Find the curves with the property that the tangent segment…
Find the curves with the property that the tangent segment between the $x$-axis and the line $y=ax+b$ is bisected at the point of tangency.
Explanation of Method
Express the tangent line's intersections with the $x$-axis and with $y=ax+b$ in terms of the point of tangency $(x_0,y_0)$ and slope $y'$, impose the midpoint condition, simplify to a first-order ODE, and recognize $y\,dx+x\,dy=d(xy)$ to integrate it directly.
Step 1: Find the tangent's two intersection points
The tangent at $(x_0,y_0)$ is $y=y_0+y'(x-x_0)$. It meets the $x$-axis at $A=\left(x_0-\dfrac{y_0}{y'},\,0\right)$, and meets $y=ax+b$ at $B=\left(\dfrac{y_0-y'x_0-b}{a-y'},\,\dfrac{ay_0-y'(ax_0+b)}{a-y'}\right)$.
Step 2: Impose the midpoint (bisection) condition
Requiring $(x_0,y_0)=\tfrac12(A+B)$ and simplifying both the $x$- and $y$-coordinate equations (they turn out to be equivalent) yields, after dropping subscripts:
\[y'=\frac{ay}{2y-ax-b}.\]
Step 3: Recognize an exact differential
Cross-multiplying: $(2y-ax-b)\,dy=ay\,dx$, i.e. $2y\,dy-b\,dy=a(y\,dx+x\,dy)=a\,d(xy)$, since $y\,dx+x\,dy=d(xy)$.
Step 4: Integrate
$\displaystyle\int(2y-b)\,dy=\int a\,d(xy)$ gives $y^2-by=axy+C$.
Result
Final answer:
\[\boxed{\,y^2-axy-by=C,\qquad C\in\mathbb{R}\,}\]
a family of conics (hyperbolas in general; if $a=0$ these are the rectangular hyperbolas $y(y-b)=C$, and if $a=b=0$ they degenerate to the horizontal lines $y=\pm\sqrt C$). Verified with sympy: implicitly differentiating the boxed relation reproduces $y'=ay/(2y-ax-b)$ exactly.