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$The minimum distance of the curve \frac{a^{2}}{x^{2}} + \frac{b^{2}}{y^{2}} = 1 from origin is (a, b > 0)$

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a

a-b

b

a+b

c

ab

d

0

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detailed solution

Correct option is B

Given curve is a2x2+b2y2=1 here x≤-a or x≥a & y⩽-b or y⩾b compare with cos2⁡θ+sin2⁡θ=1⇒ax=cos⁡θ,by=sin⁡θ⇒x=asec⁡θ and y=bcosec⁡θ Minimum distance of the curve from the origin =x2+y2=a2sec2⁡θ+b2cosec2⁡θ=a2+b2+a2tan2⁡θ+b2cot2⁡θ=a2+b2+(atan⁡θ−bcot⁡θ)2+2ab=(a+b)2+(atan⁡θ−bcot⁡θ)2≥a+b Therefore, the correct answer is (2).

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