Question

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 a hyperbola, then eccentricity of the hyperbola is

• A. $\frac{e+1}{e-1}$
• B. $\frac{e}{\sqrt{e^2-1}}$
• C. $e$
• D. none of these

Answer 1

Answer: (B)

$\frac{e}{\sqrt{e^2-1}}$

The equation of the normal at $P(a\sec \theta ,b\tan \theta )$ to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is
$ax\cos \theta +by\cot \theta =a^2+b^2$
This intersects the transverse and conjugate axes at $L(\frac{a^2+b^2}{a}\sec \theta ,0)$ and
$M(0,\frac{a^2+b^2}{b^{}}\tan \theta )$ respectively
Let $N(h,k)$. then $h=\frac{a^2+b^2}{b}\sec \theta$ and
$k=\frac{a^2+b^2}{b^{}}\tan \theta$
$\Rightarrow \sec \theta =\frac{2ah}{a^2+b^2},\tan \theta =\frac{2bk}{a^2+b^2}$
$\therefore {\sec }^2\theta -{\tan }^2\theta =1\Rightarrow 4a^2{h}^2-4b^2{k}^2={(a^2+b^2)}^2$
Thus the locus of
$(h,k)$ is $\Rightarrow 4a^2{x}^2-4b^2{y}^2={(a^2+b^2)}^2$
Let $e_1$ be the eccentricity of this hyperbola. Then
${e_1}^2=1+\frac{a^2}{b^2}=\frac{a^2+b^2}{b^2}=\frac{a^2{e}^2}{a^2(e^2-1)}$
$\Rightarrow e_1=\frac{e}{\sqrt{e^2-1}}$