Tangent at which point of the curve $$x=ct^3, y=\frac{c}{t^3}$$ is parallel to the line $$y=2x-1?$$​​

Correct option is D

​$$\textbf{Given:} \quad x = ct^3, \quad y = \frac{c}{t^3} \\[8pt]\text{We are to find the point where the tangent is parallel to the line } y = 2x - 1 \\[8pt]\text{The slope of the line is: } m = 2 \\[8pt]\text{Differentiate } x \text{ and } y \text{ with respect to } t: \\[4pt]\frac{dx}{dt} = 3ct^2, \quad \frac{dy}{dt} = -\frac{3c}{t^4} \\[8pt]\text{Slope of tangent to the curve:} \\[4pt]\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-\frac{3c}{t^4}}{3ct^2} = -\frac{1}{t^6} \\[8pt]\text{Set slope equal to that of the line:} \\[4pt]-\frac{1}{t^6} = 2 \Rightarrow t^6 = -\frac{1}{2} \\[8pt]\text{This equation has no real solution, as } t^6 > 0 \text{ for all real } t \\[8pt]\boxed{\text{Therefore, the tangent is never parallel to the line } y = 2x - 1}$$

Final Answer:
Option D – at no point​​