10. Given the parabola $y^{2}=4 p(x+p)$ (where $p>0$), draw any two mutually perpendicular chords $A B$ and $C D$ through the origin $O$. Find the minimum value of $|A B|+|C D|$.
10. Since the coefficient of $x$ in the parabola equation $y^{2}=4 p(x+p)$ is $4 p$, the distance from the focus to the vertex is $p$. Therefore, the origin $O$ is precisely the focus of the parabola.

Taking a polar coordinate system with the origin $O$ as the pole and the positive half-axis of the $x$-axis as the polar axis, the polar equation of this parabola is $\rho=\frac{2 p}{1-\cos \theta}$. Given that $A B \perp C D$, we can set the polar coordinates of points $A, B, C, D$ to be $\left(\rho_{1}, \theta\right), \left(\rho_{2}, \theta+\pi\right), \left(\rho_{3}, \theta+\frac{\pi}{2}\right), \left(\rho_{4}, \theta+\frac{3 \pi}{2}\right)$. Thus, $\rho_{1}=\frac{2 p}{1-\cos \theta}, \rho_{2}=\frac{2 p}{1-\cos (\theta+\pi)}=\frac{2 p}{1+\cos \theta}, \rho_{3}=\frac{2 p}{1-\cos \left(\theta+\frac{\pi}{2}\right)}=\frac{2 p}{1+\sin \theta}, \rho_{4}=\frac{2 p}{1-\cos \left(\theta+\frac{3 \pi}{2}\right)}=\frac{2 p}{1-\sin \theta}$. Therefore, $|A B|=|O A|+|O B|=\rho_{1}+\rho_{2}=\frac{2 p}{1-\cos \theta}+\frac{2 p}{1+\cos \theta}=\frac{4 p}{\sin ^{2} \theta}$. Similarly, we get $|C D|=\rho_{3}+\rho_{4}=\frac{4 p}{\cos ^{2} \theta}$. Hence, $|A B|+|C D|=\frac{4 p}{\sin ^{2} \theta}+\frac{4 p}{\cos ^{2} \theta}=\frac{4 p}{\sin ^{2} \theta \cos ^{2} \theta}=\frac{16 p}{\left|\sin ^{2} 2 \theta\right|} \geqslant 16 p$. When $\theta=\frac{\pi}{4}$ or $\theta=\frac{3 \pi}{4}$, $|A B|+|C D|$ achieves its minimum value of $16 p$.