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

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

(b) Equation of the auxiliary circle is 

                                     x^2+y^2=a^2\: \: \: \: \: \: \: \: \: \: \: \: \: \: ...(i)

Equation of the tangent at point P(a \cos \phi, b \sin \phi) is 

                              \left(\frac{x}{a}\right) \cos \phi+\left(\frac{y}{b}\right) \sin \phi=1\: \: \: \: \: \: \: \: \: \: \: \: \: ...(ii)

which meets the auxiliary circle at points A and B.  \therefore  Equation of pair of lines OA and OB is obtained by making homogenous Eq. (i) with the heln of Eq (ii) as 

\begin{gathered} x^2+y^2=a^2\left(\frac{x}{a} \cos \phi+\frac{y}{b} \sin \phi\right)^2 \\ \\\left(1-\cos ^2 \phi\right) x^2-\frac{2 x y a \sin \phi \cos \phi}{b}+\left(1-\frac{a^2}{b^2} \sin ^2 \phi\right) y^2=0 \end{gathered}

But                                  \angle A O B=90^{\circ}

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

\begin{array}{llrl} \Rightarrow & 1-\cos ^2 \phi+1-\left(\frac{a^2}{b^2}\right) \sin ^2 \phi=0 \\ \\\Rightarrow & \sin ^2 \phi\left(1-\frac{a^2}{b^2}\right)+1=0 \\ \\\Rightarrow & \left(a^2-b^2\right) \sin ^2 \phi=b^2 \end{array}

\Rightarrow\: \: a^2 e^2 \sin ^2 \phi=a^2\left(1-e^2\right)

\begin{array}{ll} \Rightarrow & \left(1+\sin ^2 \phi\right) e^2=1 \\ \\\Rightarrow & e=\frac{1}{\sqrt{1+\sin ^2 \phi}}=\left(1+\sin ^2 \phi\right)^{-1 / 2} \end{array}

Posted by

Ramraj Saini

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