Question

If the normal at angle $\varphi$ on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ meets the transverse axis at $G$ such that $AG\cdot A^{\prime }G=a^m(e^n{\sec }^p\varphi -1)$, where $A,A^{\prime }$ are the vertices of the hyperbola, $e$ is the eccentricity and $m,n$ and $p$ are positive integers, then value of $(m+n+p)$ is

Answer 1

The equation of any normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at $\varphi$ is
$ax\cos \varphi +by\cot \varphi =a^2+b^2$


Thus the coordinates of $G=(\frac{a^2+b^2}{a\cos \varphi },0)$
Clearly $A=(a,0)$ and $A^{\prime }=(-a,0)$
$AG=(\frac{a^2+b^2}{a\cos \varphi }-a)$
$A^{\prime }G=(\frac{a^2+b^2}{a\cos \varphi }+a)$
$AG\cdot A^{\prime }G=\left[(\frac{a^2+b^2}{a\cos \varphi }-a)(\frac{a^2+b^2}{a\cos \varphi }+a)\right]$
$=a^2(e^4{\sec }^2\varphi -1)$
$\Rightarrow m=2;n=4;p=2$
$\therefore (m+n+p)=8$