The curve $C$ has parametric equations
$x = 2t^2, \quad y = 4t.$

1. Find the derivatives of $x$ and $y$ with respect to the parameter $t$.

$\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t}(2t^2) = 4t$

$\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t}(4t) = 4$

2. Use the chain rule to find the gradient $\frac{\mathrm{d}y}{\mathrm{d}x}$ at $t = p$.

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}} = \frac{4}{4t} = \frac{1}{t}$

At $t=p$, the gradient of the tangent is $m_t = \frac{1}{p}$.

3. Formulate the equation of the tangent at $t = p$.

At $t=p$, the point on the curve is $(2p^2, 4p)$.
Using $y - y_1 = m(x - x_1)$:
$y - 4p = \frac{1}{p}(x - 2p^2)$
$py - 4p^2 = x - 2p^2$
$py = x + 2p^2$

4. Find the gradient of the normal and formulate its equation.

The gradient of the normal $m_n = -\frac{1}{m_t} = -p$.
Equation of the normal:
$y - 4p = -p(x - 2p^2)$
$y - 4p = -px + 2p^3$
$y + px = 2p^3 + 4p$

1. Find the parameter corresponding to the given point.

At the point $(18, -12)$,
$4t = -12 \implies t = -3$
(Check with $x$-coordinate: $2(-3)^2 = 18$)

2. Find the equation of the normal at the point.

From part (i), or by differentiating:
$\frac{\mathrm{d}x}{\mathrm{d}t} = 4t$, $\frac{\mathrm{d}y}{\mathrm{d}t} = 4 \implies \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{t}$
At $t = -3$, the gradient of the tangent is $-\frac{1}{3}$, so the gradient of the normal is $3$.
Equation of the normal:
$y - (-12) = 3(x - 18)$
$y = 3x - 66$

3. Find the parameter of the point where the normal intersects the curve again.

Substitute $x = 2t^2$ and $y = 4t$ into the normal equation:
$4t = 3(2t^2) - 66$
$6t^2 - 4t - 66 = 0$
$3t^2 - 2t - 33 = 0$
We know $t = -3$ is a root, so we can factorise:
$(t + 3)(3t - 11) = 0$
Since $Q$ is the other intersection point, $t = \frac{11}{3}$.

4. Calculate the coordinates of $Q$.

Substitute $t = \frac{11}{3}$ back into the parametric equations:
$x = 2\left(\frac{11}{3}\right)^2 = \frac{242}{9}$
$y = 4\left(\frac{11}{3}\right) = \frac{44}{3}$

1. Derive the general equation of the tangent at a point with parameter $p$ (as found in part (i)).

$\frac{dx}{dt} = 4t$ and $\frac{dy}{dt} = 4 \implies \frac{dy}{dx} = \frac{4}{4t} = \frac{1}{t}$.
At parameter $p$, the gradient is $\frac{1}{p}$ and the point is $(2p^2, 4p)$.
The equation of the tangent is:
$y - 4p = \frac{1}{p}(x - 2p^2)$
$py - 4p^2 = x - 2p^2$
$py = x + 2p^2$

2. Substitute the point $(-2, -3)$ into the tangent equation to form a quadratic equation in $p$.

Since the tangents pass through $(-2, -3)$, substitute $x = -2$ and $y = -3$ into the general tangent equation:
$p(-3) = -2 + 2p^2$
$2p^2 + 3p - 2 = 0$

3. Solve for $p$ to find the gradients of the two individual tangents.

$(2p - 1)(p + 2) = 0$
$p = \frac{1}{2} \quad \text{or} \quad p = -2$
The gradients of the tangents are given by $m = \frac{1}{p}$:
$m_1 = \frac{1}{1/2} = 2$
$m_2 = \frac{1}{-2} = -\frac{1}{2}$

4. Show that the two tangents are perpendicular.

Product of gradients $= m_1 \times m_2 = 2 \times \left(-\frac{1}{2}\right) = -1$.
Since the product of their gradients is $-1$, the tangents are perpendicular. (Shown)