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If the product of the perpendicular distances from any point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ to its asymptotes is 6 and eccentricity of the hyperbola is $\sqrt{3}$, then the length of the conjugate axis of the hyperbola is
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The correct answer is:
$6$
$P Q=\left|\frac{a b \sec \theta-a b \tan \theta}{\sqrt{b^2+a^2}}\right|$
and $P R=\left|\frac{a b \sec \theta+a b \tan \theta}{\sqrt{b^2+a^2}}\right|$
According to question,
$\frac{a b(\sec \theta-\tan \theta)}{\sqrt{a^2+b^2}} \times \frac{a b(\sec \theta+\tan \theta)}{\sqrt{a^2+b^2}}=6$
$\Rightarrow \frac{a^2 b^2\left(\sec ^2 \theta-\tan ^2 \theta\right)}{\left(a^2+b^2\right)}=6$
$\Rightarrow \quad \frac{a^2 b^2}{a^2 e^2}=6 \quad\left[\because a^2+b^2=a^2 e^2\right]$
$\begin{array}{ll}\Rightarrow & b^2=6 \times e^2=6 \times 3=18 \\ \because & a^2+b^2=a^2 e^2\end{array}$
$\therefore \quad a^2+18=a^2 \times 3 \Rightarrow a^2=9$
$\Rightarrow \quad a=3$
Length of transverse axis $=2 \times 3=6$
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