Define the collections E1,E2,E3,,,, of ellipses and R1,R2,R3,... of rectangles as follows : E1:x29+y24=1 R1 : rectangle of largest area, with sides parallel to the axes, inscribed in E1En : ellipse x2an2+y2bn2=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>1Then which of the following options is/are correct?

E1:x29+y24=1( ellipse ),e=53R1: Max. Rectangle inscribed having sides parallel to axes For this A=(3cosθ,2sinθ), where θ=π4 i.e A=32,32 Sides: 32,22 Area of R1=(32)(22) If a.b are semi axes of ellipse then a2,b2 are sides. For E2:x2322+y2222=1 i.e. semi axes are a2,b2 . ∴E9:x23(2)82+y22(2)82⇒ distance from centre to foucs of E9=ae3(2)8×53=516 L.L. R=2b2a=2×2(2)823(2)8=2×4256316=132×163=16 Eccentricities for all ellipses are equal. Also sum of areas =R1+R2+R3………∞ =4 Areas sum in Q1=43222+3(2)22(2)2+3(2)3223+……=4612+14+18+…….=24×121-12=24∴∑n=1N area of Rn<24

Define the collections E1,E2,E3,,,, of ellipses and R1,R2,R3,... of rectangles as follows : E1:x29+y24=1 R1 : rectangle of largest area, with sides parallel to the axes, inscribed in E1En : ellipse x2an2+y2bn2=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>1Then which of the following options is/are correct?

E1:x29+y24=1( ellipse ),e=53R1: Max. Rectangle inscribed having sides parallel to axes For this A=(3cosθ,2sinθ), where θ=π4 i.e A=32,32 Sides: 32,22 Area of R1=(32)(22) If a.b are semi axes of ellipse then a2,b2 are sides. For E2:x2322+y2222=1 i.e. semi axes are a2,b2 . ∴E9:x23(2)82+y22(2)82⇒ distance from centre to foucs of E9=ae3(2)8×53=516 L.L. R=2b2a=2×2(2)823(2)8=2×4256316=132×163=16 Eccentricities for all ellipses are equal. Also sum of areas =R1+R2+R3………∞ =4 Areas sum in Q1=43222+3(2)22(2)2+3(2)3223+……=4612+14+18+…….=24×121-12=24∴∑n=1N area of Rn<24

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Define the collections $\{E_{1}, E_{2}, E_{3},,,,\}$ of ellipses and $\{R_{1}, R_{2}, R_{3}, . . .\}$ 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 which of the following options is/are correct?

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The distance of a focus from the center $E_{9}$ is $\frac{\sqrt{5}}{32}$

$\sum_{n = 1}^{N} (a r e a o f R_{n}) < 24$ for each integer

The length of latus rectum of $E_{9}$ is $\frac{1}{6}$

The eccentricities of $E_{18}$ and $E_{19}$ are NOT equal

answer is B, C.

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Detailed Solution

$E_{1} : \frac{x^{2}}{9} + \frac{y^{2}}{4} = 1 (ellipse), e = \frac{\sqrt{5}}{3} R_{1} : Max. Rectangle inscribed having sides parallel to axes$

$\begin{array}{l} For this A = (3 \cos θ, 2 \sin θ), where θ = \frac{π}{4} i.e A = (\frac{3}{\sqrt{2}}, \frac{3}{\sqrt{2}}) \\ Sides: 3 \sqrt{2}, 2 \sqrt{2} Area of R_{1} = (3 \sqrt{2}) (2 \sqrt{2}) \\ If a.b are semi axes of ellipse then a \sqrt{2}, b \sqrt{2} are sides. \end{array}$

$\begin{array}{l} For E_{2} : \frac{x^{2}}{{(\frac{3}{\sqrt{2}})}^{2}} + \frac{y^{2}}{{(\frac{2}{\sqrt{2}})}^{2}} = 1 i.e. semi axes are \frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}} . \\ ∴ E_{9} : \frac{x^{2}}{{(\frac{3}{(\sqrt{2})^{8}})}^{2}} + \frac{y^{2}}{{(\frac{2}{(\sqrt{2})^{8}})}^{2}} \Rightarrow distance from centre to foucs of E_{9} = ae \end{array}$

$\frac{3}{(\sqrt{2})^{8}} \times \frac{\sqrt{5}}{3} = \frac{\sqrt{5}}{16} L.L. R = \frac{2 b^{2}}{a} = \frac{2 \times {(\frac{2}{(\sqrt{2})^{8}})}^{2}}{\frac{3}{(\sqrt{2})^{8}}} = \frac{2 \times \frac{4}{256}}{\frac{3}{16}} = \frac{1}{32} \times \frac{16}{3} = \frac{1}{6}$

$\begin{array}{l} Eccentricities for all ellipses are equal. \\ Also sum of areas = R_{1} + R_{2} + R_{3} \dots \dots \dots \infty = 4 [Areas sum in Q_{1}] \\ = 4 [\frac{3}{\sqrt{2}} \frac{2}{\sqrt{2}} + \frac{3}{(\sqrt{2})^{2}} \frac{2}{(\sqrt{2})^{2}} + \frac{3}{(\sqrt{2})^{3}} \frac{2}{({\sqrt{2}}^{3})} + \dots \dots] = (4) (6) [\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots \dots .] \\ = 24 \times \frac{\frac{1}{2}}{1 - \frac{1}{2}} = 24 ∴ \sum_{n = 1}^{N} (area of R_{n}) < 24 \end{array}$