24. Given $a \in \mathbf{Z}_{+}$. The area of the quadrilateral enclosed by the curves
$$
(x+a y)^{2}=4 a^{2} \text { and }(a x-y)^{2}=a^{2}
$$

is ( ).
(A) $\frac{8 a^{2}}{(a+1)^{2}}$
(B) $\frac{4 a}{a+1}$
(C) $\frac{8 a}{a+1}$
(D) $\frac{8 a^{2}}{a^{2}+1}$
(E) $\frac{8 a}{a^{2}+1}$
24. D.

From $(x+a y)^{2}=4 a^{2} \Rightarrow x+a y= \pm 2 a$.
From $(a x-y)^{2}=a^{2} \Rightarrow a x-y= \pm a$.
Thus, the curve represents four lines:
$$
\begin{array}{l}
x+a y \pm 2 a=0, \\
a x-y \pm a=0 .
\end{array}
$$

Let the quadrilateral formed by the four lines be rectangle $A B C D$, as shown in Figure 8.
Let the inclination angle of the line $x+a y \pm 2 a=0$ be $\theta$. Then
$$
\begin{array}{l}
\tan \theta=-\frac{1}{a} . \\
\text { Therefore, } S_{\text {rectangle } A B C D}=|2 a \cos \theta \cdot 4 \sin \theta| \\
=4 a|\sin 2 \theta|=4 a\left|\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right| \\
=4 a\left|\frac{-\frac{2}{a}}{1+\frac{1}{a^{2}}}\right|=\frac{8 a^{2}}{a^{2}+1} .
\end{array}
$$