Two concentric ellipses are such that the foci of one are on the other and their major axes are equal. Let e and e' be their eccentricities. Then,

Difficult

A

the quadrilateral formed by joining the foci of the two ellipses is a parallelogram
B

the angle $θ$ between their axes is given by $θ = \cos^{- 1} \sqrt{\frac{1}{e^{2}} + \frac{1}{e^{' 2}} - \frac{1}{e^{2} e^{' 2}}}$
C

If e² + e'² = 1, then the angle between the axes of the two ellipses is 90°
D

none of these

Solution

Clearly, O is the midpoint of SS', and HH'.
Therefore, the diagonals of quadrilateral HSH'S' bisect each other. So it is a parallelogram.
Let H'OH = 2r. Then, OH = r = ae'.
The point H lies on
$\begin{array}{ll} \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (Suppose) \\ ∴ \frac{r^{2} \cos^{2} θ}{a^{2}} + \frac{r^{2} \sin^{2} θ}{b^{2}} = 1 \\ e^{2} \cos^{2} θ + \frac{e^{2} \sin^{2} θ}{1 - e^{2}} = 1 [∵ b^{2} = a^{2} (1 - e^{2})] \\ or e^{2} \cos^{2} θ - \frac{e^{2} \cos^{2} θ}{1 - e^{2}} = 1 - \frac{e^{2}}{1 - e^{2}} \end{array}$
$or \cos^{2} θ = \frac{1}{e^{2}} + \frac{1}{e^{' 2}} - \frac{1}{e^{2} e^{' 2}} or θ = \cos^{- 1} \sqrt{\frac{1}{e^{2}} + \frac{1}{e^{2}} - \frac{1}{e^{2} e^{' 2}}} For θ = 90^{\circ}, \begin{array}{l} \frac{e^{2} + e^{2}}{e^{2} e^{2}} = \frac{1}{e^{2} e^{2}} \\ or e^{2} + {e^{'}}^{2} = 1 \end{array}$

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