The equation of a parabola is given: $y^{2}=18 x$ and the coordinates of a point: $x_{1}=4, y_{1}=3$. Determine the equation of the chord that is bisected by the point, as well as the area of the region bounded by the chord and the parabola.
The equation of the line passing through the point $\mathrm{Az}\left(x_{1}, y_{1}\right)$:

$$
y-3=a(x-4)
$$

If the point $\left(x_{1}, y_{1}\right)$ bisects the chord $A B$, then

$$
a=\frac{p}{y_{1}}=\frac{9}{3}=3
$$

Thus, the equation of the chord $A B$ is:

$$
y=3 x-9
$$

The coordinates of points $A$ and $B$ are:

$$
\begin{gathered}
A\left(x_{2}=4+\sqrt{7}, y_{2}=3+3 \sqrt{7}\right), B\left(x_{3}=4-\sqrt{7}, y_{3}=3-3 \sqrt{7}\right) \\
A B=\sqrt{\left(x_{2}-x_{3}\right)^{2}+\left(y_{2}-y_{3}\right)^{2}}=2 \sqrt{70}
\end{gathered}
$$

The intersection point of the diameter passing through $\left(x_{1}, y_{1}\right)$ with the parabola is: $D\left(\frac{1}{2}, 3\right)$. The distance from point $D$ to the line $A B$ is:

$$
m=\frac{y_{4}-3 x_{4}+9}{\sqrt{1+9}}=\frac{21}{2 \sqrt{10}}
$$

The parabolic segment $A O B = \frac{4}{3} \times \text{Area of } \triangle A B D = \frac{4}{3} \cdot \frac{m \cdot A B}{2} = 14 \sqrt{7}$.

(Krisztián György, Pécs.)

The problem was also solved by: Bayer B., Filkorn J., Holzmann M., Lukhaub Gy., Póka Gy., Sasvári G., Smodics K.