Question

If $P_1$ and $P_2$ are the perpendiculars from any point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ on its asymptotes, then :

• A. $P_1{P}_2=a^2+b^2$
• B. $P_1+P_2=ab$
• C. $P_1+P_2=(ab{)}^2$
• D. $\frac{1}{P_1{P}_2}=\frac{1}{a^2}+\frac{1}{b^2}$

Answer 1

The correct option is D $\frac{1}{P_1{P}_2}=\frac{1}{a^2}+\frac{1}{b^2}$

The equations of the asymptotes of the given hyperbola
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ are
$bx-ay=0$ and $bx+ay=0$

Let $P$ be any point on the given hyperbola be
$(a\sec \varphi ,b\tan \varphi ).$
It is given that, $P_1$ and $P_2$ be the lengths of perpendiculars from the point $P$ to the asymptotes
Thus, $P_1=\left|\frac{ab\sec \varphi -ab\tan \varphi }{\sqrt{b^2+a^2}}\right|$
and $P_2=\left|\frac{ab\sec \varphi +ab\tan \varphi }{\sqrt{b^2+a^2}}\right|$
$\therefore P_1{P}_2=\left|\frac{(ab\sec \varphi -ab\tan \varphi )(ab\sec \varphi +ab\tan \varphi )}{a^2+b^2}\right|$
$\Rightarrow \frac{1}{P_1{P}_2}=\frac{1}{a^2}+\frac{1}{b^2}$