Find the area of the greatest isosceles triangle that can be inscribed in the ellipse `((x^2)/(a^2))+((y^2)/(b^2))=1` having its vertex coincident with one extremity of the major axis.

Explanation

Consider the isosceles triangle ABC:
`A(a,0),B(acostheta,bsintheta) and C(acostheta,−bsintheta)`
Area of `triangleABC= 1/2×BC× height of triangleABC`
Height of `triangleABC=a(1+costheta)`
`BC=2bsintheta`
or, triangle=`1/2×2bsintheta×a(1+costheta)`=`ab(sintheta)(1+costheta)`
For maximum area of the triangle,

`(d(triangle ))/(d(theta))`=`abcostheta(1+costheta)−ab(sin^2theta)=0`
or, `costheta(1+costheta)−sin^2theta=0`
or, `costheta(1+costheta)−(1+costheta)(1−costheta)=0`
or, `(1+costheta)(costheta−1+costheta)=0`
or,` (1+costheta)(2costheta−1)=0`
or, `costheta=−1 or, costheta= 1/2`
For maximum value, we take
`costheta= 1/2` and `sintheta= sqrt(3) /2`
or, Maximum area =`ab× sqrt(3) /2 ( 1+ 1/2) = 3 sqrt(3) / (4) ab`

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