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Question: The normal at P to a hyperbola of eccentricity e, intersects its transverse and conjugate axis at L ...

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

A

ee2−1\frac{e}{\sqrt{e^{2} - 1}}

B

ee4−1\frac{e}{\sqrt{e^{4} - 1}}

C

ea2e2−1\frac{e}{\sqrt{a^{2}e^{2} - 1}}

D

None

Answer

ee2−1\frac{e}{\sqrt{e^{2} - 1}}

Explanation

Solution

The equation of the normal at P(asec⁡φ,btan⁡φ)P(a\sec\varphi,b\tan\varphi) to the hyperbola is axcos⁡φ+bycot⁡φ=a2+b2=a2e2ax\cos\varphi + by\cot\varphi = a^{2} + b^{2} = a^{2}e^{2}It meets the transverse and conjugate axes at L and M, then L(ae2sec⁡φ,0)L(ae^{2}\sec\varphi,0); M(0,a2e2tan⁡φb)M\left( 0,\frac{a^{2}e^{2}\tan\varphi}{b} \right)

Let the middle point of LM is (α,β)(\alpha,\beta); then α=ae2sec⁡φ2\alpha = \frac{ae^{2}\sec\varphi}{2}

⇒ sec⁡φ=2αae2\sec\varphi = \frac{2\alpha}{ae^{2}} .....(i)

and β=a2e2tan⁡φ2b\beta = \frac{a^{2}e^{2}\tan\varphi}{2b} ⇒ tan⁡φ=2bβa2e2\tan\varphi = \frac{2b\beta}{a^{2}e^{2}} ......(ii)

∵\because 1=sec⁡2φ−tan⁡2φ1 = \sec^{2}\varphi - \tan^{2}\varphi; 1=4α2a2e4−4b2β2a4e41 = \frac{4\alpha^{2}}{a^{2}e^{4}} - \frac{4b^{2}\beta^{2}}{a^{4}e^{4}}, ∴\therefore Locus of (α,β)(\alpha,\beta) is x2(a2e44)−y2(a4e44b2)=1\frac{x^{2}}{\left( \frac{a^{2}e^{4}}{4} \right)} - \frac{y^{2}}{\left( \frac{a^{4}e^{4}}{4b^{2}} \right)} = 1

It is a hyperbola, let its eccentricity

e1=(a2e44+a4e44b2)(a2e44)=1+a2b2=a2+b2b2=a2e2a2(e2−1)e_{1} = \frac{\sqrt{\left( \frac{a^{2}e^{4}}{4} + \frac{a^{4}e^{4}}{4b^{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}(e^{2} - 1)}};

∴\therefore e1=ee2−1e_{1} = \frac{e}{\sqrt{e^{2} - 1}}