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

1. Find the parameter $t$ when the curve cuts the x-axis and show the corresponding $x$-coordinate is 8.

At the $x$-axis, $y = 0$:
$4t - t^3 + 15 = 0$
$t^3 - 4t - 15 = 0$
By observation or using a calculator, $t = 3$ is a root. Factorising gives:
$(t - 3)(t^2 + 3t + 5) = 0$
For the quadratic factor $t^2 + 3t + 5 = 0$, the discriminant is $3^2 - 4(1)(5) = -11 < 0$. Therefore, $t = 3$ is the only real root.
When $t = 3$:
$x = (3)^2 - 1 = 8$
Thus, the curve cuts the $x$-axis at $(8, 0)$.

2. Find the parameters $t$ when the curve cuts the y-axis and show the corresponding $y$-coordinates are 12 and 18.

At the $y$-axis, $x = 0$:
$t^2 - 1 = 0$
$t = 1 \text{ or } t = -1$
Substitute these values of $t$ into the equation for $y$:
When $t = 1$:
$y = 4(1) - (1)^3 + 15 = 18$
When $t = -1$:
$y = 4(-1) - (-1)^3 + 15 = -4 + 1 + 15 = 12$
Thus, the curve cuts the $y$-axis at $(0, 18)$ and $(0, 12)$.

1. Deduce the equation of the line of symmetry by comparing the coordinates for parameters $t$ and $-t$.

Since $x = t^2 - 1$, the $x$-coordinate is the same for both $t$ and $-t$.
Let us evaluate the sum of their corresponding $y$-coordinates:
$y(t) + y(-t) = (4t - t^3 + 15) + (-4t + t^3 + 15) = 30$
$\frac{y(t) + y(-t)}{2} = 15$
This means the midpoint of the two points $(x(t), y(t))$ and $(x(-t), y(-t))$ always lies on the horizontal line $y = 15$.
Thus, the equation of the line of symmetry is $y = 15$.

2. Identify key features and sketch the curve using a graphing calculator.

Using a graphing calculator, sketch the shape of $C$.
Key features to clearly label on the sketch:

• The $x$-intercept at $(8, 0)$ (found in part i)
• The $y$-intercepts at $(0, 12)$ and $(0, 18)$ (found in part i)
• The leftmost vertex where $\frac{\mathrm{d}x}{\mathrm{d}t} = 0 \implies 2t = 0 \implies t = 0$, giving the point $(-1, 15)$
• A dashed horizontal line for the line of symmetry $y = 15$
• The curve forming a loop opening towards the right, visually crossing itself on the line of symmetry.

1. Establish the relationship between the two parameter values at the point of intersection.

Let the curve cross itself at a point with parameters $t_1$ and $t_2$, where $t_1 \neq t_2$.
Equating the $x$-coordinates:
$t_1^2 - 1 = t_2^2 - 1$
$t_1^2 = t_2^2$
Since $t_1 \neq t_2$, we must have $t_1 = -t_2$.

2. Equate the $y$-coordinates to find the exact parameter values.

Equating the $y$-coordinates and substituting $t_2 = -t_1$:
$4t_1 - t_1^3 + 15 = 4t_2 - t_2^3 + 15$
$4t_1 - t_1^3 = 4(-t_1) - (-t_1)^3$
$4t_1 - t_1^3 = -4t_1 + t_1^3$
$2t_1^3 - 8t_1 = 0$
$2t_1(t_1^2 - 4) = 0$
Since $t_1 \neq 0$ (otherwise $t_1 = t_2 = 0$), we have:
$t_1^2 - 4 = 0 \implies t_1 = 2 \text{ or } -2$

3. Calculate the coordinates of the intersection point.

Substitute $t = 2$ (or $t = -2$) into the parametric equations:
$x = (2)^2 - 1 = 3$
$y = 4(2) - (2)^3 + 15 = 15$
The coordinates are $(3, 15)$.

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

\text{At } (8, 0),
$x = t^2 - 1 = 8 \implies t^2 = 9 \implies t = \pm 3$
\text{Substitute into } y = 4t - t^3 + 15:
\text{If } t = 3,, y = 4(3) - 3^3 + 15 = 0 \text{ (matches)}
\text{If } t = -3,, y = 4(-3) - (-3)^3 + 15 = 30 \neq 0
\text{Hence, the point } (8, 0) \text{ corresponds to } t = 3.

2. Find the gradient of the curve at $t = 3$.

$\frac{\mathrm{d}x}{\mathrm{d}t} = 2t$
$\frac{\mathrm{d}y}{\mathrm{d}t} = 4 - 3t^2$
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}} = \frac{4 - 3t^2}{2t}$
\text{At } t = 3,, \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{4 - 3(3)^2}{2(3)} = \frac{4 - 27}{6} = -\frac{23}{6}

3. Formulate the equation of the tangent at $(8, 0)$.

\text{Using the point-gradient form } y - y_1 = m(x - x_1):
$y - 0 = -\frac{23}{6}(x - 8)$
$6y = -23x + 184$
$23x + 6y - 184 = 0 \quad \text{(Shown)}$

4. Set up an equation to find where the tangent cuts the curve again.

\text{Substitute the parametric equations } x = t^2 - 1 \text{ and } y = 4t - t^3 + 15 \text{ into the tangent equation:}
$23(t^2 - 1) + 6(4t - t^3 + 15) - 184 = 0$
$23t^2 - 23 + 24t - 6t^3 + 90 - 184 = 0$
$-6t^3 + 23t^2 + 24t - 117 = 0$
$6t^3 - 23t^2 - 24t + 117 = 0$

5. Solve the cubic equation to find the new value of $t$.

\text{Since the line is tangent to the curve at } t = 3\text{, } (t - 3)^2 \text{ is a factor of the cubic polynomial.}
$6t^3 - 23t^2 - 24t + 117 = (t - 3)^2(6t + 13) = 0$
\text{For the point where the tangent cuts the curve again, we solve:}
$6t + 13 = 0 \implies t = -\frac{13}{6}$

6. Calculate the exact coordinates at the new parameter $t$.

\text{When } t = -\frac{13}{6},
$x = \left(-\frac{13}{6}\right)^2 - 1 = \frac{169}{36} - 1 = \frac{133}{36}$
$y = 4\left(-\frac{13}{6}\right) - \left(-\frac{13}{6}\right)^3 + 15 = -\frac{26}{3} + \frac{2197}{216} + 15 = \frac{3565}{216}$
\text{The exact coordinates are } \left(\frac{133}{36}, \frac{3565}{216}\right).