Subject: Mathematics — Parabola
Question: Define the collections
$\left\{ E_{1},E_{2},E_{3},\ldots\ldots \right\}$ of ellipses and
$\left\{ R_{1},R_{2},R_{3} \right.\ $, $\_\_\_\_$ } of rectangles as
follows:\
$E_{1}:\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$;\
$R_{1}$ : rectangle of largest area, with sides parallel to the axes,
inscribed in $E_{1}$;\
$E_{n}$ : ellipse
$\frac{x^{2}}{a_{n}^{2}} + \frac{y^{2}}{b_{n}^{2}} = 1$ of largest area
inscribed in $R_{n - 1}$, $n > 1$;\
$R_{n}$ : rectangle of largest area, with sides parallel to the axes,
inscribed in $E_{n},n > 1$.\
Then find the length of latus rectum of $E_{9}$ and value of $n = 1$.
Answer: $\frac{1}{6},\ 24$
Solution: $E_{1}:\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1$
$R_{1}$ : rectangle $ABCD$ with largest area.
[IMAGE] Area of $R_{1} = A = 2x \times 2y$\
$${\Rightarrow A = 4x \times \frac{2}{3}\sqrt{9 - x^{2}} = \frac{8}{3}x\sqrt{9 - x^{2}}
}{\therefore\ \frac{d\text{ }A}{dx} = \frac{8}{3}\left\lbrack \sqrt{9 - x^{2}} - \frac{x^{2}}{\sqrt{9 - x^{2}}} \right\rbrack
}{\Rightarrow \frac{8}{3}\left\lbrack \frac{9 - 2x^{2}}{\sqrt{9 - x^{2}}} \right\rbrack = 0 \Rightarrow x = \frac{3}{\sqrt{2}},y = \frac{2}{3}\sqrt{9 - \frac{9}{2}} = \frac{2}{\sqrt{2}}
}$$⇒ For $E_{2}:a = \frac{3}{\sqrt{2}},b = \frac{2}{\sqrt{2}}$\
Similarly for
$E_{3}:a = \frac{3}{(\sqrt{2})^{2}},b = \frac{2}{(\sqrt{2})^{2}}$ and
so on.\
Now eccentricity depends on $\frac{b}{a}$ which is same for all
$E_{n}$, therefore eccentricity for all the $E_{n}$ \'s will remain
$\sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3}$.\
For $E_{9}:a = \frac{3}{(\sqrt{2})^{8}},b = \frac{2}{(\sqrt{2})^{8}}$\
⇒ Length of latus rectum
$= \frac{2b^{2}}{a} = \frac{2 \times \frac{4}{256}}{\frac{3}{16}} = \frac{1}{6}$\
Area of
$R_{1} = 4 \times \frac{3}{\sqrt{2}} \times \frac{2}{\sqrt{2}} = \frac{24}{2}$\
Area of
$R_{2} = 4 \times \frac{3}{(\sqrt{2})^{2}} \times \frac{2}{(\sqrt{2})^{2}} = \frac{24}{2^{2}}$\
Area of
$R_{3} = 4 \times \frac{3}{(\sqrt{2})^{3}} \times \frac{2}{(\sqrt{2})^{3}} = \frac{2}{(2)^{3}}$
and so on\
$\sum_{n = 1}^{\infty}\mspace{2mu}$ area of
$R_{n} = \frac{24}{2} + \frac{24}{2^{2}} + \frac{24}{2^{3}} + \ldots = \frac{12}{1 - \frac{1}{2}} = 24$\
$\Rightarrow \sum_{n = 1}^{N}\mspace{2mu}$ (area of
$\left. \ R_{n} \right) < 24$ for each positive integer $N$.
For
$E_{9}:a = \frac{3}{(\sqrt{2})^{8}},b = \frac{2}{(\sqrt{2})^{8}},e = \frac{\sqrt{5}}{3}$\
Thus, focus $= (ae,0) = \left( \frac{\sqrt{5}}{16},0 \right)$\
Hence, distance of focus from centre $= \frac{\sqrt{5}}{16}$