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?

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- March 18th 2008, 04:07 PMSocratesFind 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, 07:44 PMtopher0805

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

Can you solve it from there? - March 18th 2008, 07:52 PMtopher0805
Hint: is the distance from the center to the vertex.

- March 18th 2008, 07:52 PMSocrates
What would I be solving it for? What variable?

- March 18th 2008, 07:54 PMtopher0805
I gave you this equation:

And I gave you this hint:

is the distance from the center to the vertex.

Now all you have to do is plug in values for and . - March 18th 2008, 07:56 PMSocrates

Then Solve for ? - March 18th 2008, 07:58 PMtopher0805
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, 08:04 PMSocrates

This the right equation so far?

Edit: I'm lost because that would make it which is 0. What did I do wrong? - March 18th 2008, 08:26 PMtopher0805
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:

- March 18th 2008, 08:30 PMSocrates
Hrmm...That is some complex stuff. I hate graphs so much. :)

- March 18th 2008, 08:31 PMtopher0805
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, 08:32 PMSocrates

Final Equation. Right? - March 18th 2008, 08:36 PMtopher0805
Yes that is correct.

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