图中显示了一个中心在原点的椭圆，其半短轴为 $a$，半长轴为 $2a$，其中 $a > 0$。矩形 $PQRS$ 内接于该椭圆，其四个顶点 $P$、$Q$、$R$ 和 $S$ 均在椭圆上。$P$ 点的坐标为 $(x, y)$，其中 $x$ 和 $y$ 均为正数。

1. State the standard equation using the given intercepts

The ellipse is centred at the origin with a horizontal semi-axis of $a$ and a vertical semi-axis of $2a$. Substituting these into the standard ellipse equation $\frac{x^2}{a_x^2} + \frac{y^2}{a_y^2} = 1$ gives:

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

1. Re-derive the Cartesian equation of the ellipse (from part i) to use in this part.

The ellipse is centred at the origin with x-intercepts $\pm a$ and y-intercepts $\pm 2a$.
Its Cartesian equation is:
$\frac{x^2}{a^2} + \frac{y^2}{(2a)^2} = 1 \implies \frac{x^2}{a^2} + \frac{y^2}{4a^2} = 1$

2. Express the area of rectangle $PQRS$ in terms of $x$ and $y$.

Since $P(x, y)$ is in the first quadrant and the inscribed rectangle $PQRS$ has vertices on the ellipse, it is symmetric about both coordinate axes.
The width of the rectangle is $2x$ and its height is $2y$.
The area $A$ is given by:
$A = (2x)(2y) = 4xy$

3. Express $y$ in terms of $x$ and substitute it into the area equation to show the required result.

From the equation of the ellipse:
$\frac{y^2}{4a^2} = 1 - \frac{x^2}{a^2} = \frac{a^2 - x^2}{a^2}$
$y^2 = 4(a^2 - x^2)$
Since $P(x, y)$ lies in the first quadrant, $y > 0$, so we take the positive square root:
$y = 2\sqrt{a^2 - x^2}$
Substitute $y$ into the area equation:
$A = 4x \left( 2\sqrt{a^2 - x^2} \right)$
$A = 8x\sqrt{a^2 - x^2}$

1. Differentiate the area function $A$ with respect to $x$.

Given $A = 8x(a^2 - x^2)^{\frac{1}{2}}$.
Using the product rule and chain rule:
$\frac{\mathrm{d}A}{\mathrm{d}x} = 8(a^2 - x^2)^{\frac{1}{2}} + 8x \left[ \frac{1}{2}(a^2 - x^2)^{-\frac{1}{2}}(-2x) \right]$
$\frac{\mathrm{d}A}{\mathrm{d}x} = 8\sqrt{a^2 - x^2} - \frac{8x^2}{\sqrt{a^2 - x^2}}$

2. Equate the derivative to zero and solve for $x$ in terms of $a$.

For $A$ to be a maximum, set $\frac{\mathrm{d}A}{\mathrm{d}x} = 0$:
$8\sqrt{a^2 - x^2} - \frac{8x^2}{\sqrt{a^2 - x^2}} = 0$
Multiply throughout by $\sqrt{a^2 - x^2}$ (since $x < a$ and the denominator is non-zero):
$8(a^2 - x^2) - 8x^2 = 0$
$8a^2 - 16x^2 = 0$
$16x^2 = 8a^2$
$x^2 = \frac{a^2}{2}$
Since it is given that $x$ and $a$ are positive,
$x = \frac{a}{\sqrt{2}}$

1. Find the value of $x$ that maximizes the area $A$ (as required from part iii).

From part (iii), to find $x$ when $A$ is a maximum, we set $\frac{dA}{dx} = 0$:
$\frac{dA}{dx} = 8\sqrt{a^2 - x^2} - \frac{8x^2}{\sqrt{a^2 - x^2}} = 0$
$\implies a^2 - 2x^2 = 0$
$\implies x = \frac{a}{\sqrt{2}}$

2. Substitute $x = \frac{a}{\sqrt{2}}$ into the area expression to find the maximum area in terms of $a$.

Substitute $x = \frac{a}{\sqrt{2}}$ into $A = 8x\sqrt{a^2 - x^2}$:
$A_{\text{max}} = 8\left(\frac{a}{\sqrt{2}}\right)\sqrt{a^2 - \left(\frac{a}{\sqrt{2}}\right)^2}$
$A_{\text{max}} = \frac{8a}{\sqrt{2}}\sqrt{a^2 - \frac{a^2}{2}}$
$A_{\text{max}} = \frac{8a}{\sqrt{2}}\left(\frac{a}{\sqrt{2}}\right)$
$A_{\text{max}} = \frac{8a^2}{2} = 4a^2$

3. Equate the maximum area to 100 and solve for $a$.

Given that the maximum value of $A$ is 100:
$4a^2 = 100$
$a^2 = 25$
Since $a > 0$, we reject the negative root:
$a = 5$