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Question: If w is one of the angles between the normals to the ellipse \(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^...

If w is one of the angles between the normals to the ellipse x2a2+y2b2\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 at the points whose eccentric angles are q and π2\frac{\pi}{2}+ q, then 2cot⁡θsin⁡2θ\frac{2\cot\theta}{\sin 2\theta} is-

A

e21−e2\frac{e^{2}}{\sqrt{1 - e^{2}}}

B

e21+e2\frac{e^{2}}{\sqrt{1 + e^{2}}}

C

e21−e2\frac{e^{2}}{1 - e^{2}}

D

e21+e2\frac{e^{2}}{1 + e^{2}}

Answer

e21−e2\frac{e^{2}}{\sqrt{1 - e^{2}}}

Explanation

Solution

The equations of the normals to the ellipse x2a2+y2b2\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 at the points whose eccentric angles are q and π2\frac{\pi}{2} + q are

ax sec q – by cosec q = a2 – b2

and, ax cosec q – by sec q = a2 – b2 respectively.

Since w is the angle between these two normals. Therefore,

tan w = ∣abtan⁡θ+abcot⁡θ1−a2b2∣\left| \frac{\frac{a}{b}\tan\theta + \frac{a}{b}\cot\theta}{1 - \frac{a^{2}}{b^{2}}} \right|

Ž tan w = ∣ab(tan⁡θ+cot⁡θ)b2−a2∣\left| \frac{ab(\tan\theta + \cot\theta)}{b^{2} - a^{2}} \right|

Ž tan w = ∣2ab(sin⁡2θ)(b2−a2)∣\left| \frac{2ab}{(\sin 2\theta)(b^{2} - a^{2})} \right|

Ž tan w = 2ab(a2−b2)sin⁡2θ\frac{2ab}{(a^{2} - \mathbf{b}^{2})\sin 2\theta}

Ž tan w = 2a21−e2a2e2sin⁡2θ\frac{2a^{2}\sqrt{1 - e^{2}}}{a^{2}e^{2}\sin 2\theta}

Ž 2cot⁡ωsin⁡2θ\frac{2\cot\omega}{\sin 2\theta} = e21−e2\frac{e^{2}}{\sqrt{1 - e^{2}}}.