The minimum distance of the curve a2x2+b2y2=1 from origin is (a,b>0)

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).