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, $E_{1}$ and $E_{2}$ 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)

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$e_{1}^{2} + e_{2}^{2} = \frac{43}{40}$

$e_{1} e_{2} = \frac{\sqrt{3}}{4}$

$e_{1} e_{2} = \frac{\sqrt{7}}{2 \sqrt{10}}$

$|e_{1}^{2} - e_{2}^{2}| = \frac{5}{8}$

answer is A, B.

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Detailed Solution

The line $x + y = 3$ touches the circle $x^{2} + {(y - 1)}^{2} = 2$ at P

P is foot of perpendicular from $(0, 1)$ to $x + y = 3$

$\Rightarrow$ $P (1, 2)$

Since $P Q = \frac{2 \sqrt{2}}{3} = P R$ we have $Q (1 + \frac{2 \sqrt{2}}{3 \sqrt{2}}, 2 - \frac{2 \sqrt{2}}{3 \sqrt{2}})$

$E_{1} : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ with $e_{1}^{2} = \frac{a^{2} - b^{2}}{a^{2}}$

$∴ a^{2} + b^{2} = 9 (x + y = 3 t o u c h e s E_{1} \Rightarrow a^{2} m^{2} + b^{2} = c^{2})$

$S i n c e Q l i e s o n E_{1}, w e h a v e \frac{25}{9 a^{2}} + \frac{16}{9 b^{2}} = 1$

$\Rightarrow$ $a^{2} = 5, b^{2} = 4$

$∴ e_{1}^{2} = \frac{1}{5}$

similarly

$R (1 - \frac{2}{3}, 2 + \frac{2}{3}) lies on \frac{x^{2}}{c^{2}} + \frac{y^{2}}{d^{2}} = 1 c < d$

$∴ \begin{matrix} c^{2} + d^{2} = 9 \\ \frac{1}{9 c^{2}} + \frac{64}{9 d^{2}} = 1 \end{matrix}\} s o l v i n g c^{2} = 1, d^{2} = 8$

hence, $∴ e_{2}^{2} = \frac{7}{8}$

Verifying the given options, we can see that options (A),(B) are correct.

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