$4x^2-5xy+4y^2=19$ with $Z=x^2+y^2$, find $\dfrac1{Z_{\max}}+\dfrac1{Z_{\min}}$ I have no idea how to approach this question at all. I've tried to find the maximum and minimum of the quadratic but i am too confused on what to do afterwards.

Let $x=\sqrt{z}\cos B, y=\sqrt{z}\sin B$, hence we get
$$4z-\frac{5z}{2}\sin\left (2B  \right )=19$$
for $-1\leq \sin\left ( 2B \right )\leq 1$ and $\displaystyle z=\frac{19}{4-\dfrac{5}{2}\sin\left ( 2B \right )}$.
Now the answer will follow.
