Question

If the normal at $'\varphi'$ on the hyperbola $(a\sec\varphi,b\tan\varphi)$ meets transverse axis at G, then $AG.A^{'}G =$

(Where A and $A^{'}$ are the vertices of the hyperbola)

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

Answer 1

Answer: (A)

$a^{2}(e^{4}\sec^{2}\varphi - 1)$

Answer 2

The equation of normal at $(a\sec\varphi,b\tan\varphi)$ to the given hyperbola is $ax\cos\varphi + by\cot\varphi = (a^{2} + b^{2})$ This meets the transverse axis i.e., x-axis at G. So the co-ordinates of G are $\left( \left( \frac{a^{2} + b^{2}}{a} \right)\sec\varphi,0 \right)$ and the co-ordinates of the vertices A and $A^{'}$ are $A(a,0)$ and $A^{'}( - a,0)$ respectively.

$\therefore AG.A^{'}G = \left( - a + \left( \frac{a^{2} + b^{2}}{a} \right)\sec\varphi \right)\left( a + \left( \frac{a^{2} + b^{2}}{a} \right)\sec\varphi \right)$=$\left( \frac{a^{2} + b^{2}}{a} \right)^{2}\sec^{2}\varphi - a^{2}$=$(ae^{2})^{2}\sec^{2}\varphi - a^{2}$=$a^{2}(e^{4}\sec^{2}\varphi - 1)$