曲线 $C$ 具有参数方程 $x = t - \frac{a}{t}, \quad y = t + \frac{a}{t}, \text{ 其中 } t \in \mathbb{R}, \ t \ne 0 \text{ 且 } a \text{ 为正常数。}$

1. Find expressions for $y+x$ and $y-x$ to isolate components of $t$.

Given the parametric equations:
$x = t - \frac{a}{t}$
$y = t + \frac{a}{t}$

Adding both equations gives:
$y + x = \left(t + \frac{a}{t}\right) + \left(t - \frac{a}{t}\right) = 2t$

Subtracting $x$ from $y$ gives:
$y - x = \left(t + \frac{a}{t}\right) - \left(t - \frac{a}{t}\right) = \frac{2a}{t}$

2. Multiply the two expressions to eliminate the parameter $t$.

Multiplying the sum and the difference:
$(y + x)(y - x) = (2t)\left(\frac{2a}{t}\right)$

Simplifying both sides (since $t \ne 0$):
$y^2 - x^2 = 4a$

1. Differentiate the parametric equations to find $\frac{dx}{dt}$ and $\frac{dy}{dt}$.

$x = t - a t^{-1} \implies \frac{dx}{dt} = 1 + a t^{-2} = \frac{t^2 + a}{t^2}$

$y = t + a t^{-1} \implies \frac{dy}{dt} = 1 - a t^{-2} = \frac{t^2 - a}{t^2}$

2. Find the gradient of the tangent at point $P$.

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\frac{t^2 - a}{t^2}}{\frac{t^2 + a}{t^2}} = \frac{t^2 - a}{t^2 + a}$

At point $P$ where $t = p$, the gradient $m = \frac{p^2 - a}{p^2 + a}$.

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

The coordinates of $P$ are $\left(p - \frac{a}{p}, p + \frac{a}{p}\right)$.

Using the point-gradient form $y - y_1 = m(x - x_1)$, the equation of the tangent is:
$y - \left(p + \frac{a}{p}\right) = \frac{p^2 - a}{p^2 + a} \left[ x - \left(p - \frac{a}{p}\right) \right]$

4. Rearrange and simplify the equation to the required form to determine $k$.

Multiply both sides by $(p^2 + a)$:
$y(p^2 + a) - \left(p + \frac{a}{p}\right)(p^2 + a) = x(p^2 - a) - \left(p - \frac{a}{p}\right)(p^2 - a)$

Rearrange to group $x$ and $y$ terms on the left:
$y(p^2 + a) - x(p^2 - a) = \left(p + \frac{a}{p}\right)(p^2 + a) - \left(p - \frac{a}{p}\right)(p^2 - a)$

Expand the right-hand side (RHS):
$\text{RHS} = \left(p^3 + ap + ap + \frac{a^2}{p}\right) - \left(p^3 - ap - ap + \frac{a^2}{p}\right)$
$\text{RHS} = p^3 + 2ap + \frac{a^2}{p} - p^3 + 2ap - \frac{a^2}{p}$
$\text{RHS} = 4ap$

So the equation becomes:
$y(p^2 + a) - x(p^2 - a) = 4ap$

Comparing this with $y(p^2 + a) - x(p^2 - a) = kap$, we find $k = 4$.