Question

What will be the maximum area of an isosceles triangle inscribed in the ellipse x^2/a^2 + y^2/b^2 = 1 if its vertex at one end of the major axis?

(a) 3√3/4 ab sq units

(b) 3√3/2 ab sq units

(c) √3/2 ab sq units

(d) 3/4 ab sq units

The question was posed to me in an online quiz.

Enquiry is from First Order Derivative topic in division Limits and Derivatives of Mathematics – Class 11

Answer 1

The correct option is (a) 3√3/4 ab sq units

Easiest explanation: Let A, B, C be the vertices of the isosceles triangle.

Let B = (a cosθ, b sinθ)

=> C – (a cosθ. -b sinθ)  (by symmetry).

Now, area of a triangle is given by

A = ½(BC)(AD)

   = ½(2b sinθ)(1 + cosθ)

=> dA/dθ = ab[cosθ(1 + cosθ) – sin^2θ]

    = ab(1 + cosθ)(2 cosθ – 1)

Now, putting dA/dθ = 0,

We get, θ = π/3, π

Now, for θ = π, triangle ABC is not possible.

And, dA/dθ]θ = π/3 < 0,

Thus, for θ = π/3, the area of isosceles triangle will be maximum.

So, A = 3√3/4 ab sq units.