The product of the perpendicular distances from any point on the hyperbola…
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
$\frac{a^2 b^2}{a^2-b^2}$
$\frac{a^2 b^2}{a^2+b^2}$
$\frac{a^2+b^2}{a^2 b^2}$
$\frac{a^2-b^2}{a^2 b^2}$
Solution
Let $(a \sec \theta, b \tan \theta)$ be any point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
The equation of the asymptotes of the given hyperbola are
$\left(\frac{x}{a}+\frac{y}{b}\right)=0 \text { and }\left(\frac{x}{a}-\frac{y}{b}\right)=0$
Now, $P_1=$ length of the perpendicular from $(a \sec \theta, b \tan \theta$ ) on
$\left(\frac{x}{a}+\frac{y}{b}\right)=0$
$\Rightarrow \quad P_1=\frac{\sec \theta+\tan \theta}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}$...(i)
$P_2=$ length of the perpendicular from $(a \sec \theta, b \tan \theta)$ on
$\left(\frac{x}{a}-\frac{y}{b}\right)=0$
$\Rightarrow \quad P_2=\frac{\sec \theta-\tan \theta}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}$...(ii)
$\therefore \quad P_1 P_2=\frac{(\sec \theta+\tan \theta)}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}} \cdot \frac{(\sec \theta-\tan \theta)}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}$
$=\frac{\left(\sec ^2 \theta-\tan ^2 \theta\right)}{\left(\frac{1}{a^2}+\frac{1}{b^2}\right)}=\frac{1}{\frac{\left(a^2+b^2\right)}{a^2 b^2}}$
$\Rightarrow \quad P_1 P_2=\frac{a^2 b^2}{a^2+b^2}$