3. As shown in the figure, $C_{1}, C_{2}$ are ellipses with the same eccentricity $e$, points $A, B$ are the right vertex and the upper vertex of $C_{2}$, respectively. Tangents $l_{1}, l_{2}$ to $C_{1}$ are drawn through points $A, B$. If the slopes of lines $l_{1}, l_{2}$ are $k_{1}, k_{2}$, respectively, then the value of $\left|k_{1} k_{2}\right|$ is
A. $e^{2}$
B. $e^{2}-1$
C. $1-e^{2}$
D. $\frac{1}{e^{2}}$
Let $l_{1}, l_{2}$ be tangent to $C_{1}$ at points $M\left(x_{1}, y_{1}\right), N\left(x_{2}, y_{2}\right)$, respectively, then $l_{1}: \frac{x_{1} x}{a^{2}}+\frac{y_{1} y}{b^{2}}=1$ $\Rightarrow A\left(\frac{a^{2}}{x_{1}}, 0\right), k_{1}=-\frac{b^{2} x_{1}}{a^{2} y_{1}}$, similarly, $B\left(0, \frac{b^{2}}{y_{2}}\right), k_{2}=-\frac{b^{2} x_{2}}{a^{2} y_{2}}$. Since $\frac{a^{2} y_{2}}{b^{2} x_{1}}=\frac{a}{b} \Rightarrow \frac{x_{1}}{y_{2}}=\frac{a}{b}$, let $x_{1}=m a, y_{2}=m b$ and substitute into the ellipse equation to get $\frac{x_{2}^{2}}{y_{1}^{2}}=\frac{a^{2}}{b^{2}}$. Therefore, $\left|k_{1} k_{2}\right|=\frac{b^{4}}{a^{4}} \cdot\left|\frac{x_{1} x_{2}}{y_{1} y_{2}}\right|=\frac{b^{2}}{a^{2}}=1-e^{2}$, so the answer is $C$.