8. If the ellipse $\frac{x^{2}}{a_{1}^{2}}+\frac{y^{2}}{b_{1}^{2}}=1\left(a_{1}>b_{1}>0\right)$ and the hyperbola $\frac{x^{2}}{a_{2}^{2}}-\frac{y^{2}}{b_{2}^{2}}=1\left(a_{2}>0, b_{2}>0\right)$ have the same foci, $P$ is an intersection point of the two curves, and the slopes of the tangents to the ellipse and hyperbola at point $P$ are $k_{1}$, $k_{2}$, respectively, then $k_{1} k_{2}=$ $\qquad$
8. -1 Explanation: Let $P(x, y)$, and let point $F$ be one of the foci. By the focal radius formula, we have $|P F|=a_{1}-e_{1} x=e_{2} x-a_{2}$. Then $x=\frac{a_{1}+a_{2}}{e_{1}+e_{2}}=\frac{a_{1} a_{2}}{c}$, substituting into the ellipse equation yields $y=\frac{b_{1} b_{2}}{c}$. For the ellipse, we have $\frac{2 x}{a_{1}^{2}}+\frac{2 y}{b_{1}^{2}} \cdot k_{1}=0$, thus $k_{1}=-\frac{a_{2} b_{1}}{a_{1} b_{2}}$; for the hyperbola, we have $\frac{2 x}{a_{2}^{2}}-\frac{2 y}{b_{2}^{2}} \cdot k_{2}=0$, thus $k_{2}=\frac{a_{1} b_{2}}{a_{2} b_{1}}$, which means $k_{1} k_{2}=-1$.