Question

I really need help with this!
Find the vertex, focus, directrix, and focal width of the parabola.

x = 4y^2

Answer 1

@mathmale

Answer 2

Hello, BG!!
The simplest relevant form of the equation of a parabola is
4px=y^2. Compare that to your
x = 4y^2. With some relatively straightforward algebra, we can find the value of p. Can you do that? Note: p represents the distance between focus and vertex.

Answer 3

So if we just swap them, p=1?

Answer 4

We need to check that. Substituting your p=1 into 4px=y^2, 4x=y^2. Is this the same as your given equation, x=4y^2?

Answer 5

Nope, wait. y^2=x/4.

Answer 6

So what is the correct value of p?

Answer 7

?

Answer 8

I'm not catching on. Not enough sleep.

Answer 9

:)

Compare 4px=y^2 to your x=4y^2.
From the first equation, y^2=4px.
Substitute this expression for y^2 into x=4y^2.

Then solve the resulting equation for p.

Answer 10

So I plug in 4px? That gets me 16px... Right?

Answer 11

You're dealing with equations here, so your result should be an equation. Mind trying once more?

Answer 12

x=4y^2 becomes x=4(4px) after y^2=4px has been substituted.

Answer 13

solve this equation for p.

Answer 14

p=x/16x, so 1/16?