Changing form of equation of ellipse $4x^2 + 9y^2 - 32x - 36y + 64 = 0$ into standard form

Question

My homework question is:

Find the vertices and foci of the ellipse whose equation is given by: $$4x^2 + 9y^2 - 32x - 36y + 64 = 0.$$

I'm trying to convert it into the standard form so I can get $c^2$.

currently I have: $$4(x-4)^2 + 9(y-2)^2 = -164$$ (youtube told me to add $64$ and $36$ to make perfect square)

after this point I am confused because $\frac{9(y-2)^2}{-164}$ seems out of the scope of my precalc course and I have a feeling I made a mistake.

any help is appreciated :D

Answer 1

$4x^2 +9y^2 -32x -36y +64 = 4(x^2 - 8x +16-16) +9(y^2 - 4y +4 - 4) +64 = 4(x-4)^2 + 9(y-2)^2 - 64 -36 + 64 = 0$.

Finally, $4(x-4)^2 + 9(y-2)^2 = 36$, where the foci are deduced by dividing by 36:

\begin{equation} \frac{1}{9}(x-4)^2 + \frac{1}{4}(y-2)^2 = 1 \end{equation}

Answer 2

Let me follow the youtube suggestion. $$4x^2+9y^2-32x-36y=-64$$

$$4x^2-32x+64+9y^2-36y+36=-64+64+36$$

$$4(x-4)^2+9(y-2)^2=36$$