From the point $S(-3,7)$, we draw tangents to the parabola $y^{2}=5 x$. Write the equations of these two tangents and the coordinates of the points of tangency. At what angle do the two tangents intersect?
Let the equation of the parabola be

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

and the coordinates of point $S$ be $a$ and $b$. The equation of the tangent at point $(\xi, \eta)$ of the parabola is:

$$
y \eta=p(x+\xi)
$$

If we want this tangent to pass through point $S$, then the coordinates of $S$ satisfy (2), that is:

$$
b \eta=p(a+\xi)
$$

If we also consider that the point $\left(\xi, \eta_{1}\right)$ lies on the parabola, then from (1) we have

$$
\eta^{2}=2 p \xi
$$

Now, let's square (3) and substitute the value of $\eta^{2}$ from (4), then we get:

$$
2 b^{2} p \xi=p^{2}(a+\xi)^{2}
$$

Taking the square root of both sides, we obtain:

$$
p \xi-(b \sqrt{2 p}) \sqrt{\xi}+a p=0
$$

Thus,

$$
\sqrt{\xi}=\frac{b \sqrt{2 p} \pm \sqrt{2 b^{2} p-4 a p^{2}}}{2 p}=\frac{b \pm \sqrt{b^{2}-2 a p}}{\sqrt{2 p}}
$$

and hence:

$$
\xi=\frac{\left(b \pm \sqrt{b^{2}-2 a p}\right)^{2}}{2 p}=\frac{b^{2}-a p \pm b \sqrt{b^{2}-2 a p}}{p}
$$

and

$$
\eta=b \pm \sqrt{b^{2}-2 a p}
$$

Substituting the values of $\xi_{1}, \eta_{1}$ and $\xi_{2}, \eta_{2}$ into (2), we get the equations of the tangents that satisfy the problem. As can be seen, the problem generally has two solutions. However, if

$$
b^{2}-2 a p=0
$$

then $\xi_{1}=\xi_{2}$ and $\eta_{1}=\eta_{2}$, so the problem has only one solution, which is evident from comparing (1) and (6), if $S$ lies on the parabola. Finally, if

$$
b^{2}-2 a p<0
$$

then the problem has no real solutions.

In our case, $a=-3 ; b=7 ; p=\frac{5}{2}$, so:

$$
\begin{array}{cc}
\xi_{1}=45 ; & \xi_{2}=\frac{1}{5} \\
\eta_{1}=15 ; & \eta_{2}=-1
\end{array}
$$

The equations of the sought tangents are therefore:

$$
y=\frac{x}{6}+\frac{15}{2}
$$

and

$$
y=-\frac{5}{2} x-\frac{1}{2}
$$

For the angle $(\alpha)$ formed by the two tangents:

$$
\operatorname{tg} \alpha=\frac{\frac{1}{6}-\left(-\frac{5}{2}\right)}{1-\frac{5}{12}}=\frac{32}{7}
$$

and thus

$$
\alpha=77^{\circ} 39^{\prime} 38^{\prime \prime}
$$

(Kürti Imre, Eger.)

The problem was also solved by: Bartók I., Dömény I., Dömény E., Enyedi B., Haat A., Kertész G., Liebner A., Neidenbach E., Pichler S., Pivnyik I., Rássy P., Riesz K., Rosenberg J., Schwemmer I., Szombathy J., Szúcs A.