3. (25 points) In the Cartesian coordinate system $x O y$, it is known that points $A\left(-1, y_{0}\right)$ and $B(2,0)$ are both on the parabola $y=a x^{2}+b x(a>0)$, and the area of $\triangle A O B$ is 1.
(1) Find the equation of the parabola;
(2) The equation of line $l$ is $y=x-\frac{25}{4}$, and $P$ is any point on the parabola. Find the minimum distance from point $P$ to line $l$, and the coordinates of point $P$ at this time.
3. (1) From the condition, we know $y_{0}>0$, and
$$
\frac{1}{2} \times 2 y_{0}=1 \Rightarrow y_{0}=1 \text {. }
$$

Therefore, the equation of the parabola is
$$
y=\frac{1}{3} x^{2}-\frac{2}{3} x \text {. }
$$
(2) As shown in Figure 4, let the line $l_{1} / / l$, and suppose it is tangent to the parabola at point $P_{0}$.

Thus, when point $P$ coincides with $P_{0}$, the distance from $P$ to line $l$ is minimized.
Let the line $l_{1}: y=x+m$, and the parabola $y=\frac{1}{3} x^{2}-\frac{2}{3} x$.

By solving the system, we get
$$
x^{2}-5 x-3 m=0 \text {. }
$$

From $\Delta=25+12 m=0$, we know $m=-\frac{25}{12}$.
At this point, $x=\frac{5}{2}, y=\frac{5}{12}$, so $P_{0}\left(\frac{5}{2}, \frac{5}{12}\right)$.
Let the lines $l$ and $l_{1}$ intersect the $x$-axis at points
$C\left(x_{C}, y_{C}\right), D\left(x_{D}, y_{D}\right)$.
Thus, $x_{C}=\frac{25}{4}, x_{D}=\frac{25}{12}$.
(Provided by Li Changyong)