Given an ellipse's center, focus and point, find its equation.

Question

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

$a^2 = 16 + \frac{1}{2} (1+5 \sqrt{41})$

Answer 1

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.