$26 \cdot 18$ The curve $\frac{x^{2}}{16-k}-\frac{y^{2}}{k}=1$ has the same focal distance as the curve $9 x^{2}+25 y^{2}=225$ if and only if
(A) $k0$, and $k \neq 16$.
(C) $016$.
(2nd "Hope Cup" National Mathematics Invitational Competition, 1991)
[Solution] It is easy to find that the focal distance of the ellipse $9 x^{2}+25 y^{2}=225$ is 8.
From $\frac{x^{2}}{16-k}-\frac{y^{2}}{k}=1$, we know $k \neq 0$.
And when $k<0$, it represents an ellipse, and the focal distance is
$$
2 \sqrt{(16-k)+k}=2 \sqrt{16}=8 ;
$$

When $0<k<16$, the equation represents a hyperbola, and the focal distance is
$$
2 \sqrt{(16-k)+k}=2 \sqrt{16}=8 \text {. }
$$

Therefore, $k \in(-\infty, 0) \cup(0,16)$.
Hence, the answer is $(A)$.