曲线 $C$ 的参数方程为
$x = \mathrm{e}^{2t}, \ y = t^2.$

1. Differentiate $x$ and $y$ with respect to $t$.

$\frac{\mathrm{d}x}{\mathrm{d}t} = 2\mathrm{e}^{2t}$

$\frac{\mathrm{d}y}{\mathrm{d}t} = 2t$

2. Find the gradient of the tangent, $\frac{\mathrm{d}y}{\mathrm{d}x}$, using the chain rule.

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}} = \frac{2t}{2\mathrm{e}^{2t}} = \frac{t}{\mathrm{e}^{2t}}$

3. Equate the gradient to the given value and solve for $t$.

$\frac{t}{\mathrm{e}^{2t}} = \frac{1}{2\mathrm{e}}$

$\frac{t}{\mathrm{e}^{2t}} = \frac{\frac{1}{2}}{\mathrm{e}^{2\left(\frac{1}{2}\right)}}$

By comparing the form of both sides (or using a graphing calculator to find the exact intersection), we deduce:
$t = \frac{1}{2}$

1. Find the gradient of the tangent at the point $P$ where $t=p$.

$\frac{dx}{dt} = 2\mathrm{e}^{2t}, \quad \frac{dy}{dt} = 2t$
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2t}{2\mathrm{e}^{2t}} = t\mathrm{e}^{-2t}$
At $t=p$, the gradient of the tangent is $m = p\mathrm{e}^{-2p}$.

2. Formulate the equation of the tangent at $P$.

Using the point-gradient form $y - y_1 = m(x - x_1)$ with $P(\mathrm{e}^{2p}, p^2)$:
$y - p^2 = p\mathrm{e}^{-2p}(x - \mathrm{e}^{2p})$

3. Find the intersection point $D$ with the $y$-axis.

Substitute $x = 0$ into the tangent equation to find the $y$-intercept:
$y - p^2 = p\mathrm{e}^{-2p}(0 - \mathrm{e}^{2p})$
$y - p^2 = p\mathrm{e}^{-2p}(-\mathrm{e}^{2p})$
$y - p^2 = -p$
$y = p^2 - p$
Therefore, the coordinates of $D$ are $(0, p^2 - p)$.

1. Determine the coordinates of $D$ from the previous part

$\frac{\mathrm{d}x}{\mathrm{d}t} = 2\mathrm{e}^{2t}, \quad \frac{\mathrm{d}y}{\mathrm{d}t} = 2t \implies \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{t}{\mathrm{e}^{2t}}$
At $P(\mathrm{e}^{2p}, p^2)$, the gradient is $p\mathrm{e}^{-2p}$.
Equation of tangent at $P$:
$y - p^2 = p\mathrm{e}^{-2p}(x - \mathrm{e}^{2p})$
Substitute $x = 0$ to find the $y$-intercept $D$:
$y = p^2 - p$
So, $D$ is $(0, p^2 - p)$.

2. Find the parametric coordinates of $M$, the midpoint of $PD$

Using the coordinates of $P(\mathrm{e}^{2p}, p^2)$ and $D(0, p^2 - p)$:
$x_M = \frac{\mathrm{e}^{2p} + 0}{2} = \frac{1}{2}\mathrm{e}^{2p}$
$y_M = \frac{p^2 + (p^2 - p)}{2} = \frac{2p^2 - p}{2} = p^2 - \frac{1}{2}p$
Thus, the parametric equations of the curve traced by $M$ are $x = \frac{1}{2}\mathrm{e}^{2p}$ and $y = p^2 - \frac{1}{2}p$.

3. Express the parameter $p$ in terms of $x$

$x = \frac{1}{2}\mathrm{e}^{2p}$
$\mathrm{e}^{2p} = 2x$
$2p = \ln(2x)$
$p = \frac{1}{2}\ln(2x)$

4. Substitute $p$ into the $y$-equation to find the Cartesian equation

$y = \left(\frac{1}{2}\ln(2x)\right)^2 - \frac{1}{2}\left(\frac{1}{2}\ln(2x)\right)$
$y = \frac{1}{4}(\ln(2x))^2 - \frac{1}{4}\ln(2x)$