已知一曲线的参数方程为 $x=2t$，$y=e^{-t^2}$。

1. Find the coordinates of the intersection with the axes.

For the y-intercept, set $x = 0$:
$2t = 0 \implies t = 0$
Substitute $t = 0$ to find $y$:
$y = e^{-(0)^2} = e^0 = 1$
The point of intersection with the y-axis is $(0, 1)$.

For the x-intercept, set $y = 0$:
$e^{-t^2} = 0$
Since $e^{-t^2} > 0$ for all real $t$, there are no real solutions. Thus, the curve does not intersect the x-axis.

2. Determine the equations of any asymptotes.

Consider the behavior of the curve as $t$ becomes very large in magnitude ($t \to \pm\infty$):
As $t \to \pm\infty$, the x-coordinate $x = 2t \to \pm\infty$.
At the same time, the y-coordinate $y = e^{-t^2} \to 0$.
Therefore, the line $y = 0$ is a horizontal asymptote.

3. Sketch the curve based on its properties (or via its Cartesian equation).

To confirm the shape, we can find the Cartesian equation by substituting $t = \frac{x}{2}$:
$y = e^{-\left(\frac{x}{2}\right)^2} = e^{-\frac{x^2}{4}}$
The graph is an even function (symmetric about the y-axis), passing through the maximum point $(0, 1)$ and forming a bell shape that approaches $y = 0$ as $x \to \pm\infty$.

(The student draws this bell-shaped curve on the axes, labelling the intercept $(0,1)$ and the asymptote $y=0$.)

1. Differentiate parametrically to find the gradient of the tangent at $t = p$.

$\frac{\mathrm{d}x}{\mathrm{d}t} = 2$

$\frac{\mathrm{d}y}{\mathrm{d}t} = -2te^{-t^2}$

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}} = \frac{-2te^{-t^2}}{2} = -te^{-t^2}$

At the point $(2p, e^{-p^2})$, the parameter is $t = p$.
So the gradient of the tangent is $-pe^{-p^2}$.

2. Find the gradient of the normal and formulate the equation of the normal.

The gradient of the normal is $m_n = -\frac{1}{-pe^{-p^2}} = \frac{1}{pe^{-p^2}} = \frac{e^{p^2}}{p}$.

Using the point-slope form, the equation of the normal at $(2p, e^{-p^2})$ is:
$y - y_1 = m_n(x - x_1)$

$y - e^{-p^2} = \frac{e^{p^2}}{p}(x - 2p)$

$y - e^{-p^2} = \frac{e^{p^2}}{p}x - 2e^{p^2}$

$y = \frac{e^{p^2}}{p}x - 2e^{p^2} + e^{-p^2}$

1. Find the parameter $t$ and the equation of the normal at point $C$.

At point $C\left(2, \frac{1}{e}\right)$, we use the x-coordinate to find $t$:
$x = 2t \implies 2 = 2t \implies t=1$
Using the derivatives from the parametric equations:
$\frac{dx}{dt} = 2 \quad \text{and} \quad \frac{dy}{dt} = -2te^{-t^2}$
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-2te^{-t^2}}{2} = -te^{-t^2}$
At $t=1$, the gradient of the tangent is:
$m_t = -1(e^{-1}) = -\frac{1}{e}$
The gradient of the normal $m_n$ is the negative reciprocal of the tangent gradient:
$m_n = -\frac{1}{m_t} = e$
Using the point-slope form, the equation of the normal at $C$ is:
$y - \frac{1}{e} = e(x - 2)$

2. Find the coordinates of intersection points $A$ and $B$.

The normal meets the x-axis at $A$. Set $y = 0$:
$0 - \frac{1}{e} = e(x - 2)$
$x - 2 = -\frac{1}{e^2} \implies x = 2 - \frac{1}{e^2}$
Thus, $A$ is $\left(2 - \frac{1}{e^2}, 0\right)$.

The normal meets the y-axis at $B$. Set $x = 0$:
$y - \frac{1}{e} = e(0 - 2)$
$y = \frac{1}{e} - 2e$
Thus, $B$ is $\left(0, \frac{1}{e} - 2e\right)$.

3. Calculate the distances $|OA|$ and $|OB|$ to deduce the ratio.

The distance $|OA|$ is the magnitude of the x-coordinate of $A$:
$|OA| = \left| 2 - \frac{1}{e^2} \right| = 2 - \frac{1}{e^2} \quad (\text{since } 2 > \frac{1}{e^2})$
The distance $|OB|$ is the magnitude of the y-coordinate of $B$:
$|OB| = \left| \frac{1}{e} - 2e \right| = 2e - \frac{1}{e} \quad (\text{since } 2e > \frac{1}{e})$
We can factor out $e$ from $|OB|$ to compare with $|OA|$:
$|OB| = e\left(2 - \frac{1}{e^2}\right)$
The required ratio is:
$|OA| : |OB| = \left(2 - \frac{1}{e^2}\right) : e\left(2 - \frac{1}{e^2}\right)$
$|OA| : |OB| = 1 : e$