Q23 / 54 · JEE Advanced 2019 (Paper 1) · Mathematics

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 E1;
En: 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>1.
Then which of the following options is/are correct?

  • A
    The eccentricities of E18 and E19 are NOT equal
  • B
    The distance of a focus from the centre in E9 is 532
  • Correct answer
    C
    The length of latus rectum of E9 is 16
  • Correct answer
    D
    ∑n=1Narea of Rn<24, for each positive integer N

Solution

As given:
E1=x2a2+y2b2=1 (Here a=3 and b=2 )
Semi major axis =a and semi minor axis =b
Let a vertex of R1 be acosθ,bsinθ
Then area of R1=2acosθ×2bsinθ
R1=2absin2θ
Maximum area of R1=2ab,

when sin2θ=1
or, 2θ=π2
θ=π4
Maximum area of R1=2ab ....(i)
Figure
Now, ellipse E2 will have semi major axis a2 and semi minor axis b2
∵ E2=x2a22+y2b22=1,

hence maximum area of R2=2a2b2
Similarly, ellipse E3 will have semi major axis a22 and semi minor axis b22
E3:x2a222+y2b22=1
And maximum area of
Rn=2a2n-1b2n-1
Option (1):
All ellipse will have some eccentricity as the ratios of semi major axis and semi minor axis is same for all ellipse.
e=1-bn2an2=1-b2a2=53

Option (4):
∑n = 1marea of rectangle Rn=R1+R2+…..+Rm
Area of m rectangle will be lesser than area of infinite
Rectangle ⇒R1+R2+R2+….+Rm<R1+R2+R3+…..∞
∑n=1marea of rectangle<2ab+2a2b2+.....∞=2ab1+122+.....∞
∑n=1marea of rectange Rn<2ab11-12
∑n=1marea of rectange Rn<4ab
∑n=1marea of rectange Rn<4×3×2
∑n=1marea of rectange Rn<24
Option (3):
Length of latus rectum of En=2bn2an2=2b2a2n-1
L.R. of Eg=2b2a22=2223×24=16
Option (2):
Distance between focus and centre of E9=a9e
=a(2)2×53
=316×53
=516

Hence, the correct answer is options: 'C', 'D'.

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