Question

A normal at $P\left(\theta \right)$ to the hyperbola $\frac{x^2}{4}-\frac{y^2}{1}=1$ has equal intercepts on the positive $x,y-$axes. If this normal touches the circle $x^2+y^2=r^2,$ then the value of $12r^2$ is

• A. $32$
• B. $25$
• C. $50$
• D. $64$

Answer 1

The correct option is C $50$

$\frac{x^2}{4}-\frac{y^2}{1}=1$
$\Rightarrow a=2,b=1$
$\Rightarrow$ Normal at $P\left(\theta \right)$ is $2x\cos \theta +y\cot \theta =4+1$
$\Rightarrow 2x\cos \theta +y\cot \theta =5$
Given, normal has equal intercepts on positive $x$ and $y$ axes.
$\Rightarrow$ Slope of normal $=-1$
$\Rightarrow -2\sin \theta =-1$
$\Rightarrow \sin \theta =\frac{1}{2}$
$\Rightarrow \theta =\frac{\pi }{6}$ $(\because$ normal line lies in $1^{\text{st}}$ quadrant$)$
$\therefore$ Equation of normal is $2x\cdot \frac{\sqrt{3}}{2}+y\sqrt{3}=5$
$\Rightarrow \sqrt{3}x+\sqrt{3}y=5$
The above line touches the circle $x^2+y^2=r^2$
$\Rightarrow$ Perpendicular distance from centre $(0,0)$ is $r$
$\Rightarrow r=\frac{5}{\sqrt{3+3}}=\frac{5}{\sqrt{6}}$
$\therefore 12r^2=12\times \frac{25}{6}=50$