Q. The product of perpendiculars drawn from any point of a hyperbola to its asymptotes is

A) $\frac{a^{2}b^{2}}{a^{2} + b^{2}}$
B) $\frac{a^{2} + b^{2}}{a^{2}b^{2}}$
C) $\frac{ab}{\sqrt{a} + \sqrt{b}}$
D) $\frac{ab}{a^{2} + b^{2}}$
Explanation

(a)
Let the equation of hyperbola be $\frac{x^{2}}{a} - \frac{y^{2}}{b^{2}} = 1$
Let $(x_{1,}y_{1})$be any point on the hyperbola.
$\frac{x_{1}^{2}}{a_{1}^{2}} - \frac{y_{1}^{2}}{b^{2}} = 1 \Rightarrow \ b^{2}x_{1}^{2} - a^{2}y_{1}^{2} = a^{2}b^{2}\ \ \ldots(i)$
The asymptotes of given hyperbola are
$\frac{X^{2}}{a^{2}}\ \ –\frac{y^{2}}{b^{2}}\ = 0\$
$\therefore$ Product of perpendicular form$\ (x_{1},y_{1})$ to pair of lines $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 0$ is$\$
$\frac{|Ax_{1}^{2} + 2Hx_{1}y_{1} + By_{1}^{2}|}{\sqrt{(A - B)^{2}} + 4H^{2}} = \frac{b^{2}x_{1}^{2} - a^{2}y_{1}^{2}}{\sqrt{\left( b^{2} + a^{2} \right)^{2}}}$
$= \frac{a^{2}b^{2}}{a^{2} + b^{2}}\ \ \ \ \lbrack from\ Eq.\ (i)\rbrack$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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