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JEE Advanced Mathematics Ellipse 2022 JEE Advanced 2022 (Paper 1)

JEE Advanced Mathematics Question (2022) — Solution

Question

Consider the ellipse  x 2 4 + y 2 3 = 1 . Let  H α , 0 , 0 < α < 2 , be a point. A straight line drawn through  H  parallel to the  y -axis crosses the ellipse and its auxiliary circle at points  E  and  F  respectively, in the first quadrant. The tangent to the ellipse at the point  E  intersects the positive  x -axis at a point  G . Suppose the straight line joining  F  and the origin makes an angle  ϕ  with the positive  x -axis.   List- I   List- II I If  ϕ = π 4 , then the area of the triangle  F G H  is P 3 - 1 4 8 II If  ϕ = π 3 , then the area of the triangle  F G H  is Q 1 III If  ϕ = π 6 , then the area of the triangle  F G H  is R 3 4 IV If  ϕ = π 12 , then the area of the triangle  F G H  is S 1 2 3     T 3 3 2 The correct option is:

Options

  1. A. I → R ;   II → S ; III → Q ; IV → P
  2. B. I → R ;   II → T ; III → S ; IV → P
  3. C. I → Q ;   II → T ; III → S ; IV → P
  4. D. I → Q ;   II → S ; III → Q ; IV → P

Answer

C. I → Q ;   II → T ; III → S ; IV → P

Step-by-step solution

Given  x 2 4 + y 2 3 = 1 Let  α ≡ 2 cos ϕ Tangent at  E 2 cos ϕ , 3 sin ϕ  to the ellipse is   x cos ϕ 2 + y sin ϕ 3 = 1 This intersect  x -axis at  G 2 sec ϕ , 0 Drawing the figure as per the information, we get,  Area of triangle  F G H = 1 2 2 sec ϕ - 2 cos ϕ 2 sin ϕ Δ = 2 sin 2 ϕ . tan ϕ Δ = 1 - cos 2 ϕ · tan ϕ I.  If  ϕ = π 4 , Δ = 1 → Q II. If  ϕ = π 3 , Δ = 2 · 3 2 2 · 3 = 3 3 2 → T III.  If  ϕ = π 6 , Δ = 2 · 1 2 2 · 1 3 = 1 2 3 → S IV.  If  ϕ = π 12 , Δ = 1 - 3 2 · 2 - 3 = 2 - 3 2 2 = 3 - 1 4 8 → P

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