The correct option is
C a hyperbola
Let the equation of curve be y=f(x)
Let P(x,y) be any point on the curve.
Equation of tangent at P(x,y) is
Y−y=y′(X−x)
⇒y′X−Y=xy′−y
⇒Xxy′−yy′+Yy−xy′=1
So, the intercepts of the tangent on the axis are
xy′−yy′ and y−xy′
So, the point A on x-axis is (xy′−yy′,0) and B on y-axis is (0,y−xy′).
Now, P is mid-point of AB
So coordinates of P are (xy′−y2y′,y−xy′2)
⇒x=xy′−y2y′ and y=y−xy′2
⇒xy′−y=2y′x and xy′−y=−2y
⇒2y′x=−2y
⇒dyy=−dxx
Integrating both sides, we get
logy=−logx+logC
⇒logxy=logC
⇒xy=C