曲线 $C_1$ 的方程为 $p^2x^2 - y^2 = p^2$，其中 $p > 1$。

1. Rewrite the equation into the standard form of a hyperbola.

$p^2x^2 - y^2 = p^2$
Dividing through by $p^2$:
$x^2 - \frac{y^2}{p^2} = 1$
This represents a standard hyperbola with a horizontal transverse axis.

2. Find the coordinates of the points of intersection with the axes.

For $x$-intercepts, set $y = 0$:
$p^2x^2 = p^2 \implies x^2 = 1 \implies x = \pm 1$
The intersections with the $x$-axis are $(1, 0)$ and $(-1, 0)$.

For $y$-intercepts, set $x = 0$:
$-y^2 = p^2$
Since $p > 1$, $p^2 > 0$, giving $y^2 < 0$ which has no real solutions. There are no intersections with the $y$-axis.

3. Determine the coordinates of any stationary points using differentiation.

Differentiating implicitly with respect to $x$:
$2p^2x - 2y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{p^2x}{y}$
For stationary points, $\frac{dy}{dx} = 0 \implies x = 0$.
However, as shown in the previous step, when $x = 0$, there are no real values for $y$. Thus, there are no stationary points on $C_1$.

4. Find the equations of the asymptotes.

The asymptotes for the hyperbola $x^2 - \frac{y^2}{p^2} = 1$ are given by replacing the constant $1$ with $0$:
$x^2 - \frac{y^2}{p^2} = 0$
$y^2 = p^2x^2 \implies y = px \text{ or } y = -px$

5. Synthesize the findings to describe the sketch of C1.

Draw the coordinate axes and the two diagonal asymptotes $y = px$ and $y = -px$. Since $p > 1$, these lines have a steeper gradient than $y=x$.
Draw the two branches of the curve opening outwards to the left and right:

• The left branch has a vertex at $(-1, 0)$.
• The right branch has a vertex at $(1, 0)$.
  The curves must smoothly approach the asymptotes as $x \to \pm\infty$.

1. Determine the equation of $C_2$ based on the change in the line of symmetry.

The curve $C_1$, given by $p^2x^2 - y^2 = p^2$, is centered at the origin and has a vertical line of symmetry at $x = 0$.
For $C_2$ to have a line of symmetry at $x = 2$ through a single transformation (based on standard transformations $y = \mathrm{f}(x+a)$), $C_1$ must be translated 2 units in the positive $x$-direction.
Replacing $x$ with $(x - 2)$ yields the equation of $C_2$:
$p^2(x - 2)^2 - y^2 = p^2$

2. Substitute the given point $(4,3)$ to solve for $p$.

Substitute $x = 4$ and $y = 3$ into the equation of $C_2$:
$p^2(4 - 2)^2 - 3^2 = p^2$
$p^2(2)^2 - 9 = p^2$
$4p^2 - 9 = p^2$
$3p^2 = 9$
$p^2 = 3$
Since it is given that $p > 1$, we reject the negative root ($p = -\sqrt{3}$) to get:
$p = \sqrt{3}$

1. Identify the key features to sketch the graph of $y = f'(x)$.

Based on the given graph of $y = f(x)$:

• $f(x)$ has a maximum at $x = -1$, meaning its gradient is zero. The graph of $f'(x)$ must pass through the $x$-axis at $(-1, 0)$.
• For $x < -1$, $f(x)$ is increasing, so $f'(x) > 0$ (the graph is above the $x$-axis).
• For $x > -1$, $f(x)$ is decreasing, so $f'(x) < 0$ (the graph is below the $x$-axis).
• As $x \to \infty$, $f(x)$ approaches a horizontal asymptote $y = 0$. The gradient approaches $0$, so $f'(x) \to 0$ from below the $x$-axis.

2. Relate the given equation to the graphs of $y = f'(x)$ and $C_1$.

We are given the equation:
$p^2x^2 - [f'(x)]^2 = p^2$
Substituting $y = f'(x)$, this can be rewritten as:
$p^2x^2 - y^2 = p^2$
This is exactly the equation of the hyperbola $C_1$. Thus, the roots of the given equation are exactly the $x$-coordinates of the points of intersection between the graph of $y = f'(x)$ and the curve $C_1$.

3. Determine the number of intersections between the two graphs.

By visualizing both graphs on the same diagram:

• At $x = -1$: The left branch of $C_1$ has a vertex at $(-1, 0)$. The graph of $f'(x)$ also passes through $(-1, 0)$. This gives 1 point of intersection.
• For $x < -1$: $f'(x) > 0$. The upper left branch of $C_1$ increases from $y=0$ towards $+\infty$ with an initially infinite gradient, remaining above $f'(x)$. There is no intersection here.
• For $-1 < x < 1$: $C_1$ is undefined (no real points), so there are no intersections.
• For $x \ge 1$: The lower right branch of $C_1$ starts at $(1, 0)$ and decreases to $-\infty$. At $x = 1$, $f'(1) < 0$ (meaning $f'(x)$ is below $C_1$). As $x \to \infty$, $f'(x)$ approaches $0$ and stays bounded, while $C_1$ goes to $-\infty$. Because $f'(x)$ starts below $C_1$ and ends up above it, the two curves must cross. This gives 1 point of intersection.

Total number of intersection points $= 1 + 1 = 2$.