Question

Let $x,y\in R$ satisfying $x^2+y^2=4$. If range of $x^2-xy+3y^2$ is $[a,b]$ then find the value of $\left(\frac{a+b}{2}\right)$.

Answer 1

Answer: 8

$\because x^2+y^2=4\Rightarrow x=2\cos \theta \text{ and }y=2\sin \theta$
$\therefore x^2-xy+3y^2=4{\cos }^2\theta -4\cos \theta \sin \theta +12{\sin }^2\theta$
$=2(1+\cos 2\theta )-2\sin 2\theta +6(1-\cos 2\theta )$
$=8-4\cos 2\theta -2\sin 2\theta$
$\because -\sqrt{20}\le 4\cos 2\theta +2\sin 2\theta \le \sqrt{20}$
$8-\sqrt{20}\le 8-(4\cos 2\theta +2\sin 2\theta )\le 8+\sqrt{20}$
$\therefore \left(\frac{a+b}{2}\right)=8$