Given an ellipse's center, focus and point, find its equation. Given an ellipse's center is $(2,1)$, focus is (2,4) and point is (3,-3), we have
Plug in center:
$\frac{(x-2)^2}{a^2}+\frac{(y-1)^2}{b^2} = 1$
Use focus:
$4^2=a^2-b^2$
$16=a^2-b^2$
Use point:
$\frac{(3-2)^2}{a^2}+\frac{(-3-1)^2}{b^2} = 1$
$\frac{1}{a^2}+\frac{16}{b^2} = 1$
$b^2+16a^2 = a^2b^2$
Is this right?
Combine equations:
$b^2+16a^2 = a^2b^2$
$16=a^2-b^2$
So, we have:
$0 = b^2 - b^2 - 16^2$
$b^2 = \frac{1}{2} (1+5 \sqrt{41})$
and
$a^2 = 16 + \frac{1}{2} (1+5 \sqrt{41})$

This is wrong. The answer is $\frac{(x-2)^2}{9}+\frac{(y-1)^2}{18}=1$. Thanks for the WolframAlpha suggestion, @alexqwx . I never knew WA could do that.
