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If  \mathrm e_1  and  \mathrm e_2  be the eccentricities of two concentric ellipses such that foci of one lie on the other and they have same length of major axes. If \mathrm {e_1^2+e_2^2=p}  then
 

Option: 1

p<1


Option: 2

p>1


Option: 3

p \geq 1


Option: 4

p \leq 1


Answers (1)

best_answer

Let \theta be the angle of inclination of major axes and equation of one of the ellipses be
\mathrm {\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}................(1)

whose foci are \mathrm {S } and  \mathrm {S^{\prime} } and eccentricity \mathrm {=e_1 } 

\therefore \mathrm {\quad b^2=a^2\left(1-e_1^2\right)}

Let \mathrm P and \mathrm P^{\prime}  be foci of other ellipse whose eccentricity \mathrm {=e_2} 
Let \mathrm {P P^{\prime}=2 R \Rightarrow O P=R=a e_2} 
\mathrm {\because \quad Diagonals \: \: P P^{\prime}\: \: and\: \: S S^{\prime}}  bisect each other
\mathrm {\therefore \quad P S^{\prime} P^{\prime} S}  is a parallelogram.
Here, co-ordinates of  \mathrm {P \: \: are\: \: (R \cos \theta, R \sin \theta)} 
\mathrm {\therefore From (i), \frac{R^2 \cos ^2 \theta}{a^2}+\frac{R^2 \sin ^2 \theta}{b^2}=1}

\begin{aligned} & \mathrm {\Rightarrow \frac{a^2 e_2^2\left(1-\sin ^2 \theta\right)}{a^2}+\frac{a^2 e_2^2 \sin ^2 \theta}{a^2\left(1-e_1^2\right)}=1 \quad \text { [Using (ii)]} } \\ & \mathrm {\Rightarrow \sin ^2 \theta\left(\frac{e_2^2}{1-e_1^2}-e_2^2\right)=1-e_2^2} \\ & \mathrm {\Rightarrow \sin ^2 \theta\left(\frac{e_1^2 e_2^2}{1-e_1^2}\right)=1-e_2^2} \\ & \mathrm {\Rightarrow \quad \frac{\left(1-e_1^2\right)\left(1-e_2^2\right)}{e_1^2 e_2^2}=\sin ^2 \theta \leq 1 }\\\\ & \mathrm {\Rightarrow 1-e_1^2-e_2^2+e_1^2 e_2^2 \leq e_1^2 e_2^2} \\\\ & \mathrm {\Rightarrow e_1^2+e_2^2 \geq 1 \Rightarrow p \geq 1} \end{aligned}

 

 

Posted by

himanshu.meshram

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