Define the collections \ E _ 1 , E _ 2 , E _ 3 , . . . .\ of...

As given:
${E}_{1} = \dfrac{{x}^{2}}{{a}^{2}} + \dfrac{{y}^{2}}{{b}^{2}} = 1$ (Here $a = 3$ and $b = 2$ )
Semi major axis $= a$ and semi minor axis $= b$
Let a vertex of ${R}_{1}$ be \left(acos\theta, bsin\theta\right)$<br >Then area of${R}_{1} = 2acos\theta \times 2bsin\theta$<br >${R}_{1} = 2absin2\theta$<br >Maximum area of${R}_{1} = 2ab,$</p><p>when$\sin 2\theta = 1$<br >or,$2\theta = \dfrac{\pi}{2}$<br >$\theta = \dfrac{\pi}{4}$<br >Maximum area of${R}_{1} = 2ab$....(i)<br ><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_advanced/akloqbgqpbohzxzokfhiqtzcxhgxaiegqspfcliktkbppqtbvz.png"/><br >Now, ellipse${E}_{2}$will have semi major axis$\dfrac{a}{\sqrt{2}}$and semi minor axis$\dfrac{b}{\sqrt{2}}$<br >$∵ {E}_{2} = \dfrac{{x}^{2}}{{\left(\dfrac{a}{\sqrt{2}}\right)}^{2}} + \dfrac{{y}^{2}}{{\left(\dfrac{b}{\sqrt{2}}\right)}^{2}} = 1,$</p><p>hence maximum area of${R}_{2} = 2\left(\dfrac{a}{\sqrt{2}}\right)$\left(\dfrac{b}{\sqrt{2}}\right)$
Similarly, ellipse ${E}_{3}$ will have semi major axis $\dfrac{a}{{\left(\sqrt{2}\right)}^{2}}$ and semi minor axis $\dfrac{b}{{\left(\sqrt{2}\right)}^{2}}$
${E}_{3}: \dfrac{{x}^{2}}{{\left(\dfrac{a}{{\left(\sqrt{2}\right)}^{2}}\right)}^{2}} + \dfrac{{y}^{2}}{{\left(\dfrac{b}{\left(\sqrt{2}\right)}\right)}^{2}} = 1$
And maximum area of
${R}_{n} = 2\left(\dfrac{a}{{\left(\sqrt{2}\right)}^{n - 1}}\right)\left(\dfrac{b}{{\left(\sqrt{2}\right)}^{n - 1}}\right)$
Option $( 1 ) :$
All ellipse will have some eccentricity as the ratios of semi major axis and semi minor axis is same for all ellipse.
$e = \sqrt{1 - \dfrac{{b}_{n}^{2}}{{a}_{n}^{2}}} = \sqrt{1 - \dfrac{{b}^{2}}{{a}^{2}}} = \dfrac{\sqrt{5}}{3}$

Option $( 4 ) :$
$\mathop{∑}_{n = 1}^{m}\text{area of rectangle}{R}_{n} = {R}_{1} + {R}_{2} + \ldots . . + {R}_{m}$
Area of $m$ rectangle will be lesser than area of infinite
Rectangle $⇒ {R}_{1} + {R}_{2} + {R}_{2} + \ldots . + {R}_{m} < {R}_{1} + {R}_{2} + {R}_{3} + \ldots . . \infty$
$\mathop{∑}_{n = 1}^{m}\text{area of rectangle} < 2ab + 2\left(\dfrac{a}{\sqrt{2}}\right)\left(\dfrac{b}{\sqrt{2}}\right) + . . . . . \infty = 2ab\left[1 + \dfrac{1}{{\left(\sqrt{2}\right)}^{2}} + . . . . . \infty\right]$
$\mathop{∑}_{n = 1}^{m}areaofrectange{R}_{n} < 2ab\left[\dfrac{\text{1}}{\text{1-}\dfrac{\text{1}}{\text{2}}}\right]$
$\mathop{∑}_{n = 1}^{m}areaofrectange{R}_{n} < 4ab$
$\mathop{∑}_{n = 1}^{m}areaofrectange{R}_{n} < 4 \times 3 \times 2$
$\mathop{∑}_{n = 1}^{m}areaofrectange{R}_{n} < 24$
Option $( 3 ) :$
Length of latus rectum of ${E}_{n} = \dfrac{2{b}_{n}^{2}}{{a}_{n}^{2}} = \dfrac{2{b}^{2}}{a{\left(\sqrt{2}\right)}^{n - 1}}$
$L . R . of{E}_{g} = \dfrac{2{b}^{2}}{a{\left(\sqrt{2}\right)}^{2}} = \dfrac{2{\left(2\right)}^{2}}{3 \times {2}^{4}} = \dfrac{1}{6}$
Option $( 2 ) :$
Distance between focus and centre of ${E}_{9} = {a}_{9}e$
$= \dfrac{a}{{( \sqrt{2} )}^{2}} \times \dfrac{\sqrt{5}}{3}$
$= \dfrac{3}{16} \times \dfrac{\sqrt{5}}{3}$
$= \dfrac{\sqrt{5}}{16}$