$26 \cdot 20$ Let $a, c$ be positive real constants. For each real number $t$, let $\left(x_{t}, y_{t}\right)$ be the coordinates of the vertex of the parabola $y=a x^{2}+t x+c$. Then for all real numbers $t$, the set of points $\left(x_{t}, y_{t}\right)$ in the plane is
(A) a straight line.
(B) a parabola.
(C) a part of a parabola but not the whole.
(D) one branch of a hyperbola.
(E) none of the above.
(35th American High School Mathematics Examination, 1984)
[Solution] $y=a x^{2}+t x+c$ is a family of parabolas containing the parameter $t$, so their vertex coordinates $\left(x_{t}, y_{t}\right)$ must also be functions of the parameter $t$.
The vertex of the parabola $y=a x^{2}+t x+c$ lies on its axis of symmetry $x=\frac{-t}{2 a}$. Thus, $x_{t}=\frac{-t}{2 a}$, and
$$
y_{t}=a\left(\frac{-t}{2 a}\right)^{2}+t\left(\frac{-t}{2 a}\right)+c=c-\frac{t^{2}}{4 a} .
$$

These are the parametric equations of $x_{t}, y_{t}$ in terms of $t$. Eliminating the parameter, we convert it to the general equation, which gives
$$
y_{t}=c-a x_{t}^{2}=-a x_{t}^{2}+c .
$$

Thus, each vertex $\left(x_{t}, y_{t}\right)$ lies on the parabola $y=-a x^{2}+c$.
Conversely, every point $(x, y)$ on the parabola $y=-a x^{2}+c$ is of the form $\left(x_{t}, y_{t}\right)$, meaning that by introducing the parameter $t, x=-\frac{t}{2 a}$, it is the vertex of the original family of parabolas. Therefore, the correct choice is $(B)$.