# Thread: Finding the foci of an ellipse from an equation.

1. ## Finding the foci of an ellipse from an equation.

I have an odd question, that I think I have worked out properly, but the answer makes no sense.

What are (a) the x-intercepts,(b) the y-intercepts and(c) the foci of the ellipse $\frac{x^2}{25}+ \frac{y^2}{36}=1$.

I have already determined that the x and y intercepts are: $\pm5$ and $\pm6$,respectively, meaning that the major axis is vertical as the y-intercept is greater than the x-intercept.
My problem is that when I use the relation: $c^2=a^2-b^2$
the result is -9, so c is the square root of -9--which doesn't exist and cant be graphed, meaning I did something wrong.....
What was it that I did wrong?

2. Originally Posted by MathBlaster47
I have an odd question, that I think I have worked out properly, but the answer makes no sense.

What are (a) the x-intercepts,(b) the y-intercepts and(c) the foci of the ellipse $\frac{x^2}{25}+ \frac{y^2}{36}=1$.

I have already determined that the x and y intercepts are: $\pm5$ and $\pm6$,respectively, meaning that the major axis is vertical as the y-intercept is greater than the x-intercept.
My problem is that when I use the relation: $c^2=a^2-b^2$
the result is -9, so c is the square root of -9--which doesn't exist and cant be graphed, meaning I did something wrong.....
What was it that I did wrong?
the foci are on the y-axis ... $c^2 = b^2-a^2$

also, check you arithmetic when calculating ... (it's not 9)

3. Originally Posted by skeeter
the foci are on the y-axis ... $c^2 = b^2-a^2$

also, check you arithmetic when calculating ... (it's not 9)
GYAHH....(insert expletive), 1+1=2 and 36-25=11.....typing too fast is killing me....Must lay off the sugar.....
Ok! I think I understand what is going on, since the axis is on y then the equation for establishing the foci changes to reflect that, meaning that I have to turn the numbers around.
Meaning that the foci are $(0,\sqrt{11})$ and $(0,-\sqrt{11})$?

4. Originally Posted by MathBlaster47
GYAHH....(insert expletive), 1+1=2 and 36-25=11.....typing too fast is killing me....Must lay off the sugar.....
Ok! I think I understand what is going on, since the axis is on y then the equation for establishing the foci changes to reflect that, meaning that I have to turn the numbers around.
Meaning that the foci are $(0,\sqrt{11})$ and $(0,-\sqrt{11})$?
Yup, that's what I got;
The x-intercepts are $5$ and $-5$
The y-intercepts are $6$ and $-6$
And the foci of the ellipse are $(0,\sqrt{11})$ and $(0,-\sqrt{11})$