10. Let $F$ be the left focus of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, and $A$ be a point on the ellipse in the first quadrant. Draw a tangent line from $A$ to the circle $x^{2}+y^{2}=b^{2}$, and let the point of tangency be $P$. Then $|A F|-|A P|=$ $\qquad$.
Given $A\left(x_{0}, y_{0}\right)$, then $\frac{x_{0}^{2}}{a^{2}}+\frac{y_{0}^{2}}{b^{2}}=1$ $\Rightarrow y_{0}^{2}=b^{2}\left(1-\frac{x_{0}^{2}}{a^{2}}\right)=b^{2}-\frac{b^{2}}{a^{2}} x_{0}^{2}$, thus $|A P|^{2}=|O A|^{2}-|O P|^{2}=x_{0}^{2}+y_{0}^{2}-b^{2}$ $=x_{0}^{2}-\frac{b^{2}}{a^{2}} x_{0}^{2}=e^{2} x_{0}^{2} \Rightarrow|A P|=e x_{0}$.
Therefore, $|A F|-|A P|=a+e x_{0}-e x_{0}=a$.