Let the co-ordinates of the vertices of rectangle ABCD are A(a cos θ , b sin θ ), B(–a cos θ , b sin θ ), C (–a cos θ , – b sin θ ) and D (a cos θ , – b sin θ ), then length of rectangle, AB = 2 cos θ and breadth of rectangle, AD = 2b sin θ .

∴ Area of rectangle = AB × AD

⇒ Area of rectangle = 2a cos θ . 2b sin θ

⇒ Area of rectangle = 2 ab sin 2 θ

∴ = 2 × 2 ab cos 2 θ

Put = 0, for maxima or minima

∴ = 0

⇒ cos 2 θ = 0 ⇒ 2 θ = ⇒ θ =

= –8ab sin 2 θ

Now, < 0 ⇒ θ =

∴ Area is maximum at θ =

⇒ Maximum area of rectangle = 2ab.

[(from (i))].