方程为 $\frac{x^2}{a^2}+\frac{y^2}{4a^2}=1$ 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 of $a$ units in the positive $x$-direction.

Replace $x$ with $(x-a)$:
$\frac{(x-a)^2}{a^2} + \frac{y^2}{4a^2} = 1$

2. Apply the stretch with scale factor 2 parallel to the $x$-axis.

Replace $x$ with $\frac{x}{2}$:
$\frac{\left(\frac{x}{2}-a\right)^2}{a^2} + \frac{y^2}{4a^2} = 1$

3. Apply the reflection in the $y$-axis and simplify the equation.

Replace $x$ with $-x$:
$\frac{\left(-\frac{x}{2}-a\right)^2}{a^2} + \frac{y^2}{4a^2} = 1$
Factor out the negative sign inside the squared expression:
$\frac{\left(-\frac{1}{2}(x+2a)\right)^2}{a^2} + \frac{y^2}{4a^2} = 1$
$\frac{\frac{1}{4}(x+2a)^2}{a^2} + \frac{y^2}{4a^2} = 1$
Multiply the entire equation by $4a^2$:
$(x+2a)^2 + y^2 = 4a^2$

1. Determine the equation of $C_2$ by applying the sequence of transformations to $C_1$ (as required from part (i))

Translation of $a$ units in the positive $x$-direction (replace $x$ with $x-a$):
$\frac{(x-a)^2}{a^2} + \frac{y^2}{4a^2} = 1$
Stretch with scale factor $2$ parallel to the $x$-axis (replace $x$ with $\frac{x}{2}$):
$\frac{(\frac{x}{2}-a)^2}{a^2} + \frac{y^2}{4a^2} = 1 \implies \frac{(x-2a)^2}{4a^2} + \frac{y^2}{4a^2} = 1$
Reflection in the $y$-axis (replace $x$ with $-x$):
$\frac{(-x-2a)^2}{4a^2} + \frac{y^2}{4a^2} = 1 \implies \frac{(x+2a)^2}{4a^2} + \frac{y^2}{4a^2} = 1$
Multiplying through by $4a^2$, the equation of $C_2$ is:
$(x+2a)^2 + y^2 = (2a)^2$

2. Interpret the equation geometrically to describe the shape

The equation $(x+2a)^2 + y^2 = (2a)^2$ is of the standard form $(x-h)^2 + (y-k)^2 = r^2$.
This represents a circle with centre $(-2a, 0)$ and radius $2a$.
Therefore, the shape is a circle.