JEE MATHEMATICS MCQ: ELLIPSE

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Application +4 | -1

\omega

is one of the angles between the normals to the ellipse

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

at the points whose eccentric angles are

\theta

and

\frac{\pi}{2}+\theta

, then

\frac{2 \cos \omega}{\sin 2 \theta}

View answer and explanation

Correct option: A

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

Option A,

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

, is the correct expression for

\frac{2 \cos \omega}{\sin 2 \theta}

derived from the angle between the normals to the ellipse.

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O