Find the area of the greatest rectangle that can be inscribed in the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1.`

Options

A. `a/b`
B. `sqrt(ab)`
C. `ab`
D. `2ab`

Answer

D

Explanation

Let the coordinates of the vertices of rectangle ABCD are `A(a cos theta, b sin theta),B(-a cos theta, b sin theta)`,
`C(-a cos theta, -b sin theta)` and `D(a cos theta, -b sin theta)`, then length of rectangle `AB=2a cos theta`
and breadth of rectangle `AD=2b sin theta`
`:.` Area of rectangle `=ABxxAD`
`=2a cos thetaxx2b sin theta`
`implies` Area of rectangle `A=2 ab sin 2 theta`
`:.(dA)/(d theta)=2xx 2 ab cos 2 theta`
For maxima or minima put `(dA)/(d theta)=0`
`implies cos 2 theta=0`
`implies 2 theta=(pi)/2 impliestheta=(pi)/4`

Now, `(d^(2)A)/(d theta^(2))=-8 ab sin 2 theta`
At `theta=pi//4, ((d^(2) a)/( d theta^(2)))_(theta=(pi)/4)lt0`
`:.` Area is maximum at `theta=(pi)/4`
`implies` Maximum area of rectangle `=2ab` sq units.
[ from Eq. (i) ]