Let $C$ be the curve with equation $y = x^2$. Let $L_m$ be the line with equation $y = mx - 1$. Determine the point, $p_m$ on the line $l_m$ closest to $C$ and hence determine a parametric equation for the curve generated as the loci of points $p_m$

The point $C_m$ is the point on the curve closest to the line. The point $C_m$ is such that the line segment $C_mp_m$ is perpendicular to the line $l_m$ and normal to the curve at $C_m$

Let the point $C_m$ on the curve be parametrised by $(t,t^2)$, then it follows that the line segment $C_mp_m$ has gradient $-\frac{1}{m}$ which can be expressed in terms of t as $-\frac{1}{2t}$

The normal line through $C_m$ therefore has equation $$y - t^2 = -\frac{1}{2t}x + \frac{1}{2} + t^2$$

The line $l_m$ has equation $y = (2t)x - 1$

The point $p_m$ is therefore given as the solution of the simultaneous equation $$ \begin{align} y &= -\frac{1}{2t}x + \frac{1}{2} + t^2 \\ y &= 2tx - 1 \end{align} $$

We can now rearrange these equations to express $x$ as a function of $t$.

$$\begin{align} -\frac{1}{2t}x + \frac{1}{2} + t^2 &=2tx - 1 \\ \frac{3}{2} + t^2 &= \left(2t + \frac{1}{2t}\right)x \\ x &= \frac{\frac{3}{2} + t^2}{\left(2t + \frac{1}{2t}\right)} \\ x &= \frac{3t + 2t^3}{4t^2 + 1} \end{align}$$

We can now substitute this expression into the equation $$y = (2t)x -1 $$ to get an equation for y in terms of $t$ only: $$ y = 2t\left(\frac{3t + 2t^3}{4t^2 + 1}\right) - 1 $$

These two equations together provide us with a parametrisation for the loci of the point $p_m$: $$ \left(\frac{3t + 2t^3}{4t^2 + 1},2t\left(\frac{3t + 2t^3}{4t^2 + 1}\right) - 1\right) $$

Plots of the original curve, $y = x^2$ and the parameterised curve are shown in the diagram below.