Question

If the normal at '$\theta$' on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ meets the transverse axis at $G$ then $AG\times A^{\prime }G$ is ( where $A$ and $A^{\prime }$ are the vertices of the hyperbola)

• A. $a^2\sec \theta$
• B. $a^2(e^4{\sec }^2\theta +1)$
• C. $a^2(e^4{\sec }^2\theta -1)$
• D. None of these

Answer 1

The correct option is C $a^2(e^4{\sec }^2\theta -1)$

The equation of the normal at $(a\sec \theta ,b\tan \theta )$ to the given hyperbola is
$ax\cos \theta +by\cot \theta =(a^2+b^2)$
This meets the transverse axis (i.e) at $G$. So, the coordinates of $G$ are $\{\frac{a^2+b^2}{a}\sec \theta ,0\}$
The coordinates of the vertices $A$ and $A^{\prime }$ are $A(a,0)$ and $A^{\prime }(-a,0)$ respectively
$\therefore AG.A^{\prime }G=(-a+\frac{a^2+b^2}{a}\sec \theta )(a+\frac{a^2+b^2}{a}\sec \theta )$
$\Rightarrow AG.A^{\prime }G=(-a+a{e}^2\sec \theta )(a+a{e}^2\sec \theta )$
$=a^2(e^4{\sec }^2\theta -1)$