109 The tangent line of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ intersects the $x$-axis at $A$ and the $y$-axis at $B$. Then the minimum value of $|A B|$ is ( )
A. $2 \sqrt{a^{2}+b^{2}}$
B. $a+b$
C. $\sqrt{2 a b}$
D. $4 \sqrt{a b}$
109 B. The tangent line at the point $\left(x_{1}, y_{1}\right)$ is $\frac{x_{1} x}{a^{2}}+\frac{y_{1} y}{b^{2}}=1$, where $x_{1}=a \cos \theta, y_{1}=b \sin \theta$ are on the ellipse. Thus, $A$ and $B$ are $\left(\frac{a^{2}}{x_{1}}, 0\right),\left(0, \frac{b^{2}}{y_{1}}\right)$.
$$
\begin{aligned}
A B^{2} & =\frac{a^{4}}{x_{1}^{2}}+\frac{b^{4}}{y_{1}^{2}}=\frac{a^{2}}{\cos ^{2} \theta}+\frac{b^{2}}{\sin ^{2} \theta} \\
& =(a+b)^{2}+(a \tan \theta-b \tan \theta)^{2} \\
& \geqslant(a+b)^{2} .
\end{aligned}
$$

Therefore, the minimum value of $A B$ is $a+b$, and it is achieved when $\tan \theta=\sqrt{\frac{b}{a}}$.