Find the equation of an ellipse

**Socrates** · March 18th 2008, 04:07 PM

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?

**topher0805** · March 18th 2008, 07:44 PM

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

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?

**topher0805** · March 18th 2008, 07:52 PM

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

**Socrates** · March 18th 2008, 07:52 PM

What would I be solving it for? What variable?

**topher0805** · March 18th 2008, 07:54 PM

I gave you this equation:

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

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 $b$ .

**Socrates** · March 18th 2008, 07:56 PM

$d = \sqrt{(-3-0)^2 + (3-0)^2}$

$d=4.24$

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

Then Solve for $b^2$ ?

**topher0805** · March 18th 2008, 07:58 PM

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.

**Socrates** · March 18th 2008, 08:04 PM

$\frac{(-3+3)^{2}}{b^{2}} + \frac{(1+2)^{2}}{5^{2}} = 1$

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?

**topher0805** · March 18th 2008, 08:26 PM

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:

$5^2 = b^2 + 4^2$

**Socrates** · March 18th 2008, 08:30 PM

Hrmm...That is some complex stuff. I hate graphs so much.

**topher0805** · March 18th 2008, 08:31 PM

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.

**Socrates** · March 18th 2008, 08:32 PM

$\frac{(x+3)^{2}}{9} + \frac{(y+2)^{2}}{25} = 1$

Final Equation. Right?

**topher0805** · March 18th 2008, 08:36 PM

Yes that is correct.

You'll be a pro in no time.

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