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.
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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
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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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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