Question

Area of the greatest rectangle that can be inscribed in an ellipse (x²/a²) + (y²/b²) = 1 is

(a) ab

(b) 2ab

(c) (a/b)

(d) √(ab)

Answer 1

The correct option (b) 2ab

Explanation:

Let PQRS be rectangle. Where P is (a cos θ, b sin θ)

∴    PS = 2b sinθ and   PQ = 2acosθ

∴ Area of rectangle = PQ ∙ PS = 2acosθ ∙ 2bsinθ

∴    A = 2absin2θ

Maximum area will occur when θ = (π/4) i.e. sin 2θ = 1.

∴ Area of maximum rectangle is 2ab(1) = 2ab.