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.

Please help!(Angry)

You can answer me right here because I check often enough.

Joanne

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- Dec 1st 2008, 11:53 AMJoanneMacNeed 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.**

**Please help!(Angry)**

**You can answer me right here because I check often enough.**

**Joanne** - Dec 1st 2008, 12:13 PMrunning-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 PMvincisonfire
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

$\displaystyle (6+y)+(6-y) = 12 $

Then from the x-intercept you know that $\displaystyle 12 = 2\sqrt{5^2+y^2} $

$\displaystyle y =\pm \sqrt(11) $