Find the locus of a tangent-segment midpoint

$dx/dt=6t$ and $dy/dt=6$, so $dy/dx=1/t$ for $t\ne0$. For $p\ne0$, the tangent is $y-6p=(x-3p^2)/p$. At $x=0$, $y=3p$, so $D=(0,3p)$.

The midpoint is $M=(3p^2/2,9p/2)$. Since $p=2y/9$, $x=\tfrac32(2y/9)^2=2y^2/27$. This is the Cartesian equation printed in the solution.

At $p=0$, the tangent is $x=0$, which coincides with the y-axis: there is no unique point $D$ determined by an intersection. Thus the locus of uniquely defined midpoints is $x=2y^2/27$ with $(0,0)$ excluded. The printed full parabola includes the origin as a limiting point; it requires an extra convention such as defining $D=(0,0)$ at $p=0$.