1. ## determine the center

determine the center , coordinates of vertices, endpoints for minor axis and foci coordinates.

(x-3)^2/4 + (y+1)^2/36=1

I believe the center is (3,-1) and i can get the vertices.. i just dont know how to get the foci and endpoints?

help me.. i know ive been asking alot of questions but my final exam is tommorow and im going through alot of things I dont remeber how to do.

2. Others are about to help you with this problem. Please note that this isn't a Calculus problem, but a algebra or perhaps geometry problem. I put it in algebra because conics are taught in Algebra II in the US.

3. Originally Posted by tiga killa
determine the center , coordinates of vertices, endpoints for minor axis and foci coordinates.

(x-3)^2/4 + (y+1)^2/36=1

I believe the center is (3,-1) and i can get the vertices.. i just dont know how to get the foci and endpoints?

help me.. i know ive been asking alot of questions but my final exam is tommorow and im going through alot of things I dont remeber how to do.
Yes, the center is at (3, -1). Now because this can be written as (x-3)^2/(2^2)+ (y+1)^3/(6^2)= 1, the "semi-axes" are 2 (parallel to the x-axis) and 6 (parallel to the y-axis). The vertices are at (3+ 2, -1)= (5, -1), (3-2, -1)= (1, -1), (3, -1+ 6)= (3, 5), and (3, -1- 6)= (3, -7).

The foci are a little harder. For an ellipse, the relation between the focal distance and the semi-axes is $a^2+ c^2= b^2$ where a is the "minor semi-axis", the smaller of the two, b is the "major semi-axis", the larger of the two, and c is the focal length. Here that gives $2^2+ c^2= 6^2$ or $4+ c^2= 36$ so $c^2= 32$. That is, $4\sqrt{3}$ so the foci are at $(3, -1+ 4\sqrt{3})$ and $(3, -1- 4\sqrt{2})$. The foci are on the longer axis, here the y- axis.

4. You're correct about the center.

As for the rest, let's look at the basic form. It's an ellipse, but you probably knew that. The equation is:

$\frac {(x - h)^2}{a^2} + \frac {(y - k)^2}{b^2} = 1$, with a center at (h, k). There's one small wrinkle here, however. Since a is supposed to be the semi-major axis, and the major axis must be larger than the minor axis, by definition. Since your equation is:

$\frac {(x-3)^2}{4} + \frac {(y+1)^2}{36} = 1$

It's clear that the major axis is in the y term, and so the basic equation here would be better described by:

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

So I'd rearrange it as:

$\frac {(y+1)^2}{36} + \frac {(x-3)^2}{4} = 1$

Ok, that means we have $a^2 = 36 \implies a = \pm 6$ and $b^2 = 4 \implies b = \pm 2$.

So the major axis along the y axis would go from the center plus and minus 6. The minor axis along the x axis would go from the center plus and minus 2. Each along it's respective axis, of course. The major axis would be 2a long, and the minor axis would be 2b long (taking the positive a and b, of course).

Working that out from a center of (3, -1), we get major axis from (3, -7) to (3, 5). Minor axis from (1, -1) to (5, -1).

Lastly, the foci are described by $c = \sqrt{a^2 - b^2}$. I prefer to think of it as $c^2 = a^2 - b^2$. Whatever, same thing. Remembering that the foci are along the major axis, we solve that:

$c = \pm \sqrt {36 - 4} = \pm \sqrt {32} = \pm \sqrt {2(16)} = \pm 4 \sqrt{2}$. That gives foci along the y axis of $(3, -1 + 4 \sqrt{2})$ and $(3, -1 - 4 \sqrt{2})$.

Hope that helps, because it involved a lot of typing... lol Good luck!

5. Originally Posted by HallsofIvy
That is, $4\sqrt{3}$ so the foci are at $(3, -1+ 4\sqrt{3})$ and $(3, -1- 4\sqrt{2})$. The foci are on the longer axis, here the y- axis.
Dead on, except that $\sqrt {32} = 4 \sqrt {2}$. I'm guessing it's just a typo since you got it right in the second focus. Just thought I'd correct it so the OP doesn't get confused.