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  • The normal at P to a hyperbola of eccentricity 2 intersects its transverse and conjugate axis at Q and R. The locus of midpoint of QR is conic of eccentricity<div class='qna-opti

The normal at P to a hyperbola of eccentricity 2 intersects its transverse and conjugate axis at Q and R. The locus of midpoint of QR is conic of eccentricity

Option: 1

\frac{2}{\sqrt{6}}


Option: 2

\frac{\sqrt{6}}{2}


Option: 3

\frac{\sqrt{3}}{4}


Option: 4

\frac{2}{\sqrt{3}}


Answers (1)

best_answer

Consider the hyperbola \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b^2}=1}

Normal at any point \mathrm{P(a \sec \theta, b \tan \theta)} on the hyperbola is

\mathrm{ \frac{a x}{\sec \theta}+\frac{b y}{\tan \theta}=a^2+b^2 }

It meets x-axis at \mathrm{Q\left(\frac{\left(a^2+b^2\right) \sec \theta}{a}, 0\right)} and y-axis at

\mathrm{ R\left(0, \frac{\left(a^2+b^2\right) \tan \theta}{b}\right) \text {. } }

Let midpoint of QR is T(h, k).

 

\mathrm{ \begin{aligned} & \therefore h=\frac{\left(a^2+b^2\right) \sec \theta}{2 a} \text { and } k=\frac{\left(a^2+b^2\right) \tan \theta}{2 b} \\\\ & \Rightarrow h=\frac{a e^2 \sec \theta}{2} \text { and } k=\frac{a^2 e^2}{2 b} \tan \theta \\\\ & \therefore \frac{4 h^2}{a^2 e^4}-\frac{4 k^2 b^2}{a^4 e^4}=1 \\\\ & \Rightarrow \frac{x^2}{\frac{a^2 e^4}{4}}-\frac{y^2}{a^4 e^4}=1 \text {, which is the required locus. } \end{aligned} }

Clearly it is a hyperbola having eccentricity ' E ' given by

\mathrm{\begin{aligned} & E^2=\frac{\frac{a^2 e^4}{4}+\frac{a^4 e^4}{4 b^2}}{\frac{a^2 e^4}{4}}=1+\frac{1}{e^2-1} \\ & \Rightarrow E=\sqrt{\frac{e^2}{e^2-1}}=\sqrt{\frac{4}{4-1}}=\frac{2}{\sqrt{3}} \quad[\because e=2 \text { (given) }] \end{aligned}}

Posted by

Ajit Kumar Dubey

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