26・10 In the Cartesian coordinate system, if the equation $m\left(x^{2}+y^{2}+2 y+1\right)=$ $(x-2 y+3)^{2}$ represents an ellipse, then the range of values for $m$ is
(A) $(0,1)$.
(B) $(1,+\infty)$.
(C) $(0,5)$.
(D) $(5,+\infty)$.
(China High School Mathematics League, 1997)
[Solution] From the given, we have
$$
\left\lvert\, \frac{\sqrt{x^{2}+(y+1)^{2}}}{\left.\frac{x-2 y+3}{\sqrt{1^{2}+(-2)^{2}}} \right\rvert\,}=\sqrt{\frac{5}{m}} .\right.
$$

This indicates that the ratio of the distance from $(x, y)$ to the fixed point $(0,-1)$ and the distance to the fixed line $x-2 y+3=0$ is a constant $\sqrt{\frac{5}{m}}$.
By the definition of an ellipse, we get $\sqrt{\frac{5}{m}}=5$. Therefore, the correct choice is $(D)$.