}{4} x\]

so to find the focal length

solve

\[\frac{1}{4} = 4a\]

the the directrix will be

x = -a

focus is (h + a, k)

or

focus (a, 0)

hope it helps