The area of the parallelogram formed by the tangents at the points whose eccentric angles are θ,θ+π2,θ+π,θ+3π2 on the ellipse x2a2+y2b2=1 is
A
ab
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B
4ab
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C
3ab
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D
2ab
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Solution
The correct option is B 4ab x2a2+y2b2=1
Area=πab
Let P =(acosθ,bsinθ)
S=(ae,0)
M(h,k) be the midpoint of PS (h,k)=(ae+acosθ2,bsinθ2) (h−ae2)2(a2)2+k2(b2)2=1 Area=πab4 Ratio=14