Mathematics - Point Question with Solution | TestHub

Let the point $P$ is $\left(h,k\right)$

Distance of $P$ from $L_1=\left|\frac{h\sqrt{2}+k-1}{\sqrt{(\sqrt{2}{)}^2+1^2}}\right|$

$=\left|\frac{h\sqrt{2}+k-1}{\sqrt{3}}\right|$

Distance of $P$ from $L_2=\left|\frac{h\sqrt{2}-k+1}{\sqrt{(\sqrt{2}{)}^2+1^2}}\right|$

$=\left|\frac{h\sqrt{2}-k+1}{\sqrt{3}}\right|$

$\therefore$ The equation of the locus of $P$ is

$\left|\frac{h\sqrt{2}+k-1}{\sqrt{3}}\right|\times \left|\frac{h\sqrt{2}-k+1}{\sqrt{3}}\right|={\lambda }^2$

$\left|\left(\frac{h\sqrt{2}+k-1}{\sqrt{3}}\right)\left(\frac{h\sqrt{2}-k+1}{\sqrt{3}}\right)\right|={\lambda }^2$

$\Rightarrow \left|2h^2-(k-1{)}^2\right|=3{\lambda }^2$

Hence, the equation of the locus is $\left|2x^2-(y-1{)}^2\right|=3{\lambda }^2$

The line is $y=2x+1$ or $y-1=2x$

By substituting the value of $y$ in the equation of the curve $C$, we get

$\left|2x^2-(y-1{)}^2\right|=3{\lambda }^2$

$\Rightarrow \left|2x^2-(2x{)}^2\right|=3{\lambda }^2$

$\Rightarrow 2x^2=3{\lambda }^2$

$\Rightarrow x=\pm \sqrt{\frac{3}{2}}\lambda$

$\Rightarrow x_2-x_1=\left|\sqrt{6}\lambda \right|$

Also, $y-1=2x$

$\Rightarrow y_2-1=2x_2$ and $y_1-1=2x_1$

$\Rightarrow y_2-y_1=2\left(x_2-x_1\right)$

$\Rightarrow y_2-y_1=\left|2\sqrt{6}\lambda \right|$

Given $RS=\sqrt{270}$

$\Rightarrow \sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}=\sqrt{270}$

$\Rightarrow (\sqrt{6}\lambda {)}^2+(2\sqrt{6}\lambda {)}^2=270$

$30{\lambda }^2=270$

${\lambda }^2=9$