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  • If the normal at  on the hyperbola <img alt="\mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}" sr

If the normal at '\phi ' on the hyperbola \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1} meets transverse axis at \mathrm{G}, then \mathrm{AG.A'G=}
(Where \mathrm{A} and \mathrm{A'} are the vertices of the hyperbola) 
 

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

None of these


Answers (1)

best_answer

The equation of normal at \mathrm{(a \sec \phi, b \tan \phi)} to the given hyperbola is
\mathrm{a x \cos \phi+b y \cot \phi=\left(a^2+b^2\right)}
This meets the transverse axis i.e., \mathrm{x}- axis at \mathrm{G}. So the co-ordinates of \mathrm{G} are 
\mathrm{\left(\left(\frac{a^2+b^2}{a}\right) \sec \phi, 0\right)} and the co-ordinates of the vertices \mathrm{A} and \mathrm{A^{\prime} \text { are } A(a, 0) \text { and } A^{\prime}(-a, 0)} respectively .
\begin{aligned} &\mathrm{ \quad A G \cdot A^{\prime} G=\left(-a+\left(\frac{a^2+b^2}{a}\right) \sec \phi\right)\left(a+\left(\frac{a^2+b^2}{a}\right) \sec \phi\right)=\left(\frac{a^2+b^2}{a}\right)^2 \sec ^2 \phi-a^2= }\\ &\mathrm{ \left(a e^2\right)^2 \sec ^2 \phi-a^2=a^2\left(e^4 \sec ^2 \phi-1\right)} \end{aligned} 

Posted by

Irshad Anwar

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