Q. The product of the perpendicular from two foci on any tangent to the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$, is

A) $a^{2}$
B) $b^{2}$
C) $- a^{2}$
D) $- b^{2}$
Explanation

(b)
Let equation of tangent to hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ is
$y = mx + \sqrt{a^{2}m^{2} - b^{2}}$
$ie,\ mx - y + \sqrt{a^{2}m^{2} - b^{2}} = 0$
$\therefore\$Required product
$= \left| \frac{mae + \sqrt{a^{2}m^{2} - b^{2}}}{\sqrt{m^{2} + 1}} \right|\left| \frac{- mae + \sqrt{a^{2}m^{2} - b^{2}}}{\sqrt{m^{2} + 1}} \right|$
$= \left| \frac{a^{2}m^{2} - b^{2} - m^{2}a^{2}e^{2}}{m^{2} + 1} \right|$
$= \left| \frac{m^{2}a^{2}\left( 1 - e^{2} \right) - b^{2}}{m^{2} + 1} \right|$
$= \left| \frac{- m^{2}b^{2} - b^{2}}{m^{2} + 1} \right|\ \ \ \ \left\lbrack \because b^{2} = a^{2}\left( e^{2} - 1 \right) \right\rbrack$
$= b^{2}$

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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