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  • The tangent to the ellipse <img alt="\mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D&plus;%5Cfrac%7By%

The tangent to the ellipse \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1} at a point whose eccentric angle is \mathrm{\theta} cuts the auxiliary circle in points L and M. If LM subtends a right angle at the centre of the ellipse, then its eccentricity at \mathrm{\theta=\pi / 4} is

Option: 1

\sqrt{2 / 3}


Option: 2

\sqrt{1 / 3}


Option: 3

\sqrt{2 / 9}


Option: 4

{1 / 3}


Answers (1)

best_answer

\mathrm{\text { Tangent at } \theta \text { is } \frac{x \cos \theta}{a}+\frac{y \sin \theta}{b}=1}

\mathrm{\text { Auxiliary circle is } x^2+y^2=a^2}

Make homogeneous and put A + B = 0 as the lines joining origin to points of intersection are perpendicular.

\mathrm{\begin{aligned} & \therefore \quad x^2+y^2=a^2\left(x \frac{\cos \theta}{a}+y \frac{\sin \theta}{b}\right)^2 \\ & \therefore \quad\left(1-\cos ^2 \theta\right)+\left(1-\sin ^2 \theta \cdot \frac{a^2}{b^2}\right)=0 \\ & \text { or } \sin ^2 \theta\left[1-\frac{a^2}{a^2\left(1-e^2\right)}\right]=-1 \\ & \text { or }-e^2 \sin ^2 \theta=-\left(1-e^2\right) \\ & \therefore \quad e^2\left(1+\sin ^2 \theta\right)=1 \\ & \therefore \quad e=\left(1+\sin ^2 \theta\right)^{-1 / 2} \\ & \text { Put } \theta=45^{\circ} . \quad \therefore e=\sqrt{\frac{2}{3}} \end{aligned}}

 

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