18. (15 points) Given the ellipse $\frac{x^{2}}{a^{2}}+y^{2}=1(a>1)$, construct an isosceles right triangle $\triangle A B C$ inscribed in the ellipse with the right angle at the endpoint $B(0,1)$ of the minor axis. If there is exactly one such triangle, find the range of values for $a$.

---

The above text has been translated into English, preserving the original formatting and line breaks.
18. Let the equation of line $AB$ be $y=kx+1 (k>0)$, then the equation of line $BC$ is $y=-\frac{1}{k}x+1$.
From $\left\{\begin{array}{l}\frac{x^{2}}{a^{2}}+y^{2}=1, \\ y=kx+1,\end{array}\right.$ we get $x_{A}=\frac{-2a^{2}k}{a^{2}k^{2}+1}$.
From $\left\{\begin{array}{l}\frac{x^{2}}{a^{2}}+y^{2}=1, \\ y=-\frac{1}{k}x+1,\end{array}\right.$ we get $x_{C}=\frac{2a^{2}k}{a^{2}+k^{2}}$.
Then $|AB|=\sqrt{1+k^{2}} \cdot \frac{|-2a^{2}k|}{a^{2}k^{2}+1}$
$=\sqrt{1+k^{2}} \cdot \frac{2a^{2}k}{a^{2}k^{2}+1}$,
$|BC|=\sqrt{1+\frac{1}{k^{2}}} \cdot \frac{|2a^{2}k|}{a^{2}+k^{2}}=\frac{\sqrt{1+k^{2}}}{k} \cdot \frac{2a^{2}k}{a^{2}+k^{2}}$.
From $|AB|=|BC|$, we get $\frac{1}{a^{2}k^{2}+1}=\frac{1}{k(a^{2}+k^{2})}$.
Simplifying, we get $k^{3}-a^{2}k^{2}+a^{2}k-1=0$.
Then $(k-1)[k^{2}+(1-a^{2})k+1]=0$.
Since there is exactly one isosceles right triangle that meets the conditions, the above equation has only one real root $k=1$. Therefore, the equation $k^{2}+(1-a^{2})k+1=0$ has no real roots or has two equal real roots $k=1$. Thus,
$\Delta=(1-a^{2})^{2}-4=a^{4}-2a^{2}-3$.
When $\Delta<0$, we have $1<a<\sqrt{3}$.
When $\Delta=0$, we have $a=\sqrt{3}$, and in this case, the equation $k^{2}-2k+1=0$ has two equal real roots $k=1$.
In summary, the range of values for $a$ is $1<a \leqslant \sqrt{3}$.