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  • The normal at  to a hyperbola of eccentricity <img alt="\mathrm{e}" sr

The normal at \mathrm{P} to a hyperbola of eccentricity \mathrm{e}, intersects its transverse and conjugate axis at \mathrm{L} and \mathrm{M} respectively, then the locus of the middle point of \mathrm{LM} is a hyperbola whose eccentricity is 

Option: 1

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


Option: 2

\mathrm{\frac{e}{\sqrt{e^4-1}}}


Option: 3

\mathrm{\frac{e}{\sqrt{a^2 e^2-1}}}


Option: 4

None of these


Answers (1)

best_answer

The equation of the normal at \mathrm{P(a \sec \phi, b \tan \phi)} to the hyperbola is \mathrm{a x \cos \phi+b y \cot \phi=a^2+b^2=a^2 e^2}
It meets the transverse and conjugate axes at \mathrm{L} and \mathrm{M}, then \mathrm{L\left(a e^2 \sec \phi, 0\right) . M\left(0, \frac{a^2 e^2 \tan \phi}{b}\right)}
Let the middle point of \mathrm{LM} is \mathrm{(\alpha, \beta)_{\text {; then }} \alpha=\frac{a e^2 \sec \phi}{2} \Rightarrow \operatorname{seo} \phi=\frac{2 \alpha}{a e^2}\ \ \ \ ...........(i)}
\mathrm{\text { and } \beta=\frac{a^2 e^2 \tan \phi}{2 b} \Rightarrow \tan \phi=\frac{2 b \beta}{a^2 e^2}\ \ \ \ ...........(ii)}
\mathrm{1=\sec ^2 \phi-\tan ^2 \phi ; \quad 1=\frac{4 \alpha^2}{a^2 e^4}-\frac{4 b^2 \beta^2}{a^4 e^4}\ , \quad \therefore \text { Locus of }(\alpha, \beta) \quad \frac{x^2}{\left(\frac{a^2 e^4}{4}\right)}-\frac{y^2}{\left(\frac{a^4 e^4}{4 b^2}\right)}=1}
It is a hyperbola, let its eccentricity
\mathrm{e_1=\frac{\sqrt{\left(\frac{a^2 e^4}{4}+\frac{a^4 e^4}{4 b^2}\right)}}{\left(\frac{a^2 e^4}{4}\right)}-\sqrt{1+\frac{a^2}{b^2}}-\sqrt{\frac{a^2+b^2}{b^2}}-\sqrt{\frac{a^2 e^2}{a^2\left(e^2-1\right)}} \quad e_1=\frac{e}{\sqrt{e^2-1}}}

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