Find the point `(alpha,)beta` on the ellipse `4x^2+3y^2=12 ,` in the first quadrant, so that the area enclosed by the lines `y=x ,y=beta,x=alpha` , and the x-axis is maximum.

Explanation

Equation of the ellipse is `x^(2)//3+y^(2)//4=1`
Let point P be `sqrt(3) cos theta , 2 sin theta) , theta` in `(0,pi//2)`

Clearly line PQ is y =2 sin `theta` ,line PR is x `=sqrt(2) cos theta line OQ is`
y=x,and Q is `(2sin theta, 2 sin theta)`
Z=Area of the region PQORP (trapezium)
`=(1)/(2)(OR+PQ)PR`
`=(1)/(2)sqrt(3)cos theta + sqrt(3)cos theta -2 sin theta 2 sin theta`
`=(1)/(2)2sqrt(3)cos theta sin theta -2 sin^(2) theta`
`=(1)/(2) sqrt(3)sin 2 theta +cos2theta-1`
`=cos(2theta-(pi)/(3))-(1)/(2)`
Which is maximum when
`cos(2theta-(pi)/(3))=1 or 2theta -(pi)/(3)=0 or theta=(pi)/(6)`
Hence point is`(3//2,1)`