The equation to the curve, which is such that portion of the axis of x cut off between the origin and the tangent at any point is proportional to the ordinate of that point, is

(k is constant of proportionality)

x = y (c - k log y)

Let the curve be y = f(x). The equation of the tangent at any point (x, y) is given by Y - y = f'(x) (X - x). The portion of the axis of X which is cut off between the origin and the tangent at any point is obtained by putting Y = 0. Therefore

$x - \frac{y}{f'(x)} = ky  ⇒ x - y \frac{dy}{dx} = ky $

$⇒ \frac{dy}{dx}  - \frac{x}{y}  = - k$

which is a linear equation in x, and its integrating factor is

$e^{-∫1/y\,dx}=y^{-1}$

Therefore, multiplying by y^-1 we have

$\frac{d}{dx}(xy^{-1}) = - ky^{-1}$

 xy^-1 = - k log y + c

 or  x = y (c - k log y).

Hence (A) is the correct answer.