A curve $C$ has parametric equation
$x = t^2 - 1, \quad y = 4t - t^3 + 15, \quad \text{where } t \in \mathbb{R}.$

1. Set $y=0$ to find the parameter $t$ at the $x$-intercept and show it is unique.

For $x$-intercepts, set $y = 0$:
$4t - t^3 + 15 = 0$
$t^3 - 4t - 15 = 0$
By factorisation (or using a graphing calculator):
$(t - 3)(t^2 + 3t + 5) = 0$
For the quadratic factor $t^2 + 3t + 5 = 0$, the discriminant is $\Delta = 3^2 - 4(1)(5) = -11 < 0$.
Thus, there are no other real solutions and $t = 3$ is the unique real root.

2. Substitute $t=3$ into the equation for $x$ to find the corresponding $x$-coordinate.

Substitute $t = 3$ into the equation for $x$:
$x = (3)^2 - 1 = 8$
Therefore, the curve cuts the $x$-axis at $(8, 0)$.

3. Set $x=0$ to find the parameters for the $y$-intercepts.

For $y$-intercepts, set $x = 0$:
$t^2 - 1 = 0$
$t^2 = 1$
$t = 1 \text{ or } t = -1$

4. Substitute the parameters into the equation for $y$ to find the $y$-coordinates.

When $t = 1$:
$y = 4(1) - (1)^3 + 15 = 18$

When $t = -1$:
$y = 4(-1) - (-1)^3 + 15 = 12$

Therefore, the curve cuts the $y$-axis at $(0, 12)$ and $(0, 18)$.

1. Find the equation of the line of symmetry by analysing the parametric equations.

Observe that $x(t) = t^2 - 1$ is an even function, meaning $x(t) = x(-t)$.
For any point on the curve with parameter $t$, there is a corresponding point with parameter $-t$ having the exact same $x$-coordinate.
Let us examine the sum of their $y$-coordinates:
$y(t) + y(-t) = (4t - t^3 + 15) + (4(-t) - (-t)^3 + 15)$
$y(t) + y(-t) = 4t - t^3 + 15 - 4t + t^3 + 15 = 30$
Since their sum is always $30$, their midpoint is:
$\frac{y(t) + y(-t)}{2} = 15$
Thus, for every $x$, the pairs of points on the curve are equidistant from the horizontal line $y = 15$. The equation of the line of symmetry is $y = 15$.

2. Identify key features to sketch the curve.

Using a graphing calculator or by plotting key points, we sketch the curve $C$ based on the following features:

1. Intercepts: The curve passes through the given axial intercepts $(8, 0)$, $(0, 12)$, and $(0, 18)$.
2. Line of Symmetry: Draw a dashed horizontal line at $y = 15$.
3. Turning Point (Leftmost bound): Since $t^2 \ge 0$, the minimum value of $x$ is $-1$ (occurring at $t = 0$). At this point, $y(0) = 15$. This gives a vertical tangent at $(-1, 15)$.
4. Shape: The curve forms a loop starting from $(-1, 15)$, passing through the $y$-axis at $(0,12)$ and $(0,18)$, crossing itself, and opening towards the right as $x \to \infty$ (since $x \to \infty$ as $t \to \pm \infty$).

1. Set up the condition for a self-intersection by using two distinct parameters.

For the curve to cross itself, there must be two distinct parameters $t$ and $s$ ($t \neq s$) that produce the same $x$ and $y$ coordinates.
Equating the $x$-coordinates:
$t^2 - 1 = s^2 - 1$
$t^2 = s^2$
Since $t \neq s$, we must have $s = -t$.

2. Equate the $y$-coordinates and solve for $t$.

Substitute $s = -t$ and equate the $y$-coordinates:
$4t - t^3 + 15 = 4(-t) - (-t)^3 + 15$
$4t - t^3 = -4t + t^3$
$2t^3 - 8t = 0$
$2t(t^2 - 4) = 0$
Since $t \neq s$ implies $t \neq -t$, we have $t \neq 0$.
Therefore, $t^2 - 4 = 0$, giving $t = \pm 2$.

3. Substitute the parameter back into the equations to find the exact coordinates.

Using $t = 2$ (or $t = -2$, which gives the same point):
$x = (2)^2 - 1 = 3$
$y = 4(2) - (2)^3 + 15 = 8 - 8 + 15 = 15$
The coordinates of the point where $C$ crosses itself are $(3, 15)$.

1. Find the value of the parameter $t$ at the point $(8, 0)$.

At the point $(8, 0)$, $x = 8$:
$t^2 - 1 = 8 \implies t^2 = 9 \implies t = 3 \text{ or } -3$
Checking the $y$-coordinate:
When $t = -3$, $y = 4(-3) - (-3)^3 + 15 = 30 \neq 0$.
When $t = 3$, $y = 4(3) - (3)^3 + 15 = 0$.
Therefore, at $(8, 0)$, $t = 3$.

2. Differentiate the parametric equations to find the expression for the gradient $\frac{dy}{dx}$.

$\frac{dx}{dt} = 2t$
$\frac{dy}{dt} = 4 - 3t^2$
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4 - 3t^2}{2t}$

3. Evaluate the gradient at $t = 3$ and formulate the equation of the tangent.

At $t = 3$, the gradient $m$ is:
$m = \frac{4 - 3(3)^2}{2(3)} = \frac{4 - 27}{6} = -\frac{23}{6}$
Using the point-gradient form for the tangent at $(8, 0)$:
$y - 0 = -\frac{23}{6}(x - 8)$
$6y = -23x + 184$
$23x + 6y - 184 = 0 \quad \text{(Shown)}$

4. Substitute the parametric equations of the curve into the tangent equation to find intersection points.

$23(t^2 - 1) + 6(4t - t^3 + 15) - 184 = 0$
$23t^2 - 23 + 24t - 6t^3 + 90 - 184 = 0$
$6t^3 - 23t^2 - 24t + 117 = 0$

5. Solve the resulting polynomial to find the new value of $t$, using the fact that tangency produces a repeated root.

Since the line is a tangent to the curve at $t = 3$, $t = 3$ is a repeated root of the cubic equation. This means $(t - 3)^2$ is a factor.
$(t - 3)^2 = t^2 - 6t + 9$
By comparison or long division:
$6t^3 - 23t^2 - 24t + 117 = (t^2 - 6t + 9)(6t + 13) = 0$
$(t - 3)^2(6t + 13) = 0$
The tangent cuts the curve again when:
$6t + 13 = 0 \implies t = -\frac{13}{6}$

6. Calculate the exact coordinates of the new intersection point.

Substitute $t = -\frac{13}{6}$ into the parametric equations:
$x = \left(-\frac{13}{6}\right)^2 - 1 = \frac{169}{36} - 1 = \frac{133}{36}$
$y = 4\left(-\frac{13}{6}\right) - \left(-\frac{13}{6}\right)^3 + 15$
$y = -\frac{26}{3} + \frac{2197}{216} + 15 = \frac{-1872 + 2197 + 3240}{216} = \frac{3565}{216}$
The coordinates are $\left(\frac{133}{36}, \frac{3565}{216}\right)$.