曲线 $C$ 由参数方程
$x = t^2, \quad y = t^3.$
定义。

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

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

2. Calculate the gradient of the curve $\frac{dy}{dx}$ using the chain rule.

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

3. Form the equation of the tangent at $(t^2, t^3)$ and rearrange it to the required form.

$y - y_1 = m(x - x_1)$
$y - t^3 = \frac{3t}{2}(x - t^2)$
$2(y - t^3) = 3t(x - t^2)$
$2y - 2t^3 = 3tx - 3t^3$
$2y - 3tx + t^3 = 0 \quad \text{(shown)}$

1. Form an equation in terms of $t$ by substituting the given fixed point into the tangent equation.

Substitute $x = a$ and $y = b$ into the equation of the tangent $2y - 3tx + t^3 = 0$:
$2b - 3a(t) + t^3 = 0$
$t^3 - 3at + 2b = 0$

2. Relate the degree of the polynomial to the maximum number of tangents.

The equation $t^3 - 3at + 2b = 0$ is a cubic equation in $t$. Since a cubic equation can have at most $3$ real roots, there are at most $3$ distinct values of $t$ that satisfy this equation for any given $(a, b)$. Each value of $t$ corresponds to a unique point of tangency on the curve $C$, meaning there can be at most $3$ tangent lines passing through the point $(a, b)$.

1. Find the equation of the tangent at $P$ where $t = 2$.

Substitute $t = 2$ into the given equation of the tangent from part (i):
$2y - 3(2)x + (2)^3 = 0$
$2y - 6x + 8 = 0$
$y - 3x + 4 = 0$

2. Set up an equation to find the point of intersection $Q$.

Substitute the parametric equations for point $Q$ ($x = k^2$, $y = k^3$) into the tangent equation:
$k^3 - 3(k^2) + 4 = 0$
$k^3 - 3k^2 + 4 = 0$

3. Solve the cubic equation for $k$, using the known point of tangency to factorise.

Since the line passes through $P$ ($t=2$), we know $k=2$ must be a root of the equation. We can factorise the cubic equation using the factor theorem:
$(k - 2)(k^2 - k - 2) = 0$
$(k - 2)(k - 2)(k + 1) = 0$
$(k - 2)^2(k + 1) = 0$
(Note: $k=2$ is a repeated root because the line is a tangent to the curve at $t=2$.)

Since $Q$ is the distinct point where the line meets the curve again, $k \neq 2$.
Therefore,
$k + 1 = 0 \implies k = -1$

1. Find the coordinates of $P$ and the gradient of the normal at $P$.

At the point $P$ where $t = 2$, the coordinates are:
$x = 2^2 = 4$
$y = 2^3 = 8$

From the tangent equation in (i), $2y - 3tx + t^3 = 0$, the gradient of the tangent is $\frac{3t}{2}$.
At $t = 2$, gradient of the tangent $= \frac{3(2)}{2} = 3$.
Therefore, the gradient of the normal is $-\frac{1}{3}$.

2. Formulate the equation of the normal.

Using the point-gradient form $y - y_1 = m(x - x_1)$:
$y - 8 = -\frac{1}{3}(x - 4)$
$3y - 24 = -x + 4$
$x + 3y = 28$

1. Find the intercepts with the axes and the domain of the curve.

To find intersections with the $y$-axis, set $x = 0$:
$t^2 = 0 \implies t = 0$
Substituting $t = 0$ into $y = t^3$, we get $y = 0$.
To find intersections with the $x$-axis, set $y = 0$:
$t^3 = 0 \implies t = 0$, giving $x = 0$.
Thus, the only intersection with the axes is the origin $(0,0)$.
Since $x = t^2$, we have $x \ge 0$ for all real $t$. The curve exists only in the right half-plane.

2. Determine the shape and sketch the curve.

By squaring $y$ and cubing $x$, we can eliminate $t$ to find the Cartesian equation:
$y^2 = (t^3)^2 = t^6$
$x^3 = (t^2)^3 = t^6$
Therefore, the Cartesian equation is $y^2 = x^3$.
This indicates that the curve is symmetrical about the $x$-axis.
Furthermore, from earlier parts (or by differentiation), $\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3}{2}t$. As $t \to 0$, $\frac{dy}{dx} \to 0$, meaning the curve is horizontal at the origin, forming a cusp.

(Student sketches a semi-cubical parabola symmetric about the positive $x$-axis, with a cusp at the origin $(0,0)$ and labels the origin)

1. Establish the right-angled triangle formed by the tangent, normal, and the $x$-axis.

From previous parts, the gradient of the tangent at $P$ (where $t=2$) is $3$ and the gradient of the normal is $-\frac{1}{3}$.
Let the tangent and normal at $P$ intersect the $x$-axis at points $A$ and $B$ respectively.
Since the tangent and normal are perpendicular, $\triangle PAB$ is a right-angled triangle with $\angle APB = \frac{\pi}{2}$.

2. Use the geometrical meaning of gradients to find the angles of the triangle and deduce the identity.

The tangent has a gradient of $3$, so the angle it makes with the positive $x$-axis is $\angle PAB = \tan^{-1}(3)$.
The normal has a gradient of $-\frac{1}{3}$, so the angle it makes with the negative $x$-axis (which lies inside the triangle) is $\angle PBA = \tan^{-1}\left(\left|-\frac{1}{3}\right|\right) = \tan^{-1}\left(\frac{1}{3}\right)$.
Since the sum of angles in $\triangle PAB$ is $\pi$:
$\angle PAB + \angle PBA + \angle APB = \pi$
$\tan^{-1}(3) + \tan^{-1}\left(\frac{1}{3}\right) + \frac{\pi}{2} = \pi$
$\therefore \tan^{-1}(3) + \tan^{-1}\left(\frac{1}{3}\right) = \frac{\pi}{2}$