$4x^2-5xy+4y^2=19$ 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.

$$4x^2-5xy+4y^2-19=0$$
By quadratic formula we have,
$$x=\frac{1}{8} (5 y\pm \sqrt{304-39y^2})$$
So,
$$z=y^2+\frac{1}{64} (5 y\pm \sqrt{304-39y^2})^2$$
Now take derivative, set it to zero, and examine sign changes.
