Find the area of the greatest rectangle that can be inscribed in an ellipse x2a2+y2b2=1.
Answers
Answer:
2ab
Step-by-step explanation:
The equation of eclipse is x²/a² + y²/b² = 1.
Let PQRS be the inscribed rectangle within the ellipse.
Co-ordinates of P = (a cosθ, b sinθ)
Co-ordinates of S = (a cosθ, - b sinθ)
Co-ordinates of Q = ( - a cosθ, b sinθ)
Co-ordinates of R = ( - a cosθ, - b sinθ)
Length of rectangle, RS = 2acosθ
Breadth of rectangle, PS = 2bsinθ.
Hence Area of rectangle, A = RS * PS
= 2acosθ * 2bsinθ
= 4abcosθsinθ.
= 2abSin2θ.
Differentiating both sides with respect to θ,
dA/dθ = 2abCos2θ (2) = 4abCos2θ.
For maximum or minimum, dA/dθ = 0
=> 4abCos2θ = 0
=> Cos2θ = 0
=> 2θ = π/2
=> θ = π/4.
When θ = π/4,
d²A/dθ² = - 8abSin2θ = - 8abSin2(π/4) = - 8absinπ/2 = - 8ab < 0.
∴ π/4 is the point of maximum.
Area is maximum when θ = π/4.
∴A max = 2ab Sin(2* π/4) = 2abSinπ/2 = 2ab.
Thus the max area of rectangle that can be inscribed = 2ab.
2ab is the area of the greatest rectangle that can be inscribed in an ellipse x²/a² + y²/b² = 1
Step-by-step explanation:
refer attached diagram
Area of rectangle = 2x * 2y = 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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