Question

If two concentric ellipses be such that the foci of one be on the other and if $\frac{\sqrt{3}}{2}$ and $\frac{1}{\sqrt{2}}$ be their eccentricities. Then angle between their axes is

• A. ${\cos }^{-1}\frac{1}{\sqrt{6}}$
• B. ${\cos }^{-1}\frac{2}{3\sqrt{3}}$
• C. ${\cos }^{-1}\frac{\sqrt{2}}{3}$
• D. ${\cos }^{-1}\sqrt{\frac{2}{3}}$

Answer 1

The correct option is D ${\cos }^{-1}\sqrt{\frac{2}{3}}$

Let $S$ and $S^{\prime }$ be the foci of one ellipse and $H$ and $H^{\prime }$ be the other, $C$ being their common centre. Then $SH{S}^{\prime }{H}^{\prime }$ is a parallelogram and
Since $SH+S^{\prime }H=H{S}^{\prime }+H^{\prime }{S}^{\prime }=2a$
For standard equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$CS=\sqrt{\frac{\sqrt{3}a}{2}},CH=\frac{a}{\sqrt{2}}$
Let $\theta$ be the angle between their axes.
Then $S{H}^2=\frac{3}{4}{a}^2+\frac{a^2}{2}-\frac{2a^2\sqrt{3}}{2}.\frac{1}{\sqrt{2}}\cos \theta$ ... (1)
$(S^{\prime }H{)}^2=\frac{3}{4}{a}^2+\frac{a^2}{2}-\frac{2a^2\sqrt{3}}{2}.\frac{1}{\sqrt{2}}\cos \theta$ ... (2)

Now, $2a=SH+S^{\prime }H$
Squaring both sides, then
$4a^2=(SH{)}^2+(S^{\prime }H{)}^2+2\left(SH\right).\left(S^{\prime }H\right)$
$\Rightarrow {\cos }^2\theta =\frac{2}{3}$
$\Rightarrow \theta ={\cos }^{-1}\sqrt{\frac{2}{3}}$