Question

If $x=\frac{4\lambda }{1+{\lambda }^2}$ and $y=\frac{2-2{\lambda }^2}{1+{\lambda }^2}$, where $\lambda$ is a real parameter such that $x^2-xy+y^2$ lies in the interval $[a,b]$, then the minimum value of $p$ for which $3\cos \varphi =p^2-(a+b)p+19$ holds true is .

Answer 1

Let $\lambda =\tan \theta$
So, $x=\frac{4\tan \theta }{1+{\tan }^2\theta }=2\sin 2\theta$
Similarly $y=2(\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta })=2\cos 2\theta$

Now, we can write
$x^2-xy+y^2$
$=4{\sin }^{2}2\theta -4\sin 2\theta \cos 2\theta +4{\cos }^{2}2\theta$
$=4-2\left(2\cos 2\theta \sin 2\theta \right)$
$=4-2\sin 4\theta$

We know
$-1\le \sin 4\theta \le 1\Rightarrow 2\le 4-2\sin 4\theta \le 6$
Thus, comparing with $[a,b]$, we get
$a=2$ and $b=6$
$\Rightarrow a+b=8$
Now, we have the other equation as:
$3\cos \varphi =p^2-(a+b)p+19$
$\Rightarrow 3\cos \varphi =p^2-8p+19$
$\Rightarrow 3\cos \varphi =(p-4{)}^2+3$
We know that
$3\cos \varphi \le 3$
So, $(p-4{)}^2=0$
$\Rightarrow p=4$