QJEEAdv15Ma22eb: JEE Advanced 2015 Mathematics - Class 11 JEEAdv15Ma22eb Solution: Let and be two ellipses whose centres are at the origin. The…

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