$26 \cdot 21$ Draw a chord $P_{1} P_{2}$ perpendicular to the axis of symmetry of the parabola $y^{2}+p^{2}=2 p x(p>0)$ through $A(p, 0)$. If $O$ is the origin, then $\angle P_{1} O P_{2}$ is
(A) a right angle.
(B) an obtuse angle.
(C) an acute angle.
(D) indeterminate.
（2nd “Hope Cup” National Mathematics Invitational, 1991）
[Solution] Notice that $y^{2}+p^{2}=2 p x$ can be transformed into $y^{2}=2 p\left(x-\frac{p}{2}\right)$. From the given, we know $\quad\left|P_{1} P_{2}\right|=2 p$.

Thus, $\quad\left|A P_{1}\right|=\left|A P_{2}\right|=p$.
Also, $|O A|=p$, so $\mathrm{Rt} \triangle O A P_{1}, \mathrm{Rt} \triangle O A P_{2}$
are both isosceles right triangles. Therefore,
$$
\begin{aligned}
\angle P_{1} O P_{2} & =\angle P_{1} O A+\angle P_{2} O A \\
& =45^{\circ}+45^{\circ}=90^{\circ} .
\end{aligned}
$$

Hence, the answer is $(A)$.