Question

Area of the greatest rectangle that can be inscribed in the ellipse = 1, is –

• A.
• B.
• C. ab
• D. ) 2ab

Answer 1

)

Sol. Let the co-ordinates of the vertices of rectangle ABCD are A(a cos q, b sin q), B(–a cos q, b sin q),

C (–a cos q, – b sin q) and D (a cos q, – b sin q), then length of rectangle, AB = 2 cos q and breadth of rectangle, AD = 2b sin q.

\ Area of rectangle = AB × AD

Ž Area of rectangle = 2a cos q . 2b sin q

Ž Area of rectangle = 2 ab sin 2q

\ $\frac { \mathrm { dA } } { \mathrm { d } \theta }$ = 2 × 2 ab cos 2q

Put $\frac { \mathrm { dA } } { \mathrm { d } \theta }$ = 0, for maxima or minima

\ $\frac { \mathrm { dA } } { \mathrm { d } \theta }$ = 0

Ž cos 2q = 0 Ž 2q = $\frac { \pi } { 2 }$ Ž q = $\frac { \pi } { 4 }$

= –8ab sin 2q

Now, < 0 Ž q = $\frac { \pi } { 4 }$

\ Area is maximum at q = $\frac { \pi } { 4 }$

Ž Maximum area of rectangle = 2ab.

[(from (i))].