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.

Answer

(3/2,1)

Explanation

The 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=sin theta` , line PR is `x=sqrt(3)cos theta`, line OQ is y=x, and Q is `(2 sin theta, 2 sin theta)`,
Z= Area of region PQORP (tranpezium)
`=(1)/(2)(OR+PQ)PR`
`=(1)/(2)(sqrt(3)cos theta+sqrt(3)cos theta-2sin theta)2 sin theta`
`=(2 sqrt(3)cos theta sin theta-2sin^(2)theta)`
`=(sqrt(3)sin 2theta+ cos 2theta-1)`
`=2 cos theta(2theta-(pi)/(3))-1`
which is maximum when `cos(theta-(pi)/(3))` is maximum
`:. 2 theta-(pi)/(3)=0`
or `theta=(pi)/(6)`
Hence, point P is (3/2,1)