The curve $C$ has parametric equations
$x = 5a\sec\theta, \quad y = 3a\tan\theta, \quad \text{where } -\frac{\pi}{2} < \theta < \frac{\pi}{2}$
and $a$ is a positive constant.

1. Express the trigonometric functions in terms of $x$ and $y$.

$x = 5a\sec\theta \implies \sec\theta = \frac{x}{5a}$
$y = 3a\tan\theta \implies \tan\theta = \frac{y}{3a}$

2. Eliminate $\theta$ using the Pythagorean identity for $\sec$ and $\tan$.

We know the identity $\sec^2\theta - \tan^2\theta = 1$.
Substituting the expressions:
$\left(\frac{x}{5a}\right)^2 - \left(\frac{y}{3a}\right)^2 = 1$
$\frac{x^2}{25a^2} - \frac{y^2}{9a^2} = 1$

3. Determine the domain restriction of the curve based on the parameter's given interval.

For $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$, we have $\sec\theta \ge 1$.
Since $a$ is a positive constant, this implies $x \ge 5a$.
(The range of $\tan\theta$ is $\mathbb{R}$, so $y$ can take any real value).

1. Find the cartesian equation (from part i) and determine the valid domain of $x$ to identify the correct portion of the curve.

$x = 5a\sec\theta \implies \sec\theta = \frac{x}{5a}$
$y = 3a\tan\theta \implies \tan\theta = \frac{y}{3a}$
Using the identity $\sec^2\theta - \tan^2\theta = 1$:
$\frac{x^2}{25a^2} - \frac{y^2}{9a^2} = 1$
Given the parameter domain $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$, we have $\sec\theta \ge 1$.
Since $a > 0$, this restricts $x$ to $x \ge 5a$.
Thus, $C$ represents only the right branch of the hyperbola.

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

For $y$-intercepts, substitute $x = 0$. Since $x \ge 5a$ and $a > 0$, there are no intersections with the $y$-axis.
For $x$-intercepts, substitute $y = 0$:
$\frac{x^2}{25a^2} - 0 = 1 \implies x^2 = 25a^2$
Since $x \ge 5a$, we take the positive root: $x = 5a$.
The intersection with the axes is $(5a, 0)$.

3. Determine the equations of the asymptotes.

For a standard hyperbola $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$, the asymptotes are given by $y = \pm \frac{B}{A}x$.
Here, $A = 5a$ and $B = 3a$.
The asymptotes are $y = \pm \frac{3a}{5a}x$, which simplifies to:
$y = \frac{3}{5}x \quad \text{and} \quad y = -\frac{3}{5}x$

4. Sketch the curve $C$ incorporating all the established features.

Sketch requirements:

• Draw dashed lines for the asymptotes $y = \frac{3}{5}x$ and $y = -\frac{3}{5}x$ passing through the origin.
• Mark the $x$-intercept at $(5a, 0)$.
• Draw a smooth continuous curve representing the right branch of the hyperbola, opening towards the right.
• The curve must originate from the vertex $(5a, 0)$ and approach the asymptotes as $x \to \infty$.