# Need help finding foci of an ellipse

• Dec 1st 2008, 11:53 AM
JoanneMac
Need help finding foci of an ellipse
of this equation: x^2/25+y^2/36=1

I know the x-intercepts are (5, -5) and the y-intercepts are (6, -6), but don't know how to find the foci.

You can answer me right here because I check often enough.
Joanne
• Dec 1st 2008, 12:13 PM
running-gag
Hello

Your equation is x²/a² + y²/b² = 1 with a = 5 and b = 6
Have a look to the picture above but exchanging a and b since a < b (or rotate the ellipse of 90°)
c² = b² - a² = 11
Foci are therefore placed at (0, sqrt(11)) and (0,-sqrt(11))

http://imageshack-france.com/out.php...69_Ellipse.JPG
• Dec 1st 2008, 12:15 PM
vincisonfire
The definition of an ellipse a line. The sum of the distance to each foci of each point is the same.
You know because of the y-intercept that this is equal to
$(6+y)+(6-y) = 12$
Then from the x-intercept you know that $12 = 2\sqrt{5^2+y^2}$
$y =\pm \sqrt(11)$