The focus of the parabola $y^{2}=7 x$ is crossed by a line which intersects the $Y$ axis at (-1). What is the area of the parabolic segment thus obtained?
The parabola $y^{2}=2 p x$, whose vertex is $A$, is intersected by a certain line at points $B\left(x_{1}, y_{1}\right)$ and $C\left(x_{2}, y_{2}\right)$. If $BC$ intersects the x-axis at $P(\xi, \eta)$ and the projections of $B$ and $C$ onto the x-axis are denoted by $B_{1}$ and $C_{1}$ respectively, then the areas within the ordinates $y_{1}$ and $y_{2}$ ($t_{1}$ and $t_{2}$) are:

$$
t_{1}=\frac{2}{3} x_{1} y_{1}
$$

and

$$
t_{2}=\frac{-2}{3} x_{2} y_{2}
$$

We also need to calculate the areas of triangles $B B_{1} P$ and $C C_{1} P$:

$$
\begin{gathered}
B B_{1} P \Delta=\frac{y_{1}}{2}\left(x_{1}-\xi\right) \\
C C_{1} P \Delta=\frac{-y_{2}}{2}\left(\xi-x_{2}\right)
\end{gathered}
$$

Thus, the desired area, if $\left|y_{2}\right|<\left|y_{1}\right|$:

$$
T=t_{1}+t_{2}+C C_{1} P-B B_{1} P
$$

or

$$
=\frac{1}{6} x_{1} y_{1}-\frac{1}{6} x_{2} y_{2}+\frac{1}{2} \xi\left(y_{1}-y_{2}\right)=\frac{1}{6}\left(x_{1} y_{1}-x_{2} y_{2}\right)+\frac{1}{2} \xi\left(y_{1}-y_{2}\right)
$$

In our case, the line $BC$ passes through the focus $\left(\xi=\frac{7}{4}, 0\right)$ and the point $(0,-1)$, so its equation is

$$
y=\frac{4}{7} x-1
$$

To find the coordinates of points $B$ and $C$, we solve the common roots of equation (2) and the equation

$$
y^{2}=7 x
$$

Then we obtain:

$$
\begin{aligned}
& y_{1}=\frac{7}{8}(7+\sqrt{65}), x_{1}=\frac{7}{32}(57+7 \sqrt{65}) \\
& y_{2}=\frac{7}{8}(7-\sqrt{65}), x_{2}=\frac{7}{32}(57-7 \sqrt{65})
\end{aligned}
$$

Substituting the found values into (1):

$$
T=\frac{49 \cdot 65 \cdot \sqrt{65}}{32 \cdot 12}=66.87 \text { square units }
$$

(Author: Haar Alfréd, Budapest.)

The problem was also solved by: Bartók I., Dömény E., Enyedi B., Kertész G., Kürti I., Pivnyik I., Pichler S., Rássy P., Riesz K., Rosenberg J., Szücs A.