The normal at P to a hyperbola of eccentricity e, intersects its transverse and conjugate axes at L and M respectively. If locus of the mid point of LM is hyperbola, then eccentricity of the hyperbola is

Let $\mathrm{P}(\mathrm{a} \sec \mathrm{\theta }, \mathrm{b} \tan \mathrm{\theta })$, Equation of normal at P is 

$\begin{array}{l}\mathrm{ax} \cos \mathrm{\theta }+\mathrm{by} \cot \mathrm{\theta }={\mathrm{a}}^2+{\mathrm{b}}^2\\ \therefore \mathrm{L}=\left(\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{\mathrm{a}}\sec \mathrm{\theta }, 0\right), \mathrm{M}=\left(0, \frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{\mathrm{b}}\tan \mathrm{\theta }\right)\end{array}$

Let midpoint of $\overline{\mathrm{LM}}=\mathrm{Q}(\mathrm{h}, \mathrm{k})$

$\therefore \sec \mathrm{\theta }=\frac{2ah}{{\mathrm{a}}^2+{\mathrm{b}}^2}, \tan \mathrm{\theta }=\frac{2bk}{{\mathrm{a}}^2+{\mathrm{b}}^2}$,

Now ${\sec }^2\mathrm{\theta }-{\tan }^2\mathrm{\theta }=1$, Locus is

$\frac{{\mathrm{x}}^2}{{\left(\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{2\mathrm{a}}\right)}^2}-\frac{{\mathrm{y}}^2}{{\left(\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{2\mathrm{b}}\right)}^2}=1$

$$\begin{gathered} eccentricity=\sqrt{\frac{{\left(\frac{a^2+b^2}{2a}\right)}^2+{\left(\frac{a^2+b^2}{2b}\right)}^2}{{\left(\frac{a^2+b^2}{2a}\right)}^2}} \\ =\sqrt{\frac{b^2+a^2}{b^2}} \\ =\frac{e}{\sqrt{e^2-1}} \end{gathered}$$