Originally Posted by

**bumpjump** Hi, this is my first time posting. I need help with this problem:

Suppose the equation for an ellipse is given by

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

Find $\displaystyle b$ so that the eccentricity is 0.75

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I know that eccentricity is defined as $\displaystyle \frac{c}{a}$, making $\displaystyle c=3$ and $\displaystyle a=4$. However, if I plug these values into the Pythagorean theorem, I get $\displaystyle b^2=-7$ (which is obviously incorrect, right? We're dealing with real numbers).

How do I solve this problem? Thanks in advance.