Step-by-step Solution

1. Analyze the given information

• Ellipse E₁: Center (0,0), major axis along the x-axis. Equation: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a > b$. Eccentricity is $e_1$, where $e_1^2 = 1 - \frac{b^2}{a^2}$.
• Ellipse E₂: Center (0,0), major axis along the y-axis. Equation: $\frac{x^2}{c^2} + \frac{y^2}{d^2} = 1$ with $d > c$. Eccentricity is $e_2$, where $e_2^2 = 1 - \frac{c^2}{d^2}$.
• Circle S: Equation: $x^2 + (y-1)^2 = 2$. Center is C(0,1) and radius is $r = \sqrt{2}$.
• Line L: Equation: $x+y=3$. This line is tangent to S, E₁, and E₂.
• Points of Tangency: P on S, Q on E₁, R on E₂.
• Given Distances: $PQ = PR = \frac{2\sqrt{2}}{3}$.

2. Find the coordinates of the point of tangency P on the circle S

The point of tangency P is the foot of the perpendicular from the center of the circle C(0,1) to the tangent line $x+y-3=0$.

The equation of the line perpendicular to $x+y=3$ and passing through C(0,1) is of the form $x-y+k=0$. Since it passes through (0,1), we have $0-1+k=0$, so $k=1$. The line is $x-y+1=0$, or $y=x+1$.

The point P is the intersection of the tangent line $x+y=3$ and the normal line $y=x+1$. Substituting $y=x+1$ into $x+y=3$ gives: $x + (x+1) = 3$ $2x+1 = 3 \implies 2x = 2 \implies x = 1$ Then $y = 1+1 = 2$. So, the coordinates of P are $(1,2)$.

3. Find the coordinates of points Q and R

Points Q and R lie on the line $x+y=3$ at a distance of $d = \frac{2\sqrt{2}}{3}$ from P(1,2). We can parameterize the line $x+y=3$. The slope is -1, which corresponds to an angle of 135° with the positive x-axis. The direction vector is $(\cos{135°}, \sin{135°}) = (-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$. The coordinates of a point on the line at a distance $t$ from P(1,2) are given by: $(x,y) = (1 + t\cos{135°}, 2 + t\sin{135°}) = (1 - \frac{t}{\sqrt{2}}, 2 + \frac{t}{\sqrt{2}})$ For Q and R, $t = \pm \frac{2\sqrt{2}}{3}$.

• For $t = \frac{2\sqrt{2}}{3}$: $x = 1 - \frac{1}{\sqrt{2}}\left(\frac{2\sqrt{2}}{3}\right) = 1 - \frac{2}{3} = \frac{1}{3}$ $y = 2 + \frac{1}{\sqrt{2}}\left(\frac{2\sqrt{2}}{3}\right) = 2 + \frac{2}{3} = \frac{8}{3}$ One point is $(\frac{1}{3}, \frac{8}{3})$.

• For $t = -\frac{2\sqrt{2}}{3}$: $x = 1 - \frac{1}{\sqrt{2}}\left(-\frac{2\sqrt{2}}{3}\right) = 1 + \frac{2}{3} = \frac{5}{3}$ $y = 2 + \frac{1}{\sqrt{2}}\left(-\frac{2\sqrt{2}}{3}\right) = 2 - \frac{2}{3} = \frac{4}{3}$ The other point is $(\frac{5}{3}, \frac{4}{3})$. So, the set of points {Q, R} is ${(\frac{5}{3}, \frac{4}{3}), (\frac{1}{3}, \frac{8}{3})}$.

4. Determine the parameters of the ellipses E₁ and E₂

The line $y = -x+3$ is tangent to the ellipse $\frac{x^2}{\alpha^2} + \frac{y^2}{\beta^2} = 1$. The condition for tangency is $c^2 = \alpha^2 m^2 + \beta^2$. Here, $m=-1$ and the y-intercept is $c=3$. So, $3^2 = \alpha^2(-1)^2 + \beta^2 \implies \alpha^2 + \beta^2 = 9$. The point of tangency is given by $(-\frac{\alpha^2 m}{c}, \frac{\beta^2}{c}) = (\frac{\alpha^2}{3}, \frac{\beta^2}{3})$.

• For Ellipse E₁: Point of tangency is $Q(x_Q, y_Q) = (\frac{a^2}{3}, \frac{b^2}{3})$. Since the major axis is along the x-axis, we must have $a^2 > b^2$, which implies $x_Q > y_Q$. Comparing the two possible points, $(\frac{5}{3}, \frac{4}{3})$ has $x > y$. So, $Q = (\frac{5}{3}, \frac{4}{3})$. $\frac{a^2}{3} = \frac{5}{3} \implies a^2 = 5$ $\frac{b^2}{3} = \frac{4}{3} \implies b^2 = 4$ We check: $a^2=5 > b^2=4$ (consistent) and $a^2+b^2=5+4=9$ (consistent).

• For Ellipse E₂: Point of tangency is $R(x_R, y_R) = (\frac{c^2}{3}, \frac{d^2}{3})$. Since the major axis is along the y-axis, we must have $d^2 > c^2$, which implies $y_R > x_R$. The remaining point is $(\frac{1}{3}, \frac{8}{3})$, which has $y > x$. So, $R = (\frac{1}{3}, \frac{8}{3})$. $\frac{c^2}{3} = \frac{1}{3} \implies c^2 = 1$ $\frac{d^2}{3} = \frac{8}{3} \implies d^2 = 8$ We check: $d^2=8 > c^2=1$ (consistent) and $c^2+d^2=1+8=9$ (consistent).

5. Calculate the eccentricities e₁ and e₂

• For E₁: $a^2 = 5, b^2 = 4$. $e_1^2 = 1 - \frac{b^2}{a^2} = 1 - \frac{4}{5} = \frac{1}{5}$

• For E₂: $c^2 = 1, d^2 = 8$. $e_2^2 = 1 - \frac{c^2}{d^2} = 1 - \frac{1}{8} = \frac{7}{8}$

6. Evaluate the given options

• A: $e_1^2 + e_2^2 = \frac{43}{40}$ $e_1^2 + e_2^2 = \frac{1}{5} + \frac{7}{8} = \frac{8 \times 1 + 5 \times 7}{40} = \frac{8+35}{40} = \frac{43}{40}$ This option is correct.

• B: $e_1 e_2 = \frac{\sqrt{7}}{2\sqrt{10}}$ $e_1 = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}}$ $e_2 = \sqrt{\frac{7}{8}} = \frac{\sqrt{7}}{2\sqrt{2}}$ $e_1 e_2 = \frac{1}{\sqrt{5}} \times \frac{\sqrt{7}}{2\sqrt{2}} = \frac{\sqrt{7}}{2\sqrt{10}}$ This option is correct.

• C: $|e_1^2 + e_2^2| = \frac{5}{8}$ We found $e_1^2 + e_2^2 = \frac{43}{40}$. Since $\frac{5}{8} = \frac{25}{40}$, this option is incorrect.

• D: $e_1 e_2 = \frac{\sqrt{3}}{4}$ We found $e_1 e_2 = \frac{\sqrt{7}}{2\sqrt{10}}$. Squaring both gives $(e_1 e_2)^2 = \frac{7}{40}$ and $(\frac{\sqrt{3}}{4})^2 = \frac{3}{16}$. Since $\frac{7}{40} \neq \frac{3}{16}$, this option is incorrect.

The correct expressions are (A) and (B).