Question

Find the maximum area of an isosceles triangle inscribed in the ellipsex^2/a^2+y^2/b^2=1 with its vertex at one end of the major axis.

Answer 1

it is given that equation of ellipse $\bf{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}$

Let the major axis be along the x – axis in which ABC be the triangle inscribed in the ellipse where vertex C is at (a,0).
Since, the ellipse is symmetrical w.r.t. x - axis and y - axis, we can assume the coordinates of A to be ( acosθ,bsinθ) and the coordinates of B to be ( acosθ,-bsinθ).

here , A = ( acosθ,bsinθ), B = ( acosθ,-bsinθ)
and C = (a, 0)
use area of triangle formula,
area of triangle , A = 1/2 [ acosθ(-bsinθ) + acosθ(0- bsinθ) + a(bsinθ + bsinθ) ]
A = 1/2[-2absinθ.cosθ + 2absinθ]
A = absinθ - absinθ.cosθ
A = ab(sinθ - sinθ.cosθ)
differentiate A with respect to θ,
dA/dθ = ab(cosθ - cos²θ + sin²θ)
put dA/dθ = 0 ,
1 - cos²θ - cos²θ + cosθ = 0
-2cos²θ + cosθ + 1 = 0
2cos²θ - cosθ - 1 = 0
2cos²θ - 2cosθ + cosθ - 1 = 0
(2cosθ + 1)(cosθ - 1) = 0
cosθ = 1 , -1/2
θ = 0, 2π/3

again differentiate dA/dθ with respect to θ,
d²A/dθ² = ab(-sinθ + 2cosθ.sinθ + 2sinθ.cosθ)
= ab(4sinθ.cosθ - sinθ)

at θ = 0, d²A/dθ² = 0
and at θ = 2π/3 , d²A/dθ² < 0
hence,at θ = 2π/3 , A will be maximum.

A = ab(sin2π/3 - sin2π/3.cos2π/3)
A = ab(√3/2 - √3/4)
A = √3ab/4