求 $k$ 的取值范围，使得直线 $y = kx + 2$ 与曲线 $xy - y = 2x$ 相交于两个不同的点。 由此，写出使直线 $y = kx + 2$ 为曲线 $xy - y = 2x$ 的切线时的所有可能的 $k$ 值。

1. Substitute the equation of the line into the equation of the curve

Substitute $y = kx + 2$ into $xy - y = 2x$:
$x(kx + 2) - (kx + 2) = 2x$

2. Simplify the equation into a standard quadratic form $ax^2 + bx + c = 0$

$kx^2 + 2x - kx - 2 = 2x$
$kx^2 - kx - 2 = 0$

3. Apply the discriminant condition for two distinct real roots

For the line to intersect the curve at two distinct points, the quadratic equation must have two distinct real roots, so $b^2 - 4ac > 0$.
$(-k)^2 - 4(k)(-2) > 0$
$k^2 + 8k > 0$

4. Solve the quadratic inequality for $k$

$k(k + 8) > 0$
The critical values are $k = 0$ and $k = -8$.
Since the quadratic curve $y = k^2 + 8k$ opens upwards, the inequality is satisfied for:
$k < -8 \text{ or } k > 0$

1. Identify the condition for the line to be a tangent using the discriminant from part (i)

From part (i), the intersection of $y = kx + 2$ and $xy - y = 2x$ gives the quadratic equation $kx^2 - kx - 2 = 0$.
For the line to be a tangent, the discriminant must be zero:
$(-k)^2 - 4(k)(-2) = 0$
$k^2 + 8k = 0$

2. Solve for $k$ and reject the invalid value

$k(k + 8) = 0$
$k = 0 \text{ or } k = -8$
If $k = 0$, the equation $kx^2 - kx - 2 = 0$ becomes $-2 = 0$, which has no solution (the line $y=2$ is an asymptote, not a tangent).
Therefore, $k = -8$.