曲线 $C$ 的参数方程为
$x = t^2, \quad y = t^3 + 1 \quad \text{for } t \in \mathbb{R}.$

1. Find the coordinates of the $y$-intercept.

For the $y$-intercept, set $x = 0$:
$t^2 = 0 \implies t = 0$
Substitute $t = 0$ into the equation for $y$:
$y = (0)^3 + 1 = 1$
The $y$-intercept is $(0, 1)$.

2. Find the coordinates of the $x$-intercept.

For the $x$-intercept, set $y = 0$:
$t^3 + 1 = 0 \implies t^3 = -1 \implies t = -1$
Substitute $t = -1$ into the equation for $x$:
$x = (-1)^2 = 1$
The $x$-intercept is $(1, 0)$.

3. Determine the shape and sketch the curve.

Observe that $x = t^2 \ge 0$ for all real $t$, so the curve lies entirely on or to the right of the $y$-axis.
At $t = 0$, the curve is at $(0, 1)$.
As $t \to \infty$, $x \to \infty$ and $y \to \infty$ (upper branch, concave up).
As $t \to -\infty$, $x \to \infty$ and $y \to -\infty$ (lower branch, concave down, passing through $(1, 0)$).

Sketching instructions for the student:
Draw axes and plot $(0,1)$ and $(1,0)$. Starting from a cusp at $(0,1)$ with a horizontal tangent, draw two branches extending to the right. The upper branch curves up, and the lower branch curves down through $(1,0)$.

1. Find the coordinates of the point $P$ when $t = 1$.

When $t = 1$,
$x = 1^2 = 1$
$y = 1^3 + 1 = 2$
So, the coordinates of $P$ are $(1, 2)$.

2. Find the gradient of the tangent at $P$ using parametric differentiation.

$\frac{dx}{dt} = 2t$
$\frac{dy}{dt} = 3t^2$
Using the chain rule,
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3t^2}{2t} = \frac{3}{2}t$
At $P$ where $t = 1$, the gradient $m$ is:
$m = \frac{3}{2}(1) = \frac{3}{2}$

3. Form the equation of the tangent at $P$.

Using the point $(1, 2)$ and gradient $m = \frac{3}{2}$:
$y - y_1 = m(x - x_1)$
$y - 2 = \frac{3}{2}(x - 1)$
$2y - 4 = 3x - 3$
$2y = 3x + 1$
$y = \frac{3}{2}x + \frac{1}{2}$

1. Retrieve and state the equation of the tangent at $P$ (from part b).

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}} = \frac{3t^2}{2t} = \frac{3}{2}t$
At $t=1$, $x=1$ and $y=2$. The gradient is $m = \frac{3}{2}$.
The equation of the tangent at $P$ is:
$y - 2 = \frac{3}{2}(x - 1) \implies 2y = 3x + 1$

2. Substitute the parametric equations of $C$ into the tangent equation to form an equation in terms of $t$.

Substitute $x = t^2$ and $y = t^3 + 1$ into $2y = 3x + 1$ to find the intersections:
$2(t^3 + 1) = 3(t^2) + 1$
$2t^3 + 2 = 3t^2 + 1$
$2t^3 - 3t^2 + 1 = 0$

3. Solve the cubic equation to find the value of $t$ at $Q$.

Since the line meets the curve at $P$ where $t=1$, $t-1$ is a factor of the cubic polynomial.
Using factorisation (e.g. via long division or synthetic division):
$2t^3 - 3t^2 + 1 = (t-1)(2t^2 - t - 1) = 0$
$(t-1)(t-1)(2t + 1) = 0$
(Alternatively, since the line is a tangent at $P$, $(t-1)^2$ is a repeated factor).

For the other point $Q$, we take the remaining factor:
$2t + 1 = 0 \implies t = -\frac{1}{2}$

4. Calculate the exact coordinates of $Q$.

Substitute $t = -\frac{1}{2}$ into the parametric equations for $x$ and $y$:
$x = \left(-\frac{1}{2}\right)^2 = \frac{1}{4}$
$y = \left(-\frac{1}{2}\right)^3 + 1 = -\frac{1}{8} + 1 = \frac{7}{8}$

The exact coordinates of $Q$ are $\left(\frac{1}{4}, \frac{7}{8}\right)$.

1. Recall the parameters of the points and the tangent equation from earlier parts.

From parts (b) and (c):

• The equation of the tangent at $P(t = 1)$ is $y = \frac{3}{2}x + \frac{1}{2}$.
• Point $P$ has coordinates $(1, 2)$.
• The tangent meets the curve again at $Q$ where $t = -\frac{1}{2}$, giving $x = \frac{1}{4}$ and $y = \frac{7}{8}$.

2. Evaluate the parametric integral for the area under the curve.

The parametric integral computes the net area bounded by the curve and the $x$-axis between $x=\frac{1}{4}$ and $x=1$ (accounting for the curve folding back on itself):
[
\begin{aligned} A_1 &= \int_{-1/2}^{1} y \frac{\mathrm{d}x}{\mathrm{d}t} , \mathrm{d}t \ &= \int_{-1/2}^{1} (t^3 + 1)(2t) , \mathrm{d}t \ &= \int_{-1/2}^{1} (2t^4 + 2t) , \mathrm{d}t \ &= \left[ \frac{2}{5}t^5 + t^2 \right]_{-1/2}^{1} \ &= \left( \frac{2}{5}(1)^5 + (1)^2 \right) - \left( \frac{2}{5}\left(-\frac{1}{2}\right)^5 + \left(-\frac{1}{2}\right)^2 \right) \ &= \frac{7}{5} - \left( -\frac{1}{80} + \frac{1}{4} \right) \ &= \frac{7}{5} - \frac{19}{80} \ &= \frac{112}{80} - \frac{19}{80} = \frac{93}{80} \end{aligned}
]

3. Calculate the area under the tangent line.

The area under the tangent line from $x=\frac{1}{4}$ to $x=1$ is a trapezium:
[
\begin{aligned} A_2 &= \int_{1/4}^{1} \left( \frac{3}{2}x + \frac{1}{2} \right) , \mathrm{d}x \ &= \frac{1}{2} \times \left( y_Q + y_P \right) \times \left( x_P - x_Q \right) \ &= \frac{1}{2} \left( \frac{7}{8} + 2 \right) \left( 1 - \frac{1}{4} \right) \ &= \frac{1}{2} \left( \frac{23}{8} \right) \left( \frac{3}{4} \right) \ &= \frac{69}{64} \end{aligned}
]

4. Subtract the tangent area from the curve's net area to obtain the exact bounded area.

Subtracting $A_2$ from $A_1$ leaves exactly the area enclosed by the curve $C$ and the tangent line:
[
\begin{aligned} \text{Required Area} &= A_1 - A_2 \ &= \frac{93}{80} - \frac{69}{64} \ &= \frac{372}{320} - \frac{345}{320} \ &= \frac{27}{320} \end{aligned}
]