Math, asked by Sweetyhoty6465, 1 year ago

Find the area of the greatest rectangle that can be inscribed in an ellipse x2a2+y2b2=1.; (Note: This question has been split into 2 questions) This question appeared in 65-1,65-2 and 65-3 versions of the paper in 2013.

Answers

Answered by Anonymous
13
I have used general equation of ellipse
That is x^2/a^2 + y^2/b^2=1
Greatest area of rectangle is 2 ab.

Which paper are you talking about? 2013
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Answered by amitnrw
4

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

Step-by-step explanation:

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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