14. (20 points) In the ellipse $x^{2}+4 y^{2}=8$, $AB$ is a moving chord of length $\frac{5}{2}$, and $O$ is the origin. Find the range of the area of $\triangle AOB$.

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14. Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, and the equation of line $AB$ be $y=k x+b$. Substituting into the ellipse equation and rearranging, we get
$$
\left(4 k^{2}+1\right) x^{2}+8 k b x+4\left(b^{2}-2\right)=0 .
$$

Thus, $x_{1}+x_{2}=-\frac{8 k b}{4 k^{2}+1}, x_{1} x_{2}=\frac{4\left(b^{2}-2\right)}{4 k^{2}+1}$.
Given $\frac{25}{4}=A B^{2}=\left(k^{2}+1\right)\left(x_{2}-x_{1}\right)^{2}$
$=\left(k^{2}+1\right)\left[\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}\right]$
$=\frac{16\left(k^{2}+1\right)}{\left(4 k^{2}+1\right)^{2}}\left[2\left(4 k^{2}+1\right)-b^{2}\right]$,
we have $b^{2}=2\left(4 k^{2}+1\right)-\frac{25\left(4 k^{2}+1\right)^{2}}{64\left(k^{2}+1\right)}$.
The distance from the origin $O$ to $AB$ is $\frac{|b|}{\sqrt{k^{2}+1}}, S_{\triangle A O B}=$ $\frac{5}{4} \cdot \frac{|b|}{\sqrt{k^{2}+1}}$, and let $u=\frac{4 k^{2}+1}{k^{2}+1}$, then we have
$$
\begin{array}{l}
S_{\triangle A O B}^{2}=-\frac{625}{1024}\left(u^{2}-\frac{128}{25} u\right) \\
=4-\frac{625}{1024}\left(u-\frac{64}{25}\right)^{2} .
\end{array}
$$
Since $u=4-\frac{3}{k^{2}-1}$, the range of $u$ is $[1,4]$ (with $u=4$ for vertical chords).
Therefore, when $u=\frac{64}{25}$, the maximum value of $S_{\triangle A O B}^{2}$ is 4;
when $u=1$, the minimum value of $S_{\triangle A O B}^{2}$ is $\frac{2575}{1024}$.
Thus, the range of $S_{\triangle A O B}$ is $\left[\frac{5 \sqrt{103}}{32}, 2\right]$.