曲线 C 由参数方程定义： $x = a\left(1 + \dfrac{1}{t}\right)$ 且 $y = a\left(t - \dfrac{1}{t^2}\right)$ 其中 $a$ 是正常数且 $t \ne 0$。

1. Differentiate $x$ with respect to $t$.

$x = a(1 + t^{-1})$
$\frac{dx}{dt} = a(-t^{-2}) = -\frac{a}{t^2}$

2. Differentiate $y$ with respect to $t$.

$y = a(t - t^{-2})$
$\frac{dy}{dt} = a(1 - (-2)t^{-3}) = a(1 + 2t^{-3}) = a\left(\frac{t^3 + 2}{t^3}\right)$

3. Use the chain rule for parametric equations to find $\frac{dy}{dx}$ and simplify to the required form.

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
$\frac{dy}{dx} = \frac{a\left(\frac{t^3+2}{t^3}\right)}{-\frac{a}{t^2}}$
$\frac{dy}{dx} = \left(\frac{t^3+2}{t^3}\right) \times \left(-t^2\right)$
$\frac{dy}{dx} = -\frac{t^3+2}{t}$
$\frac{dy}{dx} = -\left(\frac{2+t^3}{t}\right)$

1. Set the first derivative to zero to find the parameter $t$ at the turning point.

For a turning point, $\frac{dy}{dx} = 0$.
$-\left(\frac{2 + t^3}{t}\right) = 0$
$2 + t^3 = 0 \implies t^3 = -2 \implies t = -2^{\frac{1}{3}}$

2. Find the $x$ and $y$ coordinates of the turning point.

Substitute $t = -2^{\frac{1}{3}}$ into the parametric equations:
$x = a\left(1 + \frac{1}{-2^{1/3}}\right) = a\left(1 - 2^{-\frac{1}{3}}\right)$
For $y$, we can rewrite the equation as $y = a\left(\frac{t^3 - 1}{t^2}\right)$.
Substituting $t^3 = -2$ and $t^2 = (-2^{1/3})^2 = 2^{2/3}$:
$y = a\left(\frac{-2 - 1}{2^{2/3}}\right) = -\frac{3a}{2^{2/3}} = -3a(2^{-\frac{2}{3}})$
The coordinates are $\left(a(1 - 2^{-\frac{1}{3}}), -3a(2^{-\frac{2}{3}})\right)$.

3. Find the second derivative parametrically to test the nature of the turning point.

Using the chain rule for parametric equations, $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$.
First, differentiate $x$ with respect to $t$:
$\frac{dx}{dt} = -at^{-2} = -\frac{a}{t^2}$
Next, differentiate $\frac{dy}{dx}$ with respect to $t$:
$\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(-2t^{-1} - t^2\right) = 2t^{-2} - 2t = \frac{2 - 2t^3}{t^2}$
Now, divide the two to find $\frac{d^2y}{dx^2}$:
$\frac{d^2y}{dx^2} = \frac{\frac{2 - 2t^3}{t^2}}{-\frac{a}{t^2}} = \frac{2t^3 - 2}{a}$

4. Evaluate the second derivative at the turning point to conclude it is a maximum.

Substitute $t^3 = -2$ into the second derivative:
$\frac{d^2y}{dx^2} = \frac{2(-2) - 2}{a} = -\frac{6}{a}$
Since $a$ is a positive constant ($a > 0$), we have $\frac{d^2y}{dx^2} < 0$.
Because the second derivative is negative at the turning point, it is a maximum.

1. Find the $x$ and $y$-intercepts of the curve.

To find the $y$-intercept, set $x = 0$:
$a\left(1 + \frac{1}{t}\right) = 0 \implies t = -1$
Substitute $t = -1$ into $y$:
$y = a\left(-1 - \frac{1}{(-1)^2}\right) = -2a$
The $y$-intercept is $(0, -2a)$.

To find the $x$-intercept, set $y = 0$:
$a\left(t - \frac{1}{t^2}\right) = 0 \implies t^3 = 1 \implies t = 1$
Substitute $t = 1$ into $x$:
$x = a\left(1 + \frac{1}{1}\right) = 2a$
The $x$-intercept is $(2a, 0)$.

2. Determine the equations of any asymptotes.

Consider the behavior of $x$ and $y$ as $t \to \pm\infty$:
As $t \to \pm\infty$, $\frac{1}{t} \to 0$ and $\frac{1}{t^2} \to 0$.
$x = a\left(1 + \frac{1}{t}\right) \to a$
$y = a\left(t - \frac{1}{t^2}\right) \to \pm\infty$
Because $y \to \pm\infty$ as $x \to a$, the line $x = a$ is a vertical asymptote.

3. Deduce the overall shape of the curve to complete the sketch.

From part (b), there is a maximum turning point at $\left(a(1 - 2^{-1/3}), -a(2^{1/3} + 2^{-2/3})\right)$ when $t = -2^{1/3}$.
The domain $t \neq 0$ divides the curve into two continuous branches separated by the vertical asymptote $x = a$ (which corresponds to $t \to \pm\infty$).

• Left branch ($t < 0 \implies x < a$): Comes from the third quadrant (as $t \to 0^-$, $x \to -\infty$ and $y \to -\infty$), crosses the $y$-axis at $(0, -2a)$, rises to the maximum turning point, and then falls towards $-\infty$ approaching the asymptote $x=a$.
• Right branch ($t > 0 \implies x > a$): Falls from $+\infty$ alongside the right of $x=a$, crosses the $x$-axis at $(2a, 0)$, and continues strictly decreasing towards the fourth quadrant as $x \to \infty$.