Question

Two concentric ellipses are such that the foci of each one are on the other and length of their major axes are equal. Let $e$ and $e^{\prime }$ be their eccentricities, then

• A. The quadrilateral formed by joining the foci of the two ellipses is a parallelogram
• B. The angle $\theta$ between their axes is given by $\theta ={\cos }^{-1}\sqrt{\frac{1}{e^2}+\frac{1}{e^{\prime 2}}-\frac{1}{e^2{e}^{\prime 2}}}$
• C. If $e^2+e^{\prime 2}=1$, then the angle between the axis of the two ellipses is $90^{\circ }$
• D. If $e+e^{\prime }=1$, then the angle between the axis of the two ellipses is $90^{\circ }$

Answer 1

Answer: (A), (B), (C)

Clearly $O$ is the maidpoint of $S{S}^{\prime }$ and $H{H}^{\prime }$
$\Rightarrow$ Diagonals of quadrilateral IISII 'S' bisect each other, so it is a parallelogram.
I.et $I{I}^{\prime }OII=2r\Rightarrow OII=r=a{e}^{\prime }$
$H$ lics on $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ (supposc)
$\therefore \frac{r^2{\cos }^2\theta }{a^2}+\frac{r^2{\sin }^2\theta }{b^2}=1$
$e^{\prime 2}{\cos }^2\theta +\frac{e^{\prime 2}{\sin }^{2}0}{1-e^2}=1\left[\because b^2=a^2\left(1-e^2\right)\right]$
$\Rightarrow e^{\prime 2}{\cos }^2\theta -\frac{e^{\prime 2}{\cos }^{2}0}{1-e^2}=1-\frac{e^{\prime 2}}{1-e^2}$
$\Rightarrow {\cos }^2\theta =\frac{1}{e^2}+\frac{1}{e^{\prime 2}}-\frac{1}{e^2{e}^{\prime 2}}$
$\Rightarrow \theta =\cos \sqrt{\frac{1}{e^2}+\frac{1}{e^{\prime 2}}-\frac{1}{e^2{e}^{\prime 2}}}$
For $\theta =90^{\circ },\frac{e^2+e^{\prime 2}}{e^2{e}^{\prime 2}}=\frac{1}{e^2{e}^{\prime 2}}\Rightarrow e^2+e^{\prime 2}=1$