下图显示了一条抛物线，它与 $x$ 轴交于点 $(-2a, 0)$ 和 $(2a, 0)$，且其顶点（最大值点）为 $(0, 4a^2)$，其中 $a$ 是正常数。三角形 $OPQ$ 内接于该抛物线，使得点 $P$ 和 $Q$ 对称地分布在抛物线上 $x = \pm k$ 处（其中 $0 < k < 2a$），$O$ 为原点。

1. Formulate the equation of the parabola using its roots and maximum point

Since the parabola cuts the $x$-axis at $x = -2a$ and $x = 2a$, its equation is of the form:
$y = c(x + 2a)(x - 2a)$

The maximum point is $(0, 4a^2)$. Substituting $x = 0$ and $y = 4a^2$:
$4a^2 = c(0 + 2a)(0 - 2a)$
$4a^2 = -4a^2c$
$c = -1$

Therefore, the equation is:
$y = -(x + 2a)(x - 2a)$
$y = 4a^2 - x^2$

1. Derive the equation of the parabola (from part a).

Since the parabola has $x$-intercepts at $x = \pm 2a$ and a maximum point at $(0, 4a^2)$, its equation can be written as:
$y = c(x + 2a)(x - 2a) = c(x^2 - 4a^2)$
Substituting the maximum point $(0, 4a^2)$:
$4a^2 = c(0 - 4a^2) \implies c = -1$
So, the equation of the parabola is $y = 4a^2 - x^2$.

2. Find the coordinates of point $P$.

Given that $P$ lies on the parabola in the first quadrant at $x = k$:
$y = 4a^2 - k^2$
Thus, the coordinates of $P$ are $(k, 4a^2 - k^2)$.

3. Formulate the area of triangle $OPQ$.

Due to symmetry about the $y$-axis, $Q$ lies at $x = -k$. The base $PQ$ is therefore a horizontal line segment with length:
$\text{Base} = k - (-k) = 2k$
The perpendicular height of $\triangle OPQ$ from the origin $O$ to the segment $PQ$ is simply the $y$-coordinate of $P$, which is $4a^2 - k^2$.
$A = \frac{1}{2} \times \text{base} \times \text{height}$
$A = \frac{1}{2}(2k)(4a^2 - k^2)$
$A = k(4a^2 - k^2) \text{ (Shown)}$

1. Differentiate $A$ with respect to $k$.

$A = 4a^2k - k^3$
$\frac{\mathrm{d}A}{\mathrm{d}k} = 4a^2 - 3k^2$

2. Find the stationary value of $k$ by equating the first derivative to zero.

For a maximum value, $\frac{\mathrm{d}A}{\mathrm{d}k} = 0$
$4a^2 - 3k^2 = 0$
$3k^2 = 4a^2$
$k^2 = \frac{4a^2}{3}$

Since it is given that $k > 0$ and $a > 0$, we take the positive root:
$k = \frac{2a}{\sqrt{3}}$

3. Use the second derivative test to verify the nature of the stationary point.

$\frac{\mathrm{d}^2A}{\mathrm{d}k^2} = -6k$

Substitute $k = \frac{2a}{\sqrt{3}}$:
$\frac{\mathrm{d}^2A}{\mathrm{d}k^2} = -6\left(\frac{2a}{\sqrt{3}}\right) = -4\sqrt{3}a$

Since $a > 0$, $\frac{\mathrm{d}^2A}{\mathrm{d}k^2} < 0$.
Thus, $A$ is maximized when $k = \frac{2a}{\sqrt{3}}$.

4. Substitute the value of $k$ back into $A$ to find the maximum area.

$A_{\text{max}} = \left(\frac{2a}{\sqrt{3}}\right)\left[4a^2 - \left(\frac{2a}{\sqrt{3}}\right)^2\right]$
$= \frac{2a}{\sqrt{3}}\left(4a^2 - \frac{4a^2}{3}\right)$
$= \frac{2a}{\sqrt{3}}\left(\frac{8a^2}{3}\right)$
$= \frac{16a^3}{3\sqrt{3}}$
$= \frac{16\sqrt{3}}{9}a^3$