The family of curves, the subtangent at any point of which is the arithmetic mean of the coordinates of point of tangency

$\begin{array}{l}LetP(x,y)bethepo\mathrm{int}\\ L.S.T=\frac{x+y}{2}\\ \Rightarrow \frac{y}{m}=\frac{x+y}{2}\\ \Rightarrow 2y=(x+y)\frac{dy}{dx}\\ \Rightarrow \frac{dx}{dy}=\frac{x+y}{2y}\\ \Rightarrow \frac{dx}{dy}-\frac{1}{2y}x=\frac{1}{2}\\ I.F.e^{\int -\frac{1}{2y}dy}=e^{-\frac{1}{2}\log y}\\ =e^{\log \left(\frac{1}{\sqrt{y}}\right)}\\ =\frac{1}{\sqrt{y}}\end{array}$

$\begin{array}{l}Solutionis\\ x.\left(\frac{1}{\sqrt{y}}\right)=\int \frac{1}{2}\left(\frac{1}{\sqrt{y}}\right)dy\\ =\frac{1}{2}\left(\frac{\sqrt{y}}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)\\ \frac{x}{\sqrt{y}}=\sqrt{y}+c\\ x=y+c\sqrt{y}\\ \therefore x-y=c\sqrt{y}\\ \Rightarrow {(x-y)}^2=cy\end{array}$