Mathematics · Ellipse

JEE Advanced 2019 — Paper 1 — Question 37

Define the collections $\left\{E_{1}, E_{2}, E_{3}, \ldots \ldots\right\}$ of ellipses and $\left\{R_{1}, R_{2}, R_{3}, \ldots\right\}$ 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 $R_{n-1}, n>1$ ;

$\mathrm{R}_{\mathrm{n}}$ : rectangle of largest area, with sides parallel to the axes, inscribed in $\mathrm{E}_{\mathrm{n}}, \mathrm{n}>1$ .

Then which of the following options is/are correct?

Option A:
$\quad \sum_{n=1}^{N}\left(\right.$ area of $\left.R_{n}\right)<24$ , for each positive integer $N$
Correct
Option B:
The distance of a focus from the centre in $\mathrm{E}_{9}$ is $\frac{\sqrt{5}}{32}$
Option C:
The eccentricities of $\mathrm{E}_{18}$ and $\mathrm{E}_{19}$ are NOT equal
Option D:
The length of latus rectum of $E_{9}$ is $\frac{1}{6}$
Correct

Answer: A, D

Step-by-step solution

Area of $R_n = 2a \cos\theta \times 2b \sin\theta = 2ab \sin 2\theta$ which will be maximum when $\theta = 45^\circ.$

\therefore R_{n \text{ max}} = 2ab

\begin{array}{|c|c|c|} \hline & a & b \\ \hline E_1 & 3 & 2 \\ \hline E_2 & \frac{3}{\sqrt{2}} & \frac{2}{\sqrt{2}} \\ \hline E_3 & \frac{3}{(\sqrt{2})^2} & \frac{2}{(\sqrt{2})^2} \\ \hline \vdots & \vdots & \vdots \\ \hline E_n & \frac{3}{(\sqrt{2})^{n-1}} & \frac{2}{(\sqrt{2})^{n-1}} \\ \hline \end{array}

\text{(A) Area of } R_n = 2 \times \frac{3}{(\sqrt{2})^{n-1}} \times \frac{2}{(\sqrt{2})^{n-1}} = \frac{12}{2^{n-1}}

\text{Area of } R_1 + \text{Area of } R_2 + \ldots \text{ Area of } R_n \ldots \infty

= 12 + \frac{12}{2} + \frac{12}{4} + \ldots

= 12 \left( 1 + \frac{1}{2} + \frac{1}{4} + \ldots \right)

= 12 \times \frac{1}{1 - \frac{1}{2}} = 12 \times \frac{1}{\frac{1}{2}} = 24

(B) $b^2 = a^2(1 - e^2)$ for ellipse

\left( \frac{2}{(\sqrt{2})^8} \right)^2 = \left( \frac{3}{(\sqrt{2})^8} \right)^2 (1 - e_9^2)

\frac{4}{2^{16}} = \frac{9}{2^{16}} (1 - e_9^2)

1 - e_9^2 = \frac{4}{9}

e_9^2 = 1 - \frac{4}{9} = \frac{5}{9}

e_9 = \frac{\sqrt{5}}{3}

Distance between centre and focus $= ae$

= \frac{3}{(\sqrt{2})^8} \times \frac{\sqrt{5}}{3} = \frac{\sqrt{5}}{2^4} = \frac{\sqrt{5}}{16}

\text{(C) } e_{18} = e_{19}

(D) Latus rectum (length) $= \frac{2b^2}{a} = \frac{2 \times \left( \frac{2}{(\sqrt{2})^8} \right)^2}{\frac{3}{(\sqrt{2})^8}} = \frac{1}{6}$

Answer key and solution verified before publishing.

Practise Ellipse

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Advanced 2019
Paper: Paper 1
Subject: Mathematics
Chapter: Ellipse
Topic: Introduction to Ellipse