# Math Help - Standard form of parabola

1. ## Standard form of parabola

For some reason I don't get this at all. I understand ellipses and hyperbolas, but not a simple parabola.

2. Originally Posted by dataspot

For some reason I don't get this at all. I understand ellipses and hyperbolas, but not a simple parabola.
Is it the $y^2$ screwing with you? If so then just find all the info for

$y=\frac{x^2}{-6}$

and then be sure to remember to adjust so that it is facing the proper direciton.

3. Originally Posted by Mathstud28
Is it the $y^2$ screwing with you? If so then just find all the info for

$y=\frac{x^2}{-6}$

and then be sure to remember to adjust so that it is facing the proper direciton.
The whole thing is. I'm not sure how they're getting those answers. Standard form for this one would be $(y-k)^2=4p(x-h)$. But how is that $-6x$ going to $4(\frac{-3}{2})x$?

4. Originally Posted by dataspot
The whole thing is. I'm not sure how they're getting those answers. Standard form for this one would be $(y-k)^2=4p(x-h)$. But how is that $-6x$ going to $4(\frac{-3}{2})x$?
because $4\cdot\frac{-3}{2}=-6$? Which is neccesary.

5. Originally Posted by dataspot
The whole thing is. I'm not sure how they're getting those answers. Standard form for this one would be $(y-k)^2=4p(x-h)$. But how is that $-6x$ going to $4(\frac{-3}{2})x$?
The parabola takes on the form $y^2=4px$

Since our equation is $y^2=-6x$, comparing it to the standard form we see that $4p=-6$, solving for p, we get $p=-\frac{3}{2}$. Thus, that's how they get $-6x=4\left(-\frac{3}{2}\right)x$

Does that make sense?

--Chris

6. They're flowing pretty easy now. Thanks!