# Find the equation of an ellipse

• March 18th 2008, 05:07 PM
Socrates
Find the equation of an ellipse
Find the equation of the ellipse with center (-3,-2), a focus at (-3,1), and a vertex at (-3,3).

What do I do first?
• March 18th 2008, 08:44 PM
topher0805
$\frac{(x-h)^{2}}{b^{2}} + \frac{(y-k)^{2}}{a^{2}} = 1$

This is the general form for the equation of the ellipse When the ellipse is vertical. When it is horizontal simply switch the $a$ and the $b$. The ellipse is centered at the point $(h,k)$, so plug in those values:

$\frac{(x+3)^{2}}{b^{2}} + \frac{(y+2)^{2}}{a^{2}} = 1$

Can you solve it from there?
• March 18th 2008, 08:52 PM
topher0805
Hint: $a$ is the distance from the center to the vertex.
• March 18th 2008, 08:52 PM
Socrates
What would I be solving it for? What variable?
• March 18th 2008, 08:54 PM
topher0805
I gave you this equation:

$
\frac{(x+3)^{2}}{b^{2}} + \frac{(y+2)^{2}}{a^{2}} = 1
$

And I gave you this hint:

$a$ is the distance from the center to the vertex.

Now all you have to do is plug in values for $a$ and $b$.
• March 18th 2008, 08:56 PM
Socrates
$d = \sqrt{(-3-0)^2 + (3-0)^2}$

$d=4.24$

$\frac{(x+3)^{2}}{b^{2}} + \frac{(y+2)^{2}}{4.24^{2}} = 1
$

Then Solve for $b^2$ ?
• March 18th 2008, 08:58 PM
topher0805
Hmm, how did you get 4.24?

Find the correct value for a (it should be 5) and then solve for b. You will need to know a point on the ellipse, but thankfully you already have one!

I haven't done these in a long time so forgive me if I am a bit rusty.
• March 18th 2008, 09:04 PM
Socrates
$\frac{(-3+3)^{2}}{b^{2}} + \frac{(1+2)^{2}}{5^{2}} = 1$

This the right equation so far?

Edit: I'm lost because that would make it $\frac{0}{b^2}$ which is 0. What did I do wrong?
• March 18th 2008, 09:26 PM
topher0805
Sorry, I just realized that you don't have a point on the ellipse. Instead you will have to use trig to find b.

You have a focus (-3,1). The definition of a focus is that the distance from the two focus' to any point on the ellipse is constant.

Eg. If you have point A and point B on the ellipse, the distance from f1 to A to f2 will be the same as f1 to B to f2. Am I explaining this clearly?

Since your one focus is (-3,1), your other will be (-3, -5). The distance from one focus to the vertex given, to the second focus is 10. Since the two focuses are equidistant from each other, the point b is therefore 5 away from both focuses. Use Pythagoras to solve for b in this manner:

$5^2 = b^2 + 4^2$
• March 18th 2008, 09:30 PM
Socrates
Hrmm...That is some complex stuff. I hate graphs so much. :)
• March 18th 2008, 09:31 PM
topher0805
Yeah, I can draw some graphs and show a better way to do it and post it here if you like but right now I'm trying to get my homework done so it won't be up till morning.
• March 18th 2008, 09:32 PM
Socrates
$\frac{(x+3)^{2}}{9} + \frac{(y+2)^{2}}{25} = 1$

Final Equation. Right?
• March 18th 2008, 09:36 PM
topher0805
Yes that is correct.

You'll be a pro in no time. :)