y^2=4x+4y , find vertex,latus rectum,directrix,focus of parabola

Question

accessdenied · Answer

Ideally, we'd like our equation to be in the form: $x = a(y - k)^2 + h$.  Do you know how to get the equation into this form?

anonymous · Answer

no

accessdenied · Answer

Okay, so we basically want to get the y-values on one side and the x-values on the other.
We'll subtract 4y from both sides.
  y^2 - 4y = 4x
Then, we will complete the square for "y^2 - 4y."  The term we add to create the perfect square is half of the -4, squared, so...
  y^2 - 4y + 4 - 4 = 4x
  (y - 2)^2 - 4 = 4x

Then we divide by 4.
  (1/4)(y - 2)^2 - 1 = x

Or, if we reverse it...
  x = (1/4)(y - 2)^2 - 1.

accessdenied · Answer

Then, the vertex is the point "(h,k)" from "x = a(y - [k])^2 + [h]"
  k = 2, h = -1
So, the vertext is V(-1, 2).

Does that make sense?

anonymous · Answer

no

accessdenied · Answer

Well, I'm sorry I was unable to help you.  I hope you are able to figure it out.