Question

prove that the greatest rectangle that can be inscribed in the ellipse has the area 2ab

Answer 1

Let PQRS be a rectangle inscribed in the ellipse.

Coordinate of P = (acosθ, bsinθ)

Coordinate of S = (acosθ, –bsinθ)

Coordinate of Q = (–acosθ, bsinθ)

Coordinate of R = (–acosθ, –bsinθ)

 

Length of the rectangle, RS = 2a cosθ

Breadth of the rectangle, PS = 2bsinθ

 

Let A be the area of the rectangle.

A = RS × PS

⇒ A = 2a cosθ × 2b sinθ

⇒ A = 4ab sinθ cosθ

⇒ A = 2ab sin2θ

Differentiating both sides w.r.t θ, we get

 refer pic

Thus, the maximum area of a rectangle that can be inscribed in the ellipse is 2ab sq. units.

 

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Answer 2

2ab is the area of the greatest rectangle that can be inscribed in an ellipse x²/a² + y²/b² = 1

Step-by-step explanation:

complete question :  ellipse x²/a² + y²/b² = 1

reffer attached diagram

Area of rectangle = 4xy

A = 4xy

dA/dx = 4xdy/dx  + 4y

x²/a² + y²/b² = 1

=> 2x/a²  + 2y(dy/dx)/b² = 0

=> dy/dx   = - xb²/a²y

dA/dx = 4x( - xb²/a²y)  + 4y

put dA/dx = 0

=> a²y² = b²x²

=> x²/a² = y²/b²

x²/a² + y²/b² = 1

=> x²/a² = y²/b²  = 1/2

x = a/√2  , y = b/√2

Area = 4xy  =  4 ( a/√2)(b/√2)  = 2ab

2ab is the area of the greatest rectangle that can be inscribed in an ellipse x²/a² + y²/b² = 1

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