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Let $E_1$ and $E_2$ be two ellipses whose centers are at the origin. The major axes of $E_1$ and $E_2$ lie along the $x$-axis and the $y$-axis, respectively. Let $S$ be the circle $x^2+(y-1)^2=2$. The straight line $x+y=3$ touches the curves $S, E_1$ ad $E_2$ at $P, Q$ and $R$, respectively. Suppose that $P Q=P R=\frac{2 \sqrt{2}}{3}$. If $e_1$ and $e_2$ are the eccentricities of $E_1$ and $E_2$, respectively, then the correct expression$(s)$ is(are)
$(A)$ $e_1^2+e_2^2=\frac{43}{40}$
$(B)$ $e_1 e_2=\frac{\sqrt{7}}{2 \sqrt{10}}$
$(C)$ $\left|e_1^2-e_2^2\right|=\frac{5}{8}$
$(D)$ $e_1 e_2=\frac{\sqrt{3}}{4}$
$(A,B)$
$(B,D)$
$(B,C)$
$(A,C)$
Solution
For the given line, point of contact for $E_1: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $\left(\frac{a^2}{3}, \frac{b^2}{3}\right)$ and for $E_2: \frac{x^2}{B^2}+\frac{y^2}{A^2}=1$ is $\left(\frac{B^2}{3}, \frac{A^2}{3}\right)$
Point of contact of $x+y=3$ and circle is (1,2)
Also, general point on $x+y=3$ can be taken as $\left(1 \mp \frac{r}{\sqrt{2}}, 2 \pm \frac{r}{\sqrt{2}}\right)$ where, $r=\frac{2 \sqrt{2}}{3}$
So, required points are $\left(\frac{1}{3}, \frac{8}{3}\right)$ and $\left(\frac{5}{3}, \frac{4}{3}\right)$
Comparing with points of contact of ellipse, $a^2=5, B^2=8$
$b^2=4, A^2=1$
$\therefore e_1 e_2=\frac{\sqrt{7}}{2 \sqrt{10}}$ and $e_1^2+e_2^2=\frac{43}{40}$
