Identify the parabola touched by every line in a family

Differentiation gives $2yy^{\prime}=k$, so at contact $y^{\prime}=m=k/(2y)$ and $y=k/(2m)$. For the nondegenerate parabola $k\ne0$, its equation gives $x=k/(4m^2)$. Substitute into the tangent: $k/(2m)=k/(4m)+3/m$, hence $k=12$. The contact point is $(3/m^2,6/m)$.

Substitution yields $(mx+3/m)^2=kx$, or $m^2x^2+(6-k)x+9/m^2=0$. Tangency requires $(6-k)^2-36=0$, giving $k=0$ or $12$. Reject $k=0$: $y^2=0$ is a degenerate line, not the stated parabola. With $k=12$, the quadratic is $(mx-3/m)^2=0$, so $x=3/m^2$ and $y=6/m$, as before.