26.39 Given point $A(1,2)$, a line passing through point $(5,-2)$ intersects the parabola $y^{2}=4 x$ at two other points $B, C$. Then $\triangle A B C$ is
(A) an acute triangle.
(B) an obtuse triangle.
(C) a right triangle.
(D) the answer is uncertain.
(China High School Mathematics Competition, 1999)
［Solution］Let $B\left(t^{2}, 2 t\right), C\left(s^{2}, 2 s\right), s \neq t, s \neq 1, t \neq 1$.
Then the equation of line $B C$ is $\frac{y-2 t}{2 s-2 t}=\frac{x-t^{2}}{s^{2}-t^{2}}$, i.e.,
$$
2 x-(s+t) y+2 s t=0 .
$$
$\because$ Line $B C$ passes through the point $(5,-2)$,
$$
\therefore \quad 2 \cdot 5-(s+t)(-2)+2 s t=0 \text {, }
$$

i.e.,
$$
(s+1)(t+1)=-4 \text {. }
$$

Therefore, $k_{A B} \cdot k_{A C}=\frac{2 t-2}{t^{2}-1} \cdot \frac{2 s-2}{s^{2}-1}=\frac{4}{(s+1)(t+1)}=-1$. $\therefore \angle B A C=90^{\circ}$, so $\triangle A B C$ is a right triangle. Hence, the answer is (C).