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The normal at \mathrm{P(a \sec \theta, b \tan \theta)} on the hyperbola \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}meets the transverse axis at G. If A and \mathrm{A^{\prime}} are the vertices of the hyperbola, then \mathrm{A G \cdot A^{\prime} G} is equal to

 

Option: 1

\mathrm{a^2\left(e^4 \sec ^2 \theta-1\right)}


Option: 2

\mathrm{a^2\left(e^2 \sec ^2 \theta-1\right)}


Option: 3

\mathrm{a^2\left(e^4 \sec ^2 \theta+1\right)}


Option: 4

\mathrm{\text { none of these }}


Answers (1)

The equation of normal at \mathrm{ (a \sec \theta, b \tan \theta)} is

\mathrm{ a x \cos \theta+b y \cot \theta=a^2+b^2 }

It meets the x-axis at G.

\mathrm{ \therefore \quad G \text { is }\left(\frac{a^2+b^2}{a} \sec \theta, 0\right)=\left(a e^2 \sec \theta, 0\right) }

as \mathrm{\quad b^2=a^2\left(e^2-1\right)}

\mathrm{ \begin{aligned} \therefore \quad A G \cdot A^{\prime} G & =(O G-O A)\left(O G+O A^{\prime}\right),\left(\text { But } O A=O A^{\prime}=a\right) \\\\ & =\left(a e^2 \sec \theta-a\right)\left(a e^2 \sec \theta+a\right) \\\\ & =a^2\left(e^4 \sec ^2 \theta-1\right) \end{aligned} }
The answer is (a)

Posted by

Ramraj Saini

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