Here, $${E_1}:{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1,\,(a > b)$$

$${E_2}:{{{x^2}} \over {{c^2}}} + {{{y^2}} \over {{d^2}}} = 1,\,(c < d)$$

and $$S:{x^2} + {(y - 1)^2} = 2$$

as tangent to E₁, E₂ and S is $$x + y = 3$$.

Let the point of contact of tangent be $$({x_1},{y_1})$$ to S.

$$\therefore$$ $$x\,.\,{x_1} + y\,.\,{y_1} - (y + {y_1}) + 1 = 2$$

or $$x{x_1} + y{y_1} - y = (1 + {y_1})$$, same as $$x + y = 3$$.

$$ \Rightarrow {{{x_1}} \over 1} = {{{y_1} - 1} \over 1} = {{1 + {y_1}} \over 3}$$

i.e. $${x_1} = 1$$ and $${y_1} = 2$$

$$\therefore$$ $$P = (1,2)$$

Since, $$PR = PQ = {{2\sqrt 2 } \over 3}$$. Thus, by parametric form,

$${{x - 1} \over { - 1/\sqrt 2 }} = {{y - 2} \over {1/\sqrt 2 }} = \pm {{2\sqrt 2 } \over 3}$$

$$ \Rightarrow \left( {x = {5 \over 3},y = {4 \over 3}} \right)$$

and $$\left( {x = {1 \over 3},y = {8 \over 3}} \right)$$

$$\therefore$$ $$Q = \left( {{5 \over 3},{4 \over 3}} \right)$$ and $$R = \left( {{1 \over 3},{8 \over 3}} \right)$$

Now, equation of tangent at Q on ellipse E₁ is

$${{x\,.\,5} \over {{a^2}\,.\,3}} + {{y\,.\,4} \over {{b^2}\,.\,3}} = 1$$

On comparing with x + y = 3, we get

$${a^2} = 5$$ and $${b^2} = 4$$

$$\therefore$$ $$e_1^2 = 1 - {{{b^2}} \over {{a^2}}} = 1 - {4 \over 5} = {1 \over 5}$$ ..... (i)

Also, equation of tangent at R on ellipse E₂ is

$${{x\,.\,1} \over {{a^2}\,.\,3}} + {{y\,.\,8} \over {{b^2}\,.\,3}} = 1$$

On comparing with x + y = 3, we get

$${a^2} = 1,\,{b^2} = 8$$

$$\therefore$$ $$e_2^2 = 1 - {{{a^2}} \over {{b^2}}} = 1 - {1 \over 8} = {7 \over 8}$$ ...... (ii)

Now, $$e_1^2\,.\,e_2^2 = {7 \over {40}} \Rightarrow {e_1}{e_2} = {{\sqrt 7 } \over {2\sqrt {10} }}$$

and $$e_1^2 + e_2^2 = {1 \over 5} + {7 \over 8} = {{43} \over {40}}$$

Also, $$\left| {e_1^2 - e_2^2} \right| = \left| {{1 \over 5} - {7 \over 8}} \right| = {{27} \over {40}}$$