曲线 $C$ 具有参数方程 $x = (a^2 - t^2)^{\frac{1}{2}}, \quad y = \ln t, \quad \text{for } 0 < t \le a \text{ and } a > 1.$

1. Find the value of the parameter $t$ at the $x$-axis intercept.

At the $x$-axis, $y = 0$.
$\ln t = 0$
$t = 1$

2. Substitute $t = 1$ to find the $x$-coordinate of $P$.

$x = (a^2 - 1^2)^{\frac{1}{2}}$
$x = \sqrt{a^2 - 1}$

1. Differentiate $x$ and $y$ with respect to the parameter $t$.

$x = (a^2 - t^2)^{\frac{1}{2}} \implies \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{1}{2}(a^2 - t^2)^{-\frac{1}{2}}(-2t) = -\frac{t}{\sqrt{a^2 - t^2}}$

$y = \ln t \implies \frac{\mathrm{d}y}{\mathrm{d}t} = \frac{1}{t}$

2. Apply the chain rule to find $\frac{\mathrm{d}y}{\mathrm{d}x}$.

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}} = \frac{\frac{1}{t}}{-\frac{t}{\sqrt{a^2 - t^2}}} = -\frac{\sqrt{a^2 - t^2}}{t^2}$

3. Analyse the sign of $\frac{\mathrm{d}y}{\mathrm{d}x}$ for the given domain of $t$.

Given the domain $0 < t \le a$:
$t^2 > 0$ (since $t \neq 0$)
$a^2 - t^2 \ge 0 \implies \sqrt{a^2 - t^2} \ge 0$

Since the numerator is non-negative and the denominator is strictly positive, the fraction $\frac{\sqrt{a^2 - t^2}}{t^2} \ge 0$.
Therefore, multiplying by the negative sign gives:
$\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{\sqrt{a^2 - t^2}}{t^2} \le 0$ (Shown).

1. Find the derivative $\frac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $t$ (from earlier parts).

$\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{1}{2}(a^2 - t^2)^{-\frac{1}{2}}(-2t) = -\frac{t}{\sqrt{a^2 - t^2}}$
$\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{1}{t}$
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}} = \frac{\frac{1}{t}}{-\frac{t}{\sqrt{a^2 - t^2}}} = -\frac{\sqrt{a^2 - t^2}}{t^2}$

2. Evaluate the limit of the gradient as $t \to 0$.

As $t \to 0$, $t^2 \to 0^+$ and $\sqrt{a^2 - t^2} \to a$.
Therefore, $\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{\sqrt{a^2 - t^2}}{t^2} \to -\infty$.

3. Interpret the limiting gradient to describe the behaviour of the tangent.

Since the gradient approaches negative infinity, the tangent becomes increasingly steep until it approaches a vertical line (parallel to the $y$-axis).

1. Find the coordinates of the points where the curve meets the axes.

For the $y$-axis intercept, let $x = 0$:
$(a^2 - t^2)^{\frac{1}{2}} = 0 \implies t = a$
Substituting $t = a$ into $y$, we get $y = \ln a$. So the curve meets the $y$-axis at $(0, \ln a)$.

For the $x$-axis intercept, let $y = 0$:
$\ln t = 0 \implies t = 1$
Substituting $t = 1$ into $x$ (since $a > 1$, $t=1$ is within the domain $0 < t \le a$), we get $x = (a^2 - 1^2)^{\frac{1}{2}} = \sqrt{a^2 - 1}$. So the curve meets the $x$-axis at $(\sqrt{a^2 - 1}, 0)$.

2. Determine the equations of any asymptotes.

The parameter is restricted to $0 < t \le a$. We consider the limits as $t$ approaches the boundary $0$ from the right:
$\lim_{t \to 0^+} y = \lim_{t \to 0^+} (\ln t) = -\infty$
$\lim_{t \to 0^+} x = \lim_{t \to 0^+} (a^2 - t^2)^{\frac{1}{2}} = \sqrt{a^2} = a$
Because $y \to -\infty$ as $x \to a$, the vertical line $x = a$ is an asymptote.

3. Synthesize the properties to describe the final graph sketch.

Using findings from earlier parts:

• From (b), $\frac{\mathrm{d}y}{\mathrm{d}x} \le 0$, meaning the curve is strictly decreasing.
• At $t=a$ (the point $(0, \ln a)$), $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$, so the curve starts at the $y$-axis with a horizontal tangent.
• The curve is sketched concave downwards from $(0, \ln a)$, crossing the $x$-axis at $(\sqrt{a^2 - 1}, 0)$, and continuously dropping to approach the vertical asymptote $x = a$ as $y \to -\infty$.

1. Find the gradient of the tangent to the curve at point $P$.

From part (a), the curve crosses the $x$-axis ($y=0$) when $\ln t = 0 \implies t = 1$, giving point $P(\sqrt{a^2-1}, 0)$.
To find the gradient, we first evaluate the parametric derivative:
$\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{1}{2}(a^2-t^2)^{-1/2}(-2t) = -\frac{t}{\sqrt{a^2-t^2}}$
$\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{1}{t}$
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t} = \frac{1/t}{-t/\sqrt{a^2-t^2}} = -\frac{\sqrt{a^2-t^2}}{t^2}$
Substitute $t=1$ to find the gradient of the tangent at $P$:
$m = -\frac{\sqrt{a^2-(1)^2}}{(1)^2} = -\sqrt{a^2-1}$

2. Find the coordinates of $Q$ by determining the $y$-intercept of the tangent.

The equation of the tangent at $P$ is:
$y - 0 = -\sqrt{a^2-1}(x - \sqrt{a^2-1})$
Since $Q$ is the $y$-intercept of the tangent, substitute $x = 0$:
$y = -\sqrt{a^2-1}(-\sqrt{a^2-1}) = a^2 - 1$
Therefore, $Q$ has coordinates $(0, a^2-1)$.

3. Find the coordinates of $R$ by determining the $y$-intercept of the normal.

The gradient of the normal at $P$ is $-\frac{1}{m} = \frac{1}{\sqrt{a^2-1}}$.
The equation of the normal at $P$ is:
$y - 0 = \frac{1}{\sqrt{a^2-1}}(x - \sqrt{a^2-1})$
Since $R$ is the $y$-intercept of the normal, substitute $x = 0$:
$y = \frac{1}{\sqrt{a^2-1}}(-\sqrt{a^2-1}) = -1$
Therefore, $R$ has coordinates $(0, -1)$.

4. Calculate the area of triangle $PQR$.

The points $Q(0, a^2-1)$ and $R(0, -1)$ both lie on the $y$-axis, so we can take segment $QR$ as the base of $\triangle PQR$.
The length of the base $QR$ is:
$QR = (a^2-1) - (-1) = a^2$
The corresponding perpendicular height from $P$ to the $y$-axis is the $x$-coordinate of $P$, which is $\sqrt{a^2-1}$.
The area of $\triangle PQR = \frac{1}{2} \times \text{base} \times \text{height}$
$\text{Area} = \frac{1}{2}(a^2)(\sqrt{a^2-1}) = \frac{1}{2}a^2\sqrt{a^2-1}$