Eccentricities of Ellipses Tangent to Straight Line

To determine the correct options, we first analyze the given geometric configurations step by step.

Step 1: Point of Contact $P$ on the Circle $S$

The circle $S$ has the equation: $x^2 + (y - 1)^2 = 2$

The line given is $x + y = 3$, which can be rewritten as $x = 3 - y$. Substituting this into the equation of the circle $S$: $(3 - y)^2 + (y - 1)^2 = 2$ $9 - 6y + y^2 + y^2 - 2y + 1 = 2$ $2y^2 - 8y + 8 = 0 \implies (y - 2)^2 = 0 \implies y = 2$

Correspondingly, $x = 3 - 2 = 1$. Thus, the point of contact $P$ of the line $x + y = 3$ with the circle $S$ is: $P = (1, 2)$

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Step 2: Finding the Points $Q$ and $R$

The points $Q$ and $R$ lie on the line $x + y = 3$ at a distance $PQ = PR = \frac{2\sqrt{2}}{3}$ from $P(1, 2)$.

Using the parametric equation of the line passing through $(1, 2)$ at an angle $\theta = -\frac{\pi}{4}$: $x = 1 + r \cos\left(-\frac{\pi}{4}\right) = 1 + \frac{r}{\sqrt{2}}$ $y = 2 + r \sin\left(-\frac{\pi}{4}\right) = 2 - \frac{r}{\sqrt{2}}$

For $r = \pm \frac{2\sqrt{2}}{3}$: $x = 1 \pm \frac{2}{3} \implies x_1 = \frac{5}{3}, \quad x_2 = \frac{1}{3}$ $y = 2 \mp \frac{2}{3} \implies y_1 = \frac{4}{3}, \quad y_2 = \frac{8}{3}$

So, the two points on the line are $\left(\frac{5}{3}, \frac{4}{3}\right)$ and $\left(\frac{1}{3}, \frac{8}{3}\right)$.

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Step 3: Equations and Eccentricities of Ellipses $E_1$ and $E_2$

For Ellipse $E_1$:

The equation of $E_1$ with major axis along the $x$-axis is: $E_1: \frac{x^2}{a_1^2} + \frac{y^2}{b_1^2} = 1 \quad (a_1 > b_1)$

The equation of the tangent at $(x_0, y_0)$ is: $\frac{x x_0}{a_1^2} + \frac{y y_0}{b_1^2} = 1$

Comparing this with $x + y = 3$ (or $\frac{x}{3} + \frac{y}{3} = 1$), we get: $x_0 = \frac{a_1^2}{3}, \quad y_0 = \frac{b_1^2}{3}$

Since $a_1 > b_1$, we must have $x_0 > y_0$. Therefore, $Q$ corresponds to the point $\left(\frac{5}{3}, \frac{4}{3}\right)$: $\frac{a_1^2}{3} = \frac{5}{3} \implies a_1^2 = 5$ $\frac{b_1^2}{3} = \frac{4}{3} \implies b_1^2 = 4$

The eccentricity $e_1$ of $E_1$ is: $e_1^2 = 1 - \frac{b_1^2}{a_1^2} = 1 - \frac{4}{5} = \frac{1}{5}$

For Ellipse $E_2$:

The equation of $E_2$ with major axis along the $y$-axis is: $E_2: \frac{x^2}{a_2^2} + \frac{y^2}{b_2^2} = 1 \quad (b_2 > a_2)$

The tangent at $(x_0, y_0)$ gives: $x_0 = \frac{a_2^2}{3}, \quad y_0 = \frac{b_2^2}{3}$

Since $b_2 > a_2$, we must have $y_0 > x_0$. Therefore, $R$ corresponds to the point $\left(\frac{1}{3}, \frac{8}{3}\right)$: $\frac{a_2^2}{3} = \frac{1}{3} \implies a_2^2 = 1$ $\frac{b_2^2}{3} = \frac{8}{3} \implies b_2^2 = 8$

The eccentricity $e_2$ of $E_2$ is: $e_2^2 = 1 - \frac{a_2^2}{b_2^2} = 1 - \frac{1}{8} = \frac{7}{8}$

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Step 4: Verification of the Statements

1. Checking $e_1^2 + e_2^2$: $e_1^2 + e_2^2 = \frac{1}{5} + \frac{7}{8} = \frac{8 + 35}{40} = \frac{43}{40}$ Hence, Statement (A) is correct.

2. Checking $e_1 e_2$: $e_1 e_2 = \sqrt{\frac{1}{5}} \cdot \sqrt{\frac{7}{8}} = \sqrt{\frac{7}{40}} = \frac{\sqrt{7}}{2\sqrt{10}}$ Hence, Statement (B) is correct.

3. Checking $\left| e_1^2 - e_2^2 \right|$: $\left| e_1^2 - e_2^2 \right| = \left| \frac{1}{5} - \frac{7}{8} \right| = \frac{27}{40} \neq \frac{5}{8}$ Hence, Statement (C) is incorrect.

4. Checking Statement (D): Since $e_1 e_2 = \frac{\sqrt{7}}{2\sqrt{10}} \neq \frac{\sqrt{3}}{4}$, Statement (D) is incorrect.

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Conclusion

The correct options are A and B.