If a curve passes through the point \((1,1)\) and at any point \((x, y)\) on the curve, the product of the slope of its tangent and \(x\) coordinate of the point is equal to the \(y\) coordinate of the point, then the curve also passes through the point

Let the equation of the curve by \(y=f(x)\) and
curve passes through the point \((1,1)\), So, \(f(1)=1\)
Also, we know that at any point \((x, y)\) on the curve, the product of the slope of its tangent and \(x\) co-ordinate of the point is equal to the \(y\) co-ordinate of the point.
\(\begin{aligned} \frac{y}{x} & =\frac{d y}{d x} \\ x y^{\prime} & =y\end{aligned}\)
\(\frac{y^{\prime}}{y}=\frac{1}{x}\)
On integrating both sides w.r.t. \(x\), we get
\(\begin{aligned} \ln |y| & =|\operatorname{In}| x \mid+e^c \\ y & =k x, \text { where } k=e^c\end{aligned}\)
Now, we can use the fact that the curve passes through the point \((1,1)\) to find the value of \(k\).
\(\Rightarrow \mathrm{l}=k(\mathrm{l}) \quad k=1\)
Therefore, the equation of the curve is \(y=x\) Substituting the value of \(x\) and \(y\) from the given option, we find that the curve passes through the point \((2,2)\)