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

Explanation

Let any point on the ellipse `x^(2)/a^(a)+y^(2)/b^(2) = 1" be "P (a cos theta, b sin theta)`.
` (##NTN_MATH_XII_C06_E04_270_S01.png" width="80%">
Draw `PM bot OX` from P and produce it to meet the ellipse at Q, then `Delta APQ` is an isosceles triangle, let its area be S, then
`S = 2 xx 1/2 xx AM xx MP`
` = (OA - OM) xx MP`
` = (a - a cos theta)* b sin theta`
` rArr S = ab (sin theta-1/2 sin 2 theta)`
` = ab ( sin theta - 1/2 sin 2 theta)`
` rArr (dS)/(d theta) = ab (cos theta- cos 2 theta)`
Again differentiate w.r.t.` theta`,
`(d^(2)S)/(d theta^(2)) = ab ( - sin theta + 2 sin 2 theta)`
For maxima/minima ` (dS)/(d theta) = 0 `
` rArr cos theta = cos 2 theta rArr 2 theta = 2 pi - 0 rArr theta =(2pi)/3 at theta=(2pi)/3`,
`((d^(2)S)/(d theta^(2)))_(theta=(2pi)/3)= ab [-sin(2pi)/2 + 2 sin (2 xx(2pi)/3)]`
` = ab (-sqrt3/2 - (2sqrt3)/2)=ab ((-3sqrt3)/2)`
` = (-3sqrt3 ab)/2 lt 0 `
` :." S is maximum at "theta = (2pi)/3`,
and maximum value of S
`= ab (sin (2pi)/3 - 1/2*2 sin.(2pi)/3cos. (2pi)/3)`
` = ab (sqrt3/2 + sqrt3/2 xx 1/2)`
`= ab ((2sqrt3+sqrt3)/4)= (3sqrt3)/4` ab sq. units .