Define the collections \{E₁,E₂,E₃,\} of ellipses and \{R₁,R₂,R₃,\} of rectangles as follows:E₁:\;{x²}{9}+\;{y²}{4}=1 R₁: rectangle of largest area, with sides parallel to the axes, inscribed in E₁; Eₙ ellipse \;{x²}{aₙ²}+\;{y²}{bₙ²}=1 of largest area inscribed in Rₙ-1},n > 1 Rₙ: rectangle of largest area, with sides parallel to the axes, inscribed in Eₙ,n > 1 Then which of the following options is/are correct? [JEE Adv 2019 P1]

The distance of a focus from the centre in E₉ is {{5}}{32}

The length of latus rectum of E₉ is 1/6

The eccentricities of E₁8} and E₁9} are NOT equal

Test Index

Ellipse

Section: Mathematics

Question : 6 of 25

Marks: +1, -0

Define the collections

\{E_1,E_2,E_3,\ldots\}

\{R_1,R_2,R_3,\ldots\}

E_{1}:\;\frac{x^{2}}{9}+\;\frac{y^{2}}{4}=1

R_1

E_1

E_n

\;\frac{x^{2}}{a_{n}^{2}}+\;\frac{y^{2}}{b_{n}^{2}}=1

R_{n-1},n > 1

R_n

E_n,n > 1

Solution:

E_{1}:\;\frac{x^{2}}{9}+\;\frac{y^{2}}{4}=1(\;\text{ellipse}\;),e=\;\frac{\sqrt{5}}{3}

R_1:

Max. Rectangle inscribed having sides parallel to axes

For this

A=(3\cos\theta,2\sin\theta),

where

\theta=\frac{\pi}{4}

i.e

A=\left(\frac{3}{\sqrt{2}},\frac{3}{\sqrt{2}}\right)

∴ Sides

3\sqrt{2},2\sqrt{2}

Area of

R_1=(3\sqrt{2})(2\sqrt{2})

∴ If a,b are semi axes of ellipse then

a\sqrt{2},b\sqrt{2}

are sides

For

E_{2}:\;\frac{x^{2}}{\left(\frac{3}{\sqrt{2}}\right)^{2}}+\;\frac{y^{2}}{\left(\frac{2}{\sqrt{2}}\right)^{2}}=1

i.e semi axes are

\frac{a}{\sqrt{2}},\frac{b}{\sqrt{2}}

\therefore E_{9}:\;\frac{x^{2}}{\left(\frac{3}{(\sqrt{2})^{8}}\right)^{8}}+\;\frac{y^{2}}{\left(\frac{2}{(\sqrt{2})^{8}}\right)^{8}}

⇒ distance from centre to focus of

E_9=ae=\;\frac{3}{(\sqrt{2})^{8}}\times\;\frac{\sqrt{5}}{3}=\;\frac{\sqrt{5}}{16}

L.L.R

=\;\frac{2b^{2}}{a}=\;\frac{2\times\left(\frac{2}{(\sqrt{2})^{8}}\right)^{2}}{\frac{3}{(\sqrt{2})^{8}}}=\;\frac{2\times\frac{4}{256}}{\frac{3}{16}}=\;\frac{1}{32}\times\frac{16}{3}=\frac{1}{6}

Eccentricities for all ellipses are equal.

Also sum of areas

=R_1+R_2+R_3\ldots\infty

=4[\text{Areas sum in} Q_1]

=4\left[\;\frac{3}{\sqrt{2}}\;\frac{2}{\sqrt{2}}+\;\frac{3}{(\sqrt{2})^{2}}\;\frac{2}{(\sqrt{2})^{2}}+\;\frac{3}{(\sqrt{2})^{3}}\;\frac{2}{(\sqrt{2})^{3}}+\dots\right]

=(4)(6)\left[\;\frac{1}{2}+\;\frac{1}{4}+\;\frac{1}{8}+\dots\right]

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

\therefore \sum\limits_{n=1}^{N}(\text{area of}\;R_n) < 24

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