The parametric equations of a curve C are $x = at,\; y = at^{3}$, where $a$ is a positive constant.

1. Find the gradient of the curve at point P.

Given the parametric equations $x = at$ and $y = at^3$.
$\frac{dx}{dt} = a, \quad \frac{dy}{dt} = 3at^2$
Using the chain rule,
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3at^2}{a} = 3t^2$
At point P, the parameter is $p$, so the gradient of the tangent is $m = 3p^2$.

2. Determine the equation of the tangent at P.

The coordinates of P are $(ap, ap^3)$.
Using the point-gradient form $y - y_1 = m(x - x_1)$, the equation of the tangent is:
$y - ap^3 = 3p^2(x - ap)$
$y = 3p^2x - 3ap^3 + ap^3$
$y = 3p^2x - 2ap^3$

3. Find the coordinates of the intercepts S and T, and the midpoint M.

To find the $y$-intercept S, set $x = 0$:
$y = -2ap^3 \implies S(0, -2ap^3)$
To find the $x$-intercept T, set $y = 0$:
$0 = 3p^2x - 2ap^3 \implies 3p^2x = 2ap^3$
Assuming $p \neq 0$, $x = \frac{2ap}{3} \implies T\left(\frac{2ap}{3}, 0\right)$
The coordinates of M, the midpoint of ST, are:
$x_M = \frac{0 + \frac{2ap}{3}}{2} = \frac{ap}{3}$
$y_M = \frac{-2ap^3 + 0}{2} = -ap^3$
Thus, $M\left(\frac{ap}{3}, -ap^3\right)$.

4. Eliminate the parameter $p$ to find the Cartesian equation of the locus of M.

The parametric equations for M are:
$x = \frac{ap}{3} \implies p = \frac{3x}{a}$
$y = -ap^3$
Substitute $p$ into the equation for $y$:
$y = -a\left(\frac{3x}{a}\right)^3$
$y = -a\left(\frac{27x^3}{a^3}\right)$
$y = -\frac{27x^3}{a^2}$

1. Find the limits for the parameter $t$ corresponding to the bounds $x=0$ and $x=3$.

When $x = 0$, $at = 0 \implies t = 0$
When $x = 3$, $at = 3 \implies t = \frac{3}{a}$

2. Set up the definite integral for the area under the curve in terms of $t$.

$\frac{dx}{dt} = a \implies dx = a \, dt$

$\text{Area} = \int_{x=0}^{x=3} y \, dx$
$\text{Area} = \int_{0}^{\frac{3}{a}} (at^3)(a) \, dt$
$\text{Area} = \int_{0}^{\frac{3}{a}} a^2 t^3 \, dt$

3. Evaluate the definite integral to find the exact area.

$= a^2 \left[ \frac{t^4}{4} \right]_0^{\frac{3}{a}}$

$= a^2 \left( \frac{1}{4} \left(\frac{3}{a}\right)^4 - 0 \right)$

$= a^2 \left( \frac{81}{4a^4} \right)$

$= \frac{81}{4a^2}$