# Find the graph of equation...

• Jun 23rd 2010, 07:51 PM
blackdragon190
Find the graph of equation...
Hi, I'm having trouble figuring out the answer to this question:

Find an equation whose graph consists of all points P(x,y) whose distance from the point F(0,p) is equal to its distance PQ (there should be a line above the PQ) from the horizontal line y = -p (p is a fixed positive number).

This is my effort of reproducing the graph that came with it:

http://img.photobucket.com/albums/v6...2/untitled.jpg

I thought I could do it by plugging (x,y) and (0,p) into the standard equation of a circle, but I'm not getting the same answer as is listed in the book (they say that it's x^2=4py (parabola))

Any help greatly appreciated!
• Jun 23rd 2010, 08:05 PM
Do you know the Distance Formula?

Play with that and show us what you get from it.
• Jun 23rd 2010, 09:57 PM
blackdragon190
I tried it, but I still can't seem to get it.

I did:

d = [(0-x)^2 + (p-y)^2]^1/2
d = [x^2 + p^2 - 2py +y^2]

I'm not sure what to do from there (Worried)
• Jun 23rd 2010, 10:50 PM
Yeah, you have the right idea, this is all you have to remember for the parabola;

$\overline{PF} = \overline{PQ}$

the following, which I'll show you how to do, should just be the manual work that follows from remembering the above property.

$\overline{PF} \ = \ \sqrt{(x \ - \ 0)^2 + (y \ - \ p)^2} \ = \sqrt{(x)^2 + (y \ - \ p)^2}$

$\overline{PQ} \ = \ \sqrt{(x \ - \ x)^2 + (y \ - \ (-p))^2} \ = \sqrt{(0)^2 + (y \ + \ p)^2}$

$\overline{PF} = \overline{PQ}$

$\sqrt{(x)^2 + (y \ - \ p)^2} \ = \ \sqrt{(0)^2 + (y \ + \ p)^2}$

Square both sides and solve;

$(x)^2 \ + \ (y \ - \ p)^2 \ = \ (y \ + \ p)^2$

$x^2 \ + \ y^2 \ - \ 2py \ + \ p^2 \ = \ y^2 \ + 2py \ + \ p^2$

Cancel terms;

$x^2 \ - \ 2py \ = \ 2py$

$x^2 \ \ = \ 2py \ + \ 2py$

$x^2 \ = \ 4py$

It's basically the same idea for the ellipse and hyperbola with a little change (Wink)
• Jun 24th 2010, 09:59 PM
blackdragon190
Thanks!(Rofl)