曲线 $C$ 的方程为
$\frac{(x-1)^2}{9} - y^2 = 1.$

1. Find the coordinates of the turning points and axial intercepts.

To find the $x$-intercepts, substitute $y=0$ into the equation of $C$:
$\frac{(x-1)^2}{9} - 0 = 1$
$(x-1)^2 = 9$
$x-1 = \pm 3$
$x = 4 \text{ or } x = -2$
These points $(-2, 0)$ and $(4, 0)$ are both the $x$-intercepts and the turning points of the horizontal hyperbola.
To find the $y$-intercepts, substitute $x=0$:
$\frac{1}{9} - y^2 = 1 \implies y^2 = -\frac{8}{9}$
Since $y^2 \ge 0$ for real $y$, there are no real solutions, meaning there are no $y$-intercepts.

2. Determine the equations of the asymptotes.

For a hyperbola of the form $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$, the asymptotes are obtained by setting the right-hand side to $0$:
$\frac{(x-1)^2}{9} - y^2 = 0$
$y^2 = \frac{(x-1)^2}{9}$
$y = \pm\frac{1}{3}(x-1)$
The equations of the asymptotes are $y = \frac{1}{3}x - \frac{1}{3}$ and $y = -\frac{1}{3}x + \frac{1}{3}$.

3. Sketch the curve with the identified features.

1. Draw a Cartesian coordinate system.
2. Draw the asymptotes $y = \frac{1}{3}(x-1)$ and $y = -\frac{1}{3}(x-1)$ as dotted lines passing through the center $(1, 0)$.
3. Plot the turning points (vertices) at $(-2, 0)$ and $(4, 0)$.
4. Sketch the two branches of the hyperbola opening leftwards and rightwards from the turning points, ensuring the curve approaches the asymptotes smoothly as $x \to \pm\infty$. Label all the key coordinates and equations clearly.

1. Identify the shape and $x$-intercepts of the new curve.

The equation $y^2 + \frac{x^2}{k^2} = 1$ describes an ellipse centered at the origin $(0,0)$ with $y$-intercepts at $(0, \pm 1)$ and $x$-intercepts at $(\pm k, 0)$ (since $k > 0$).

2. Relate the boundaries of the ellipse to the vertices of $C$ to deduce the range of $k$.

From part (i), the hyperbola $C$ has vertices at $(-2, 0)$ and $(4, 0)$.
For the ellipse to intersect $C$ at exactly two distinct points, its $x$-interval $[-k, k]$ must overlap with exactly one branch of the hyperbola.

• The ellipse reaches the left branch when $k > 2$, yielding exactly 2 intersection points on the left branch.
• As $k$ increases, it touches the right branch when $k = 4$, adding a 3rd intersection point at $(4, 0)$.
• For $k > 4$, the ellipse intersects both branches, yielding a total of 4 intersection points.
  Thus, to have exactly 2 distinct points of intersection, we require $2 < k < 4$.