Q66 / 160 · TS EAMCET 2010 · Mathematics

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

  • A
    $\frac{a^2 b^2}{a^2-b^2}$
  • Correct answer
    B
    $\frac{a^2 b^2}{a^2+b^2}$
  • C
    $\frac{a^2+b^2}{a^2 b^2}$
  • D
    $\frac{a^2-b^2}{a^2 b^2}$

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) $\therefore \quad P_1 P_2=\frac{(\sec \theta+\tan \theta)}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}} \cdot \frac{(\sec \theta-\tan \theta)}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}$ $=\frac{\left(\sec ^2 \theta-\tan ^2 \theta\right)}{\left(\frac{1}{a^2}+\frac{1}{b^2}\right)}=\frac{1}{\frac{\left(a^2+b^2\right)}{a^2 b^2}}$ $\Rightarrow \quad P_1 P_2=\frac{a^2 b^2}{a^2+b^2}$

Hence, the correct answer is option: 'B'.

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