Mathematics - Ellipse Question with Solution | TestHub

As given:
$E_1=\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ (Here $a=3$ and $b=2$ )
Semi major axis $=a$ and semi minor axis $=b$
Let a vertex of $R_1$ be $\left(a\mathrm{c}\mathrm{o}\mathrm{s}\theta ,b\mathrm{s}\mathrm{i}\mathrm{n}\theta \right)$
Then area of $R_1=2a\mathrm{c}\mathrm{o}\mathrm{s}\theta \times 2b\mathrm{s}\mathrm{i}\mathrm{n}\theta$
$R_1=2ab\mathrm{s}\mathrm{i}\mathrm{n}2\theta$
Maximum area of $R_1=2ab,$

when $\mathrm{s}\mathrm{i}\mathrm{n}2\theta =1$
or, $2\theta =\frac{\pi }{2}$
$\theta =\frac{\pi }{4}$
Maximum area of $R_1=2ab$ ....(i)

Now, ellipse $E_2$ will have semi major axis $\frac{a}{\sqrt{2}}$ and semi minor axis $\frac{b}{\sqrt{2}}$
$\because E_2=\frac{x^2}{{\left(\frac{a}{\sqrt{2}}\right)}^2}+\frac{y^2}{{\left(\frac{b}{\sqrt{2}}\right)}^2}=1,$

hence maximum area of $R_2=2\left(\frac{a}{\sqrt{2}}\right)\left(\frac{b}{\sqrt{2}}\right)$
Similarly, ellipse $E_3$ will have semi major axis $\frac{a}{{\left(\sqrt{2}\right)}^2}$ and semi minor axis $\frac{b}{{\left(\sqrt{2}\right)}^2}$
$E_3:\frac{x^2}{{\left(\frac{a}{{\left(\sqrt{2}\right)}^2}\right)}^2}+\frac{y^2}{{\left(\frac{b}{\left(\sqrt{2}\right)}\right)}^2}=1$
And maximum area of
$R_n=2\left(\frac{a}{{\left(\sqrt{2}\right)}^{n-1}}\right)\left(\frac{b}{{\left(\sqrt{2}\right)}^{n-1}}\right)$
Option $\left(1\right):$
All ellipse will have some eccentricity as the ratios of semi major axis and semi minor axis is same for all ellipse.
$e=\sqrt{1-\frac{b_n^2}{a_n^2}}=\sqrt{1-\frac{b^2}{a^2}}=\frac{\sqrt{5}}{3}$

Option $\left(4\right):$
$\sum _{n=1}^m\text{area\hspace{0.17em}of\hspace{0.17em}rectangle}{R}_n=R_1+R_2+\dots ..+R_m$
Area of $m$ rectangle will be lesser than area of infinite
Rectangle $\Rightarrow R_1+R_2+R_2+\dots .+R_m<R_1+R_2+R_3+\dots ..\infty$
$\sum _{n=1}^m\text{area\hspace{0.17em}of\hspace{0.17em}rectangle}<2ab+2\left(\frac{a}{\sqrt{2}}\right)\left(\frac{b}{\sqrt{2}}\right)+.....\infty =2ab\left[1+\frac{1}{{\left(\sqrt{2}\right)}^2}+.....\infty \right]$
$\sum _{n=1}^m\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e} R_n<2ab\left[\frac{\text{1}}{\text{1-}\frac{\text{1}}{\text{2}}}\right]$
$\sum _{n=1}^m\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e} R_n<4ab$
$\sum _{n=1}^m\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e} R_n<4\times 3\times 2$
$\sum _{n=1}^m\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{e} R_n<24$
Option $\left(3\right):$
Length of latus rectum of $E_n=\frac{2b_n^2}{a_n^2}=\frac{2b^2}{a{\left(\sqrt{2}\right)}^{n-1}}$
$L.R.of{E}_g=\frac{2b^2}{a{\left(\sqrt{2}\right)}^2}=\frac{2{\left(2\right)}^2}{3\times 2^4}=\frac{1}{6}$
Option $\left(2\right):$
Distance between focus and centre of $E_9=a_{9}e$
$=\frac{a}{{\left(\sqrt{2}\right)}^2}\times \frac{\sqrt{5}}{3}$
$=\frac{3}{16}\times \frac{\sqrt{5}}{3}$
$=\frac{\sqrt{5}}{16}$