Curve $C$ has parametric equations $x = 3t^2$, $y = -6t$.

1. Differentiate the parametric equations with respect to $t$.

$\frac{dx}{dt} = 6t$
$\frac{dy}{dt} = -6$

2. Find the expression for $\frac{dy}{dx}$ using the chain rule.

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-6}{6t} = -\frac{1}{t}$

3. Find the equation of the tangent at point $P$.

At $P$, $t = p$. The coordinates are $(3p^2, -6p)$ and the gradient of the tangent is $m_t = -\frac{1}{p}$.

Equation of tangent:
$y - (-6p) = -\frac{1}{p}(x - 3p^2)$
$y + 6p = -\frac{1}{p}x + 3p$
$y = -\frac{1}{p}x - 3p$

4. Find the equation of the normal at point $P$.

The gradient of the normal is $m_n = -\frac{1}{m_t} = p$.

Equation of normal:
$y - (-6p) = p(x - 3p^2)$
$y + 6p = px - 3p^3$
$y = px - 3p^3 - 6p$

1. Derive the equations of the tangent and normal to the curve at point $P$ (from part i).

Given $x = 3t^2$ and $y = -6t$, we have $\frac{dx}{dt} = 6t$ and $\frac{dy}{dt} = -6$.
$\frac{dy}{dx} = \frac{-6}{6t} = -\frac{1}{t}$.
At $P(3p^2, -6p)$, the gradient of the tangent is $-\frac{1}{p}$ and the gradient of the normal is $p$.

Equation of the tangent:
$y - (-6p) = -\frac{1}{p}(x - 3p^2)$
$y + 6p = -\frac{1}{p}(x - 3p^2)$

Equation of the normal:
$y - (-6p) = p(x - 3p^2)$
$y + 6p = p(x - 3p^2)$

2. Find the coordinates of $T$ where the tangent meets the $x$-axis.

The tangent meets the $x$-axis at $y = 0$.
Substituting $y=0$ into the tangent equation:
$0 + 6p = -\frac{1}{p}(x - 3p^2)$
$-6p^2 = x - 3p^2$
$x = -3p^2$
Thus, the coordinates of $T$ are $(-3p^2, 0)$.

3. Find the coordinates of $N$ where the normal meets the $x$-axis.

The normal meets the $x$-axis at $y = 0$.
Substituting $y=0$ into the normal equation:
$0 + 6p = p(x - 3p^2)$
Since $0 < p < 1$, $p \neq 0$. Dividing both sides by $p$ gives:
$6 = x - 3p^2$
$x = 3p^2 + 6$
Thus, the coordinates of $N$ are $(3p^2 + 6, 0)$.

4. Calculate the coordinates of the midpoint $M$ of $TN$ and show it is independent of $p$.

Using the midpoint formula for $T(-3p^2, 0)$ and $N(3p^2 + 6, 0)$:
$x_M = \frac{-3p^2 + (3p^2 + 6)}{2} = \frac{6}{2} = 3$
$y_M = \frac{0 + 0}{2} = 0$

Thus, the coordinates of $M$ are $(3, 0)$.
Since the coordinates are constant and do not contain $p$, they are independent of $p$. (Shown)

1. Concisely establish the coordinates of P, M, and N required for this part based on earlier parts.

From part (i), for $x = 3t^2, y = -6t$, we have $\frac{dy}{dx} = -\frac{1}{t}$. At $P(3p^2, -6p)$, the normal gradient is $p$.
The normal $PN$ is $y + 6p = p(x - 3p^2)$. Setting $y=0$ gives $N(3p^2+6, 0)$.
The tangent is $y + 6p = -\frac{1}{p}(x - 3p^2)$. Setting $y=0$ gives $T(-3p^2, 0)$.
From part (ii), the midpoint $M$ of $TN$ is $\left(\frac{-3p^2 + 3p^2 + 6}{2}, 0\right) = (3, 0)$.

2. Determine the gradient of PN and express the tangent of the acute angle it makes with the x-axis.

Let $\beta$ be the acute angle $PN$ makes with the $x$-axis.
The line $PN$ is the normal to $C$ at $P$, which has gradient $m_{PN} = p$.
Since $0 < p < 1$, the gradient is positive, so:
$\tan \beta = p$

3. Calculate the gradient of PM and express the tangent of its acute angle.

Let $\alpha$ be the acute angle $PM$ makes with the $x$-axis.
The gradient of $PM$ is:
$m_{PM} = \frac{-6p - 0}{3p^2 - 3} = \frac{-6p}{3(p^2 - 1)} = \frac{2p}{1 - p^2}$
Given $0 < p < 1$, we have $1 - p^2 > 0$, so $m_{PM} > 0$. Thus:
$\tan \alpha = \frac{2p}{1 - p^2}$

4. Use the double-angle identity to prove the relationship between the angles.

Using the tangent double-angle identity:
$\tan 2\beta = \frac{2\tan \beta}{1 - \tan^2 \beta} = \frac{2p}{1 - p^2}$
We observe that $\tan \alpha = \tan 2\beta$.
For $0 < p < 1$, we have $0 < \tan \beta < 1 \implies 0 < \beta < \frac{\pi}{4}$. This means $2\beta$ is strictly an acute angle ($0 < 2\beta < \frac{\pi}{2}$).
Since both $\alpha$ and $2\beta$ are acute angles with the same tangent, $\alpha = 2\beta$.
This proves the acute angle $PM$ makes with the $x$-axis is always twice the acute angle $PN$ makes.