Question

The product of the perpendicular from any point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ to its asymptotes, is equal to

• A. $\frac{ab}{a+b}$
• B. $\frac{a^2{b}^2}{a^2+b^2}$
• C. $\frac{2a^2{b}^2}{a^2+b^2}$
• D. None of these

Answer 1

The correct option is B $\frac{a^2{b}^2}{a^2+b^2}$

Let $(asec\theta ,btan\theta )$ be the any point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
The equations of the asymptotes of the given hyperbola are $\frac{x}{a}-\frac{y}{b}=0$ and $\frac{x}{a}+\frac{y}{b}=0$
Now, $p_1$ = length of the perpendicular from $(asec\theta ,btan\theta )$ on $\frac{x}{a}+\frac{y}{b}=0$
$=\frac{sec\theta -tan\theta }{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}$
and $p_2$ = length of the perpendicular from $(asec\theta ,btan\theta )$ $\frac{x}{a}+\frac{y}{b}=0$
$=\frac{sec\theta -tan\theta }{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}$
$\therefore$ $p_1{p}_2=\frac{se{c}^2\theta -ta{n}^2\theta }{\frac{1}{a^2}+\frac{1}{b^2}}=\frac{a^2{b}^2}{a^2+b^2}$