The correct option is B a2x2+b2y2=4
Let the equation of tangent to the ellipse x2a2+y2b2=1 be xacosθ+ybsinθ=1
So, the points
A=(acosθ,0), B=(0,bsinθ)
Let the mid point of AB be (h,k)
⇒a2cosθ=h, b2sinθ=k
⇒2cosθ=ah,2sinθ=bk
∴ The required locus is a2x2+b2y2=4