曲线 $C$ 的方程为 $x^2 - 2xy + 2y^2 = k$，其中 $k$ 为常数。 用 $x$ 和 $y$ 表示 $\frac{\mathrm{d}y}{\mathrm{d}x}$。 已知 $C$ 上存在两个点，其切线平行于直线 $y = x$，求 $k$ 的取值范围。 已知 $k = 4$，求曲线 $C$ 上切线平行于 $y$ 轴的每个点的精确坐标。

1. Differentiate the equation of the curve implicitly with respect to $x$.

$\frac{\mathrm{d}}{\mathrm{d}x}(x^2) - \frac{\mathrm{d}}{\mathrm{d}x}(2xy) + \frac{\mathrm{d}}{\mathrm{d}x}(2y^2) = \frac{\mathrm{d}}{\mathrm{d}x}(k)$
$2x - \left( 2y + 2x\frac{\mathrm{d}y}{\mathrm{d}x} \right) + 4y\frac{\mathrm{d}y}{\mathrm{d}x} = 0$

2. Make $\frac{\mathrm{d}y}{\mathrm{d}x}$ the subject of the formula.

$2x - 2y - 2x\frac{\mathrm{d}y}{\mathrm{d}x} + 4y\frac{\mathrm{d}y}{\mathrm{d}x} = 0$
$\frac{\mathrm{d}y}{\mathrm{d}x}(4y - 2x) = 2y - 2x$
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2(y - x)}{2(2y - x)} = \frac{y - x}{2y - x}$

3. Use the condition for tangents parallel to $y = x$ to find a relationship between $x$ and $y$.

The line $y = x$ has a gradient of $1$.
$\frac{y - x}{2y - x} = 1$
$y - x = 2y - x$
$y = 0$

4. Determine the range of $k$ by finding the condition for two distinct points.

Substitute $y = 0$ into the curve equation $x^2 - 2xy + 2y^2 = k$:
$x^2 - 0 + 0 = k$
$x^2 = k$
For the curve to have two distinct points where the tangent is parallel to $y = x$, this equation must yield two distinct real roots for $x$. Hence,
$k > 0$

5. Use the condition for tangents parallel to the $y$-axis to relate $x$ and $y$.

A tangent parallel to the $y$-axis has an undefined gradient, meaning $\frac{\mathrm{d}x}{\mathrm{d}y} = 0$.
Setting the denominator of $\frac{\mathrm{d}y}{\mathrm{d}x}$ to $0$:
$2y - x = 0$
$x = 2y$

6. Substitute $x = 2y$ and $k = 4$ into the curve equation to solve for $y$.

Substitute into $x^2 - 2xy + 2y^2 = 4$:
$(2y)^2 - 2(2y)y + 2y^2 = 4$
$4y^2 - 4y^2 + 2y^2 = 4$
$2y^2 = 4$
$y^2 = 2$
$y = \pm \sqrt{2}$

7. Calculate the corresponding $x$-coordinates and state the exact coordinates.

When $y = \sqrt{2}$, $x = 2(\sqrt{2}) = 2\sqrt{2}$.
When $y = -\sqrt{2}$, $x = 2(-\sqrt{2}) = -2\sqrt{2}$.

The exact coordinates are $(2\sqrt{2}, \sqrt{2})$ and $(-2\sqrt{2}, -\sqrt{2})$.