Question

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

Answer 1

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