SOLUTION: If the foci for an ellipse is (0,-1) and (0,1) and the major axis is 6, then is the ellipse horizontal or vertical? What would the equation of the ellipse look like?

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: If the foci for an ellipse is (0,-1) and (0,1) and the major axis is 6, then is the ellipse horizontal or vertical? What would the equation of the ellipse look like?      Log On


   



Question 623500: If the foci for an ellipse is (0,-1) and (0,1) and the major axis is 6, then is the ellipse horizontal or vertical? What would the equation of the ellipse look like?
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Picture the foci in your mind. (If you can't picture them, then plot them.) You should realize that they are two points on the y axis. Since the foci of ellipses are always on the major axis, the major axis of your ellipse is the y axis, a vertical line. So your ellipse is a vertical one. And vertical ellipses have an equation of the form:
%28x-h%29%5E2%2Fb%5E2+%2B+%28y-k%29%5E2%2Fa%5E2+=+1
The rest is finding out what specific numbers to put in for the h, k, a and b.

(h, k) is the center of the ellipse. The center of an ellipse is always halfway between the foci, the center of your ellipse is (0, 0) (halfway between (0, -1) and (0, 1)) So your h and k are both zeros.

"a" is the distance from the center to each vertex on the major axis. The length of the major axis is the distance from one vertex to the other. So "a" is half of the length of the major axis. Since your major axis has a length of 6, your "a" is 3.

"b" is the distance from the center to each vertex on the minor axis. (I've learned that these are sometimes called "co_vertices".) Since we do not know what these vertices are in your ellipse we cannot find "b" directly. But we do know the foci and we can use them to find "b".

"c" is the distance from the center to each focus. Your "c" is 1. There is an equation that connects the "a", "b" and "c" of every ellipse:
a%5E2+=+b%5E2+%2B+c%5E2
We can put our "a" and "c" into this and find "b":
3%5E2+=+b%5E2+%2B+1%5E2
Solving for b:
9+=+b%5E2+%2B+1
8+=+b%5E2
(Actually all we really need is b%5E2 but I'm going to solve for b anyway.)
sqrt%288%29+=+b

With our h, k a and b we are now ready to fill in our equation:
%28x-h%29%5E2%2Fb%5E2+%2B+%28y-k%29%5E2%2Fa%5E2+=+1
%28x-0%29%5E2%2F%28sqrt%288%29%29%5E2+%2B+%28y-0%29%5E2%2F%283%29%5E2+=+1
which simplifies to:
x%5E2%2F8+%2B+y%5E2%2F9+=+1