Find the curves for which the intercept that the tangent…

Find the curves for which the intercept that the tangent cuts on the $y$-axis equals the subnormal at the point of tangency.

Step 1: Set up the geometric condition

The tangent at $(x,y)$ meets the $y$-axis at height $y-xy'$; the subnormal is $yy'$. Equating:

\[y-xy'=yy'.\]

Step 2: Solve for $y'$ and substitute $y=vx$

Solving, $y=y'(x+y)$, so $y'=\dfrac{y}{x+y}$ — homogeneous. With $y=vx$, $y'=v+xv'$:

\[v+xv'=\frac{v}{1+v}\quad\Longrightarrow\quad xv'=\frac{-v^2}{1+v}.\]

Step 3: Separate and integrate

\[\left(\frac{1}{v^2}+\frac1v\right)dv=-\frac{dx}{x}\quad\Longrightarrow\quad -\frac1v+\ln|v|=-\ln|x|+C_1.\]

i.e. $\ln|vx|-\dfrac1v=C_1$.

Step 4: Back-substitute $v=y/x$

Since $vx=y$ and $1/v=x/y$: $\ln|y|-\dfrac{x}{y}=C_1$. Multiplying by $y$:

\[y\ln|y|-x=Cy.\]