方程为 $\dfrac{x^2}{a^2}+\dfrac{y^2}{4a^2}=1$ 的曲线 $C_1$ 经以下变换得到曲线 $C_2$：先沿 $x$ 轴正方向平移 $a$ 个单位，然后沿平行于 $x$ 轴的方向以比例因子 $2$ 进行拉伸，最后关于 $y$ 轴作反射变换，其中 $a$ 是正常数。

1. Apply the translation to the equation.

A translation of $a$ units in the positive $x$-direction means replacing $x$ with $(x - a)$.
The equation becomes:
$\frac{(x-a)^2}{a^2} + \frac{y^2}{4a^2} = 1$

2. Apply the stretch transformation to the equation.

A stretch with scale factor $2$ parallel to the $x$-axis means replacing $x$ with $\frac{x}{2}$.
The equation becomes:
$\frac{\left(\frac{x}{2}-a\right)^2}{a^2} + \frac{y^2}{4a^2} = 1$
$\frac{\left(\frac{x-2a}{2}\right)^2}{a^2} + \frac{y^2}{4a^2} = 1$
$\frac{(x-2a)^2}{4a^2} + \frac{y^2}{4a^2} = 1$

3. Apply the reflection transformation and simplify.

A reflection in the $y$-axis means replacing $x$ with $-x$.
The equation becomes:
$\frac{(-x-2a)^2}{4a^2} + \frac{y^2}{4a^2} = 1$
Since $(-x-2a)^2 = (x+2a)^2$:
$\frac{(x+2a)^2}{4a^2} + \frac{y^2}{4a^2} = 1$
Multiplying throughout by $4a^2$ yields:
$(x+2a)^2 + y^2 = 4a^2$

1. Determine the equation of $C_2$ from the given transformations

From part (i), applying the sequence of transformations to $C_1$: $\frac{x^2}{a^2}+\frac{y^2}{4a^2}=1$

1. Translation by $a$ units in the positive $x$-direction:
   $\frac{(x-a)^2}{a^2}+\frac{y^2}{4a^2}=1$

2. Stretch by scale factor $2$ parallel to the $x$-axis ($x \to \frac{x}{2}$):
   $\frac{(\frac{x}{2}-a)^2}{a^2}+\frac{y^2}{4a^2}=1 \implies \frac{(x-2a)^2}{4a^2}+\frac{y^2}{4a^2}=1$

3. Reflection in the $y$-axis ($x \to -x$):
   $\frac{(-x-2a)^2}{4a^2}+\frac{y^2}{4a^2}=1 \implies \frac{(x+2a)^2}{4a^2}+\frac{y^2}{4a^2}=1$

2. Identify the geometrical shape described by the final equation

Multiplying the equation of $C_2$ by $4a^2$ yields:
$(x+2a)^2 + y^2 = (2a)^2$
Since the coefficients of $x^2$ and $y^2$ are equal and the right-hand side is positive, this equation represents a circle (with centre $(-2a, 0)$ and radius $2a$).