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

construct the traingle as shown in figure and name it PAQ
in`/_ APQ = 1/2 PQAD`
`=1/2*2(PD)(OA-OD)`
In `/_ APQ => (B sin theta)(a- acos theta)`
`= ab sin theta - ab sin theta cos theta`
`= (ab)/2*[2sin theta - 2sin theta cos theta]`
`=(ab)/2*[2sin theta - sin2 theta]`
as, `(d/_)/(d theta) = 0`
`= ((ab)/2)[2cos theta -2cos 2 theta] = 0`
`= 2cos theta - 2cos 2 theta = 0`
`= cos theta - cos 2 theta = 0`
`= cos theta - (2cos^2 theta -1) = 0`
`= 2cos^2 theta - 2cos theta + cos theta -1 = 0`
`= 2cos theta(cos theta -1) + 1(cos theta-1) = 0`
`= (2cos theta+1)(cos theta -1) = 0`
`= cos theta = -1/2`
` at theta = (2 pi)/3 `
area = `ab[sqrt(3)/2+ sqrt(3)/2*1/2`
`=ab[sqrt3/2+ sqrt3/4]`
`=3sqrt3/4ab`
answer