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JEE Advanced Mathematics Ellipse 2019 JEE Advanced 2019 (Paper 1)

JEE Advanced Mathematics Question (2019) — Solution

Question

Define the collections E 1 ,   E 2 ,   E 3 , . . . . of ellipses and R 1 ,   R 2 ,   R 3 , . . . . of rectangles as follows: E 1 : x 2 9 + y 2 4 = 1 ; R 1 : rectangle of largest area, with sides parallel to the axes, inscribed in E 1 ; E n : ellipse x 2 a n 2 + 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?

Options

  1. A. The eccentricities of E 18 and E 19 are NOT equal
  2. B. The distance of a focus from the centre in E 9 is 5 32
  3. C. The length of latus rectum of E 9 is 1 6
  4. D. ∑ n = 1 N a r e a   o f   R n < 24 , for each positive integer N

Answer

D. ∑ n = 1 N a r e a   o f   R n < 24 , for each positive integer N

Step-by-step solution

As given: E 1 = x 2 a 2 + y 2 b 2 = 1 (Here a = 3 and b = 2 ) Semi major axis = a and semi minor axis = b Let a vertex of R 1 be a c o s θ , b s i n θ Then area of R 1 = 2 a c o s θ × 2 b s i n θ R 1 = 2 a b s i n 2 θ Maximum area of R 1 = 2 a b , when s i n 2 θ = 1 or, 2 θ = π 2 θ = π 4 Maximum area of R 1 = 2 a b ....(i) Now, ellipse E 2 will have semi major axis a 2 and semi minor axis b 2 ∵ E 2 = x 2 a 2 2 + y 2 b 2 2 = 1 , hence maximum area of R 2 = 2 a 2 b 2 Similarly, ellipse E 3 will have semi major axis a 2 2 and semi minor axis b 2 2 E 3 : x 2 a 2 2 2 + y 2 b 2 2 = 1 And maximum area of R n = 2 a 2 n - 1 b 2 n - 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 - b n 2 a n 2 = 1 - b 2 a 2 = 5 3 Option ( 4 ) : ∑ n = 1 m area of rectangle R n = R 1 + R 2 + … . . + R m Area of m rectangle will be lesser than area of infinite Rectangle ⇒ R 1 + R 2 + R 2 + … . + R m R 1 + R 2 + R 3 + … . . ∞ ∑ n = 1 m area of rectangle 2 a b + 2 a 2 b 2 + . . . . . ∞ = 2 a b 1 + 1 2 2 + . . . . . ∞ ∑ n = 1 m a r e a o f r e c t a n g e R n 2 a b 1 1- 1 2 ∑ n = 1 m a r e a o f r e c t a n g e R n 4 a b ∑ n = 1 m a r e a o f r e c t a n g e R n 4 × 3 × 2 ∑ n = 1 m a r e a o f r e c t a n g e R n 24 Option ( 3 ) : Length of latus rectum of E n = 2 b n 2 a n 2 = 2 b 2 a 2 n - 1 L . R . o f E g = 2 b 2 a 2 2 = 2 2 2 3 × 2 4 = 1 6 Option ( 2 ) : Distance between focus and centre of E 9 = a 9 e = a ( 2 ) 2 × 5 3 = 3 16 × 5 3 = 5 16

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