A curve is such that the sub tangent at any point on it is equal to the square of the abscissa. If it passes through (2, 1), then its equation is:

Concept Used:-

Sub-tangent of a curve is the projection on the x-axis or abscissa of the part of the tangent on that curve among the x-axis and the tangency point.

Explanation:-
The slope of the curve = dy/dx
So,
⇒ dy/dx= y/(subtangent)​
⇒ subtangent = ${\displaystyle \frac{y}{{\displaystyle \frac{dy}{dx}}}}$

From the given question,

⇒ subtangent = (absicssa)²​

$$\begin{gathered} \Rightarrow {\displaystyle \frac{y}{dy/dx}}=x^2 \\ \Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{y}{x^2}} \\ \Rightarrow {\displaystyle \frac{dy}{y}}={\displaystyle \frac{dx}{x^2}} \end{gathered}$$

On integrating, we get,

$$\begin{gathered} \Rightarrow \int {\displaystyle \frac{dy}{y}}=\int {\displaystyle \frac{dx}{x^2}} \\ \Rightarrow \ln y=-{\displaystyle \frac{1}{x}}+C.....(1) \\ \Rightarrow y=e^{-{\displaystyle \frac{1}{x}}+C}.....(2) \end{gathered}$$

Curve passes from (2, 1). So, put x=2 and y=1 in equation 1 to find the value of C.

$$\begin{gathered} \Rightarrow \ln (1)=-{\displaystyle \frac{1}{2}}+C \\ \Rightarrow 0=-{\displaystyle \frac{1}{2}}+C \\ \Rightarrow C={\displaystyle \frac{1}{2}} \end{gathered}$$

Put this value in equation 2,

$$\begin{gathered} \Rightarrow y=e^{-{\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{2}}} \\ \Rightarrow y=e^{{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{x}}} \end{gathered}$$

Thus, the equation of the curve is y = ${\mathrm{e}}^{\frac{1}{2}-\frac{1}{\mathrm{x}}}$

Hence, the correct option is 1.