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

*From the given equation $\,xy = \cfrac{4(x^2+y^2) - 19}{5} = \cfrac{4 Z - 19}{5}\,$. But for $\forall x,y \in \mathbb{R}$ the inequality holds $\,xy \le \cfrac{1}{2}(x^2+y^2)\,$ so $\,\cfrac{4 Z - 19}{5} \le \cfrac{Z}{2} \iff \boxed{Z \le \cfrac{38}{3}}\,$

*$0 \le (x+y)^2 = x^2+y^2+2xy = Z + 2 \cfrac{4 Z - 19}{5}=\cfrac{13 Z - 38}{5} \implies \boxed{Z \ge \cfrac{38}{13}}$
$Z_{min}=\cfrac{38}{13}$ is attained for $x=-y=\sqrt{\cfrac{19}{13}}$ and $Z_{max}=\cfrac{38}{3}$ is attained for $x=y=\sqrt{\cfrac{19}{3}}\,$.
A: Without Lagrange multipliers (but with trigonometry). Use polar coordinates:
$x=\sqrt Z\cos\theta$
$y=\sqrt Z\sin\theta$
$xy=Z\sin\theta\cos\theta=\frac{Z}{2}\sin(2\theta)$
From your equation you get $\frac{1}{z}=\frac{4}{19}-\frac{5\sin(2\theta)}{38}$
As $\sin$ takes values between $-1$ and $1$:
So $\frac{1}{z_{min}}+\frac{1}{z_{max}}=2\cdot\frac{4}{19}=\frac{8}{19}$
A: 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.
A: $$4x^2-5xy+4y^2-19=0$$
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.
