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The tangent at a point \mathrm{ P(a \cos \phi, b \sin \phi)} of an ellipse \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1} meets its auxilliary circle in two points, the chord joining which subtends a right angle at the centre, then the eccentricity of the ellipse is

Option: 1

\mathrm{\left(1+\sin ^2 \phi\right)^{-1}}


Option: 2

\mathrm{\left(1+\sin ^2 \phi\right)^{-1 / 2}}


Option: 3

\mathrm{\left(1+\sin ^2 \phi\right)^{-3 / 2}}


Option: 4

\mathrm{\left(1+\sin ^2 \phi\right)^{-2}}


Answers (1)

best_answer

\mathrm{\text { Given ellipse } \frac{x^2}{a^2}+\frac{y^2}{b^2}=1}      ...[i]

\mathrm{\text { Auxiliary circle of eq. (i) is } x^2+y^2=a^2}       ...[ii]

\mathrm{\text { Equation of tangent at a point } P(a \cos \phi, b \sin \phi) \text { on eq.(i) }}

\mathrm{\frac{x}{a} \cos \phi+\frac{y}{b} \sin \phi=1}        [iii]

Making homogeneous equation (ii) with the help of eq. (iii), then

\mathrm{\begin{aligned} & x^2+y^2=a^2\left(\frac{x}{a} \cos \phi+\frac{y}{b} \sin \phi\right)^2 \\ \Rightarrow & x^2 \sin \phi+\left(1-\frac{a^2}{b^2} \sin ^2 \phi\right) y^2-\frac{2 x y}{a b} a^2 \sin \phi \cos \phi=0 \\ \because \quad & \angle R C S=90^{\circ} \end{aligned}}

\mathrm{\therefore \quad \text { Coefficient of } x^2+\text { coefficient of } y^2=0}

\mathrm{\begin{aligned} \therefore & \sin ^2 \phi+1-\frac{a^2}{b^2} \sin ^2 \phi=0 \\ & 1=\left(\frac{a^2}{b^2}-1\right) \sin ^2 \phi \Rightarrow 1=\left(\frac{a^2-b^2}{b^2}\right) \sin ^2 \phi \\ \Rightarrow & 1=\frac{a^2 e^2}{a^2\left(1-e^2\right)} \sin ^2 \phi \Rightarrow 1-e^2=e^2 \sin ^2 \phi \\ \Rightarrow & 1=e^2\left(1+\sin ^2 \phi\right) \\ \Rightarrow & e^2=\frac{1}{\left(1+\sin ^2 \phi\right)}=\left(1+\sin ^2 \phi\right)^{-1} \\ \therefore & e=\left(1+\sin ^2 \phi\right)^{-1 / 2} \end{aligned}}

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