10. (15 points) Given the family of curves $C: 4 x^{2}+5 y^{2}-$ $8 m x-20 m y+24 m^{2}-20=0$ (where $m$ is a real parameter).
(1) Find the line that is intersected by each curve in the family $C$ to form a segment of length $\frac{5 \sqrt{5}}{3}$.
(2) Let $P$ be any point on the family of curves $C$, and $Q$ be any point on the circle $x^{2}+y^{2}-10 x+4 \sqrt{6} y+30=0$. Find the minimum value of $|P Q|$.
10. Solution: (1) The family of curves is a family of ellipses: $\frac{(x-m)^{2}}{5}+$ $\frac{(y-2 m)^{2}}{4}=1$, with the center $(m, 2 m)$ on the line $y=2 x$. To satisfy the condition, the slope of the required line is 2, so let $y=2 x+b$. By combining $\frac{x^{2}}{5}$ $+\frac{y^{2}}{4}=1$ and using the segment length equal to $\frac{5 \sqrt{5}}{3}$, we solve to get: $b=2$, or $b$ $=-2$. Therefore, the required lines are $y=2 x+2$ or $y=2 x-2$.
(2) The lines tangent to all curves are: $y=2 x \pm$ $2 \sqrt{6}$, so $d=2 \sqrt{5}-1$.