Explanation

(b)
Equation of normal at $P\ (a\sec{\phi,\ b\tan{\phi)}}$ is
$a\ x\cos{\phi + by\cot{\phi = a^{2} + b^{2}}}$
Then, coordinates of $L$ and $M$ are
$\left( \frac{a^{2} + b^{2}}{a}.\sec{\phi,\ 0} \right)\ and\ \left( 0,\frac{a^{2} + b^{2}}{b}\tan\phi \right)$ respectively.
Let mid point of $ML$ is $Q\ (h,\ k),$
Then $h = \frac{(a^{2} + b^{2})}{2a}\sec\phi$
$\therefore\sec{\phi = \frac{2ah}{(a^{2} + b^{2})}}\ \ \ \ldots(i)$
and $k = \frac{(a^{2} + b^{2})}{2b}\tan\phi$
$\therefore\ \tan{\phi = \frac{2bk}{{(a}^{2} + b^{2})}}$ …(ii)
From Eqs.(i) and (ii), we get
$\sec^{2}{\phi - \tan^{2}{\phi = \frac{4a^{2}h^{2}}{\left( a^{2} + b^{2} \right)^{2}} - \frac{4b^{2}k^{2}}{\left( a^{2} + b^{2} \right)^{2}}}}$
Hence, required locus is
$\frac{x^{2}}{\left( \frac{a^{2} + b^{2}}{2a} \right)} - \frac{y^{2}}{\left( \frac{a^{2} + b^{2}}{2b} \right)^{2}} = 1$
Let eccentricity of this curve is $e_{1}.$
$\Rightarrow \left( \frac{a^{2} + b^{2}}{2a} \right)^{2} = \left( \frac{a^{2} + b^{2}}{2a} \right)^{2}\left( e_{1}^{2} - 1 \right)$
$\Rightarrow a^{2} = b^{2}\left( e_{1}^{2} - 1 \right)\$
$\Rightarrow \ a^{2} = a^{2}\left( e^{2} - 1 \right)\left( e_{1}^{2} - 1 \right)\ \ \lbrack\because b^{2} = a^{2}\left( e^{2} - 1 \right)\rbrack$
$\Rightarrow e^{2}e_{1}^{2} - e^{2} - e_{1}^{2} + 1 = 1$
$\Rightarrow \ \ e_{1}^{2}\left( e^{2} - 1 \right) = e^{2}$
$\Rightarrow \ \ e_{1} = \frac{e}{\sqrt{e^{2} - 1}}$