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  • The product of the perpendiculars drawn from any point on the hyperbola <img alt="\mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfra

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

Option: 1

\mathrm{\frac{a^2 b^2}{a^2+b^2}}


Option: 2

\mathrm{\frac{a^2}{a^2+b^2}}


Option: 3

\mathrm{\frac{b^2}{a^2+b^2}}


Option: 4

\mathrm{\frac{a^2-b^2}{a^2+b^2}}


Answers (1)

best_answer

Any point on the hyperbola is \mathrm{P(a \sec \theta, b \tan \theta).} The equations of asymptotes are

\mathrm{ \frac{x}{a}-\frac{y}{b}=0 \text { and } \frac{x}{a}+\frac{y}{b}=0 }                       .......(1)

\mathrm{\therefore \quad}  The length of the perpendicular from P to \mathrm{\frac{x}{a}-\frac{y}{b}=0} is

\mathrm{ p_1=\frac{\sec \theta-\tan \theta}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}} }                                          .......(2)
Similarly, \mathrm{p_2=\frac{\sec \theta+\tan \theta}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}}

\mathrm{ \therefore \quad p_1 p_2=\frac{\sec ^2 \theta-\tan ^2 \theta}{\frac{1}{a^2}+\frac{1}{b^2}}=\frac{a^2 b^2}{a^2+b^2} }

The answer is (a)

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