10. (20 points) Let the ellipse $C: \frac{x^{2}}{a^{2}}+y^{2}=1(a>0)$.
(1) Given $a>0$, let $P$ be a moving point on the ellipse $C$ in the first quadrant. Draw the tangent line to the ellipse $C$ through point $P$, intersecting the two coordinate axes at points $A$ and $B$. When $|A B|$ is minimized, find the equation of the tangent line.
(2) Let the tangent line obtained in (1) be $l$, and the ellipse $C$ is symmetric about $l$ to get the ellipse $C^{\prime}$. If the ellipse $C^{\prime}$ intersects the $x$-axis, find the range of values for $a$.
10. (1) Let point $P\left(x_{0}, y_{0}\right)$. Then the equation of the tangent line $l$ is $\frac{x_{0} x}{a^{2}}+y_{0} y=1$.
Thus, $A\left(\frac{a^{2}}{x_{0}}, 0\right), B\left(0, \frac{1}{y_{0}}\right)$.
By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
|A B|=\sqrt{\left(\frac{a^{2}}{x_{0}}\right)^{2}+\left(\frac{1}{y_{0}}\right)^{2}} \\
=\sqrt{\frac{a^{2}}{1-y_{0}^{2}}+\frac{1}{y_{0}^{2}}} \\
\geqslant \sqrt{\frac{(a+1)^{2}}{1-y_{0}^{2}+y_{0}^{2}}}=a+1,
\end{array}
$$

The equality holds when $y_{0}^{4}=\frac{\left(1-y_{0}^{2}\right)^{2}}{a^{2}}$.
Given $y_{0} \in(0,1)$, solving yields
$y_{0}=\frac{1}{\sqrt{a+1}} \Rightarrow x_{0}=\frac{a \sqrt{a}}{\sqrt{a+1}}$.
Substituting these values, the equation of the tangent line $l$ is
$$
x+\sqrt{a} y=\sqrt{a(1+a)} .
$$
(2) As shown in Figure 2, consider the symmetric line $l_{1}$ of the $x$-axis with respect to $l$. We only need $l_{1}$ to intersect the ellipse $C$.
Let the slope of line $l$ be
$$
k=-\frac{1}{\sqrt{a}}=-\tan \theta\left(\theta \in\left(0, \frac{\pi}{2}\right)\right) .
$$

Then the inclination angle of line $l_{1}$ is $\pi-2 \theta$.
Now consider the symmetric line $l^{\prime}$ of $l$ with respect to the $x$-axis, which has an inclination angle of $\theta$ and is tangent to the ellipse $C$.
Thus, we only need $\theta \geqslant \pi-2 \theta$, i.e., $\theta \geqslant \frac{\pi}{3}$. At this point,
$$
\frac{1}{\sqrt{a}}=\tan \theta \geqslant \sqrt{3} \text {. }
$$

Therefore, the range of $a$ is $\left(0, \frac{1}{3}\right)$.