How Do I Solve

  (y-3)^2=8(x-5)

Let's see how Melody arrived at this......

Writing this in a slightly different form, we have

 (y-3)^2=4p(x-5)   where "p" is the distance (horizontal, in this case) between the x coordinate of the vertex and the x coordinate of the focal point.

So...we have.....

4p = 8   ..... and solving for "p"  we have

p = 2    ........i.e., the focal point is 2 units to the "right" from the vertex......thus......

The vertex is (5,3) and the focus is (7,3)......just as Melody found !!!!

For a "sideways" parabola that opens to the right, we add "p" to the x coordinate of the vertex. For one that opens to the left, we subtract "p" from the x coordinate of the vertex.

mmm let's think about that.

The vertex is (5,3)

It is a (pos) sideways parabola 8=4*2 so the focal length is 2.

So the focus must be y=3 x=5+2     

focus (7,3)

that sounds right to me.  

thanks chris,

You need to skech a graph of it then it should be clear.

I would, except I'm at a public library, and it's hard to upload/download stuff here.....maybe I'll do it later....!!!!

Your at a library? Why are you at a-

Nevermind, I won't ask.

I wasn't meaning you Chris.  I was meaning Ashton!

His computer is probably in for repair.  Maybe the lame questions he answers caused it to overheat. Computers seem to get tired of things and get burnt out.

I hope CPhill doesn’t get burnt out!

If Chris were here he would have some witty come back for you but he is not here so ALL 

you get is "yes yes, very funny 7up!"

I suppose you want some points now?

It’s not funny enough for points.

Only the CORRECT answers to questions should get points.

Maybe only REALLY funny posts or REALLY cool posts should get a point.

The points on here are already so inflated they almost aren’t worth anything!

7up

I think that you are right sevenUP.

The points should be given for high achievement.  They should not reflect mediocrity!

What? No points for my brilliant post? LOL

Yes, at your sensible request, no points!