Question

Find the vertex, focus, directrix, and focal width of the parabola.

x = 4y^2

Answer 1

@wio could you help on this one?

Answer 2

first isolate y

Answer 3

Vertex is 0, 0 , but I am getting confused on how to get the focus

Answer 4

hmm yeah i dont know too

Answer 5

@wio

Answer 6

or @satellite73 ?

Answer 7

Well, go back to the equation: \[
(x-x_0)^2=4p(y-y_0)
\]But we want to change it to: \[
(y-y_0)^2=4p(x-x_0)
\]In this case.

Answer 8

First:\[
y^2=4px
\]And then \[
x = 4y^2\implies y^2=\frac x4
\]And then \[
4px = \frac x4 \implies 4p=\frac 14\implies p=\frac 1{16}
\]

Answer 9

That helps so much! :) Thanks again

Answer 10

We'll get the nomenclature clear

It can be seen from the diagram that for a horizontal axis parabola where vertex is at the origin, y^2=4px, or p=1/16 to give x=4y^2 as @wio had it.
since p=1/16, so the directrix is at x=-1/16, and the focus is at (+1/16, 0).
The focal width, is the distance between the two intersections of the line parallel to the directrix, but passing through the focus.
In this case, this line is x=1/16, which gives
y^2=4px=4x/16=1/64
so y=\(\pm 1/8\), and the focal width, or latus rectum is 2(1/8)=1/4.