Parabolas?

Question

How do I convert the standard form of a parabola (y=ax2+bx+c ) into the general form (x-h)^2= 4p(y-k)? and then solve for the focus, vertex, P, and equation of the directrix? HELP!!! I have a test in an hour and I have no idea what I am doing!!!!!

Answer 1

It's called "Completing the square".
First divide thru by "a" with the a and b terms like so:

y = a(x^2 + (b/a)x ) +c then take 1/2 b/a and square it as follows

y= a(x^2 +(b/a)x + b^2/4a^2) + c - b^2/4a
Now we can express the first 3 terms as a perfect square

y= a(x + b/2a)^2 + c - b^2/4a

Then (x + b/2a)^2 = y/a -c/a + b^2/4a^2

h= -b/2a

4py=(1/a)y
p=1/4a

-4pk = -c/a + b^2/4a^2 and sub 1/4a for p
-k/a = -c/a + b^2/4a^2
k= c - b^2/4a

And next time don't procrastinate!!

Good luck my little sabra, la chaim

Answer 2

For (y=ax2+bx+c ) and (x-h)^2= 4p(y-k)

the relation is
a = (1/(4p)), b = (-h/(2p)), c=(h^2/(4p)) +k, h = (-b/(2a)),
k = ((4ac-b^2)/(4a))

I hope that it is clear now.

All the best.

Answer 3

http://en.wikipedia.org/wiki/Parabola go to this link ^^,