QJEEAdv19Ma7113: JEE Advanced 2019 Mathematics - Class 11 JEEAdv19Ma7113 Solution: Define the collections {E 1 , E 2 , E 3 , ...} of ellipses and…

Given equation of ellipse

${E_1}:{{{x^2}} \over 9} + {{{y^2}} \over 4} = 1$ ... (i)

Now, let a vertex of rectangle of largest area with sides parallel to the axes, inscribed in E₁ be (3cos $\theta$ , 2sin $\theta$ ).

So, area of rectangle

R₁ = 2(3cos $\theta$ ) $\times$ 2(2sin $\theta$ ) = 12sin(2 $\theta$ )



The area of R₁ will be maximum, $\theta = {\pi \over 4}$ and maximum area is 12 square units and length of sides of rectangle R₁ are $2a\cos \theta = \sqrt 2 a = 3\sqrt 2$ = length of major axis of ellipse E₂ and

$2b\sin \theta = \sqrt 2 b = 2\sqrt 2$ = length of minor axis of ellipse E₂.

So, ${E_2}:{{{x^2}} \over {{{\left( {{a \over {\sqrt 2 }}} \right)}^2}}} + {{{y^2}} \over {{{\left( {{b \over {\sqrt 2 }}} \right)}^2}}} = 1$ and

maximum area of rectangle

${R_2}:2\left( {{a \over {\sqrt 2 }}} \right)\left( {{b \over {\sqrt 2 }}} \right)$ and so on.

So, ${E_n} = {{{x^2}} \over {{{\left( {{a \over {{{(\sqrt 2 )}^{n - 1}}}}} \right)}^2}}} + {{{y^2}} \over {{{\left( {{b \over {{{(\sqrt 2 )}^{n - 1}}}}} \right)}^2}}} = 1$ ,

and maximum area of rectangle

${R_n} = 2\left( {{a \over {{{(\sqrt 2 )}^{n - 1}}}}} \right) + \left( {{b \over {{{(\sqrt 2 )}^{n - 1}}}}} \right)$

Now option (a),

Since, eccentricity of ellipse

${E_n} = e{'_n} = \sqrt {1 - {{{{({b_n})}^2}} \over {{{({a_n})}^2}}}}$

$= \sqrt {1 - {{{{\left( {{b \over {(\sqrt {2{)^{n - 1}}} }}} \right)}^2}} \over {{{\left( {{a \over {(\sqrt {2{)^{n - 1}}} }}} \right)}^2}}}}$

$= \sqrt {1 - {{{b^2}} \over {{a^2}}}} = \sqrt {1 - {4 \over 9}} = {{\sqrt 5 } \over 3}$

is independent of 'n', so eccentricity of E₁₈ and E₁₉ are equal.

Option (b),

Distance between focus and centre of

${E_9} = e.{a_9} = {a \over {{{(\sqrt 2 )}^8}}}(e)$

$= {3 \over {{2^4}}} \times {{\sqrt 5 } \over 3} = {{\sqrt 5 } \over {16}}$ unit

Option (c),

$\because$ \sum\limits_{n = 1}^N {(area\,of\,{R_n})} < (area\,of\,{R_1}) + (area\,of\,{R_2}) + .....\infty

< 2ab + 2{{ab} \over 2} + 2{{ab} \over {{2^2}}} + .......

< 2ab\left( {1 + {1 \over 2} + {1 \over {{2^2}}} + .....} \right)

< 12\left( {{1 \over {1 - 1/2}}} \right)

\Rightarrow \sum\limits_{n = 1}^N {(area\,of\,{R_n}} ) < 24 , for each positive integer N.

Option (d),

Length of latusrectum ${E_9} = {{2b_9^2} \over {{a_9}}} = {{2{b^2}} \over {a{{(\sqrt 2 )}^8}}}$

$= {{2 \times 4} \over {3 \times 16}} = {1 \over 6}$ units

Hence, options (c) and (d) are correct.