# Find an equation of the parabola?

• Mar 19th 2010, 11:29 AM
iluvmathbutitshard
Find an equation of the parabola?
Question says to find an equation of the parabola:
Vertex (-1, 2); Focus (-1, 0)

How do I do this? I am confused.

Thanks for any help.
• Mar 19th 2010, 12:11 PM
harish21
Quote:

Originally Posted by iluvmathbutitshard
Question says to find an equation of the parabola:
Vertex (-1, 2); Focus (-1, 0)

How do I do this? I am confused.

Thanks for any help.

You can find the equation by this formula:

$\displaystyle y = a{(x-h)^2}+k$

where, $\displaystyle (h,k)$ is your vertex

and $\displaystyle a= \frac{1}{4d}$ ; d is the distance from the vertex to the focus.

So use distance formula (with the coordinates of your vertex and focus) to find d. You already have (h,k) as your vertex. Plug these values in to get the equation of your parabola.

Try doing it and post a message if you come up with errors
• Mar 19th 2010, 12:16 PM
iluvmathbutitshard
so..
y = 1/8 (x - 1)^2 + 2

Is this right?
• Mar 19th 2010, 12:23 PM
harish21
Quote:

Originally Posted by iluvmathbutitshard
so..
y = 1/8 (x - 1)^2 + 2

Is this right?

Note:

$\displaystyle y = a{(x-h)^2}+k$

you have h = -1

$\displaystyle y = \frac{1}{8}{(x+1)^2}+2$
• Mar 19th 2010, 12:27 PM
masters
Quote:

Originally Posted by iluvmathbutitshard
so..
y = 1/8 (x - 1)^2 + 2

Is this right?

Hi iluvmathbutitshard,

Almost, you missed the sign on "a".

Another way to find "a" in $\displaystyle y=a(x-h)^2+k$

is by knowing that the focus has coordinates $\displaystyle \left(h, k+\frac{1}{4a}\right)$

Set the second coordinate of this ordered pair to the second coordinate of your focus and we have

$\displaystyle 2+\frac{1}{4a}=0$

Solving this, you get $\displaystyle a=-\frac{1}{8}$

$\displaystyle y=-\frac{1}{8}(x+1)^2+2$