曲线 C 的参数方程为 $x=2a\cos\left(t+\tfrac{\pi}{6}\right),\quad y=a\sin t,$ 其中 $0\le t<2\pi$，$a$ 是一个正常数。
下方给出了曲线 C 的草图，其中 C 与正 x 轴相交于点 P，与负 y 轴相交于点 Q。

1. Differentiate the parametric equations with respect to $t$.

$\frac{dx}{dt} = -2a\sin\left(t + \frac{\pi}{6}\right)$
$\frac{dy}{dt} = a\cos t$

2. Find the gradient of the normal at the point with parameter $\theta$.

The gradient of the tangent $\frac{dy}{dx}$ at $t = \theta$ is:
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{a\cos \theta}{-2a\sin\left(\theta + \frac{\pi}{6}\right)} = -\frac{\cos \theta}{2\sin\left(\theta + \frac{\pi}{6}\right)}$

Since the normal is perpendicular to the tangent ($m_1 m_2 = -1$), the gradient of the normal $m$ is:
$m = \frac{2\sin\left(\theta + \frac{\pi}{6}\right)}{\cos \theta}$

3. Expand the numerator using the compound angle formula to show the final result.

Using the identity $\sin(A+B) = \sin A\cos B + \cos A\sin B$:
$m = \frac{2\left(\sin\theta\cos\frac{\pi}{6} + \cos\theta\sin\frac{\pi}{6}\right)}{\cos \theta}$

Substitute the exact values $\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$ and $\sin\frac{\pi}{6} = \frac{1}{2}$:
$m = \frac{2\left(\frac{\sqrt{3}}{2}\sin\theta + \frac{1}{2}\cos\theta\right)}{\cos \theta}$
$m = \frac{\sqrt{3}\sin\theta + \cos\theta}{\cos \theta}$
$m = \sqrt{3}\left(\frac{\sin\theta}{\cos\theta}\right) + 1$
$m = 1 + \sqrt{3}\tan\theta \quad \text{(shown)}$

1. Set $x=0$ to find the points where the curve intersects the y-axis.

At the $y$-axis, the $x$-coordinate is 0. Thus,
$2a\cos\left(t+\frac{\pi}{6}\right) = 0$
Since $a$ is a positive constant, $\cos\left(t+\frac{\pi}{6}\right) = 0$.
Given the domain $0 \le t < 2\pi$, the angle $t+\frac{\pi}{6}$ lies in the interval $\left[\frac{\pi}{6}, \frac{13\pi}{6}\right)$.
$t+\frac{\pi}{6} = \frac{\pi}{2} \quad \text{or} \quad \frac{3\pi}{2}$
$t = \frac{\pi}{2} - \frac{\pi}{6} \quad \text{or} \quad t = \frac{3\pi}{2} - \frac{\pi}{6}$
$t = \frac{\pi}{3} \quad \text{or} \quad t = \frac{4\pi}{3}$

2. Determine which value of $t$ corresponds to the negative y-axis.

Point Q lies on the negative $y$-axis, so its $y$-coordinate must be negative ($y < 0$).
Substitute the possible values of $t$ into $y = a\sin t$:
When $t = \frac{\pi}{3}$, $y = a\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}a$. Since $a > 0$, $y > 0$ (rejected).
When $t = \frac{4\pi}{3}$, $y = a\sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2}a < 0$.
Therefore, $t = \frac{4\pi}{3}$ at the point Q.

1. Find the gradient of the normal at Q.

From part (i), the normal gradient is $m = 1 + \sqrt{3}\tan t$.
At Q, $t = \frac{4\pi}{3}$ (from part (ii)).
$m = 1 + \sqrt{3}\tan\left(\frac{4\pi}{3}\right)$
$m = 1 + \sqrt{3}(\sqrt{3}) = 1 + 3 = 4$

2. Find the coordinates of Q by substituting $t = \frac{4\pi}{3}$ into the parametric equations.

$x = 2a\cos\left(\frac{4\pi}{3} + \frac{\pi}{6}\right) = 2a\cos\left(\frac{9\pi}{6}\right) = 2a\cos\left(\frac{3\pi}{2}\right) = 0$
$y = a\sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2}a$
So, $Q$ is at $\left(0, -\frac{\sqrt{3}}{2}a\right)$.

3. Find the equation of the normal to the curve at Q.

Using the point-gradient form $y - y_1 = m(x - x_1)$:
$y - \left(-\frac{\sqrt{3}}{2}a\right) = 4(x - 0)$
$y + \frac{\sqrt{3}}{2}a = 4x$
$y = 4x - \frac{\sqrt{3}}{2}a$

1. Determine the coordinates of T by finding the equation of the normal at Q.

From part (ii), the parameter at Q is $t = \frac{4\pi}{3}$.
Using the gradient formula from (i), the gradient of the normal at Q is:
$m = 1 + \sqrt{3}\tan\left(\frac{4\pi}{3}\right) = 1 + \sqrt{3}(\sqrt{3}) = 4$

The coordinates of Q are:
$x = 2a\cos\left(\frac{4\pi}{3} + \frac{\pi}{6}\right) = 2a\cos\left(\frac{3\pi}{2}\right) = 0$
$y = a\sin\left(\frac{4\pi}{3}\right) = -\frac{a\sqrt{3}}{2}$

The equation of the normal at Q is:
$y - \left(-\frac{a\sqrt{3}}{2}\right) = 4(x - 0) \implies y = 4x - \frac{a\sqrt{3}}{2}$

The normal intersects the x-axis at T, so we set $y=0$:
$0 = 4x - \frac{a\sqrt{3}}{2} \implies x = \frac{a\sqrt{3}}{8}$
Thus, the distance $OT = \frac{a\sqrt{3}}{8}$.

2. Determine the coordinates of P on the curve.

Point P lies on the positive x-axis, so $y = 0$ and $x > 0$.
$a\sin t = 0 \implies t = 0$ or $t = \pi$ (since $0 \le t < 2\pi$).

For $t = 0$, $x = 2a\cos\left(0 + \frac{\pi}{6}\right) = 2a\left(\frac{\sqrt{3}}{2}\right) = a\sqrt{3} > 0$.
For $t = \pi$, $x = 2a\cos\left(\pi + \frac{\pi}{6}\right) = 2a\left(-\frac{\sqrt{3}}{2}\right) = -a\sqrt{3} < 0$.

Since P lies on the positive x-axis, it corresponds to $t = 0$, giving $P(a\sqrt{3}, 0)$.
Thus, the distance $OP = a\sqrt{3}$.

3. Calculate the ratio and show it is independent of a.

$\dfrac{OT}{OP} = \dfrac{\frac{a\sqrt{3}}{8}}{a\sqrt{3}} = \dfrac{1}{8}$

Since $\frac{1}{8}$ is a constant without the variable $a$, the ratio $\dfrac{OT}{OP}$ is independent of $a$.