<s>[INST] Maximum value of $x^2+y^2$ Question

Find the maximum value of $x^2+y^2$ if
  $4x^4+9y^4=64$

Now I really don't understand how to proceed or whether I should change my approach all together. Any help is appreciated.Thanks :)
 [/INST] $$4x^4+9y^4=64$$
$$\frac{(x^2)^2}{4^2}+\frac{(y^2)^2}{(8/3)^2}=1$$
Let $x=\pm \sqrt{4\cos(\theta)}$ and $y=\pm \sqrt{\frac{8}{3} \sin (\theta)}$.  With $\theta \in [0,\frac{\pi}{2}]$.
Then,
$x^2+y^2=4\cos (\theta)+\frac{8}{3} \sin (\theta)$
$$=\langle 4, \frac{8}{3} \rangle \cdot \langle \cos (\theta), \sin (\theta) \rangle$$
$$=\sqrt{4^2+(\frac{8}{3})^2} \cos \left( \theta-\arctan(\frac{\frac{8}{3}}{4}) \right)$$
Because $\arctan(\frac{8}{12}) \in [0, \frac{\pi}{2}]$. A maximum of,
$$\sqrt{4^2+(\frac{8}{3})^2}$$
Is achievable.
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