Question

The point $P(-2\sqrt{6},\sqrt{3})$ lies on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ having eccentricity $\frac{\sqrt{5}}{2}.$ If the tangent and normal at $P$ to the hyperbola intersect its conjugate axis at the points $Q$ and $R$ respectively, then $QR$ is equal to

• A. $3\sqrt{6}$
• B. $6$
• C. $6\sqrt{3}$
• D. $4\sqrt{3}$

Answer 1

The correct option is C $6\sqrt{3}$

$P(-2\sqrt{6},\sqrt{3})$ lies on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\Rightarrow \frac{24}{a^2}-\frac{3}{b^2}=1\cdots \left(i\right)$
Now, $\frac{b^2}{a^2}=e^2-1=\frac{1}{4}[\because e=\frac{\sqrt{5}}{2}]$
$\Rightarrow 4b^2=a^2\cdots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right),$ we get
$a^2=12$ and $b^2=3$
Hyperbola: $\frac{x^2}{12}-\frac{y^2}{3}=1$
Tangent at $P(-2\sqrt{6},\sqrt{3})$,
$-\frac{x}{\sqrt{6}}-\frac{y}{\sqrt{3}}=1\Rightarrow Q(0,-\sqrt{3})$
Slope of tangent $=-\frac{1}{\sqrt{2}}$
Normal at $P(-2\sqrt{6},\sqrt{3})$,
$y-\sqrt{3}=\sqrt{2}(x+2\sqrt{6})$
$\Rightarrow R(0,5\sqrt{3})$
$\therefore QR=6\sqrt{3}$

Alternate Solution:
Let $P(a\sec \theta ,b\tan \theta )$
$T:\frac{x}{a}\sec \theta -\frac{y}{b}\tan \theta =1\Rightarrow Q(0,-\frac{b}{\tan \theta })$
$N:\frac{ax}{\sec \theta }+\frac{by}{\tan \theta }=a^2+b^2\Rightarrow R(0,\frac{a^2+b^2}{b}\tan \theta )$
$QR=|\left(\frac{a^2+b^2}{b}\right)\tan \theta +\frac{b}{\tan \theta }|$
$\because \frac{b^2}{a^2}=e^2-1=\frac{1}{4}\Rightarrow 4b^2=a^2$
$\Rightarrow a^2=12$ and $b^2=3$

$\sec \theta =-\sqrt{2}\text{and}\tan \theta =1$

Now, $QR=|\left(\frac{12+3}{\sqrt{3}}\right)1+\frac{\sqrt{3}}{1}|=6\sqrt{3}$