JEE Advanced2022MathematicsEllipseActual
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
Options
- AI → R ;   II → S ; III → Q ; IV → P
- BI → R ;   II → T ; III → S ; IV → P
- CI → Q ;   II → T ; III → S ; IV → P
- DI → Q ;   II → S ; III → Q ; IV → P
Correct 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 →