Task B-4.8. A circle is described around the focus of the parabola $y^{2}=16 x$ which touches the directrix of the parabola. At what angle do these two curves intersect?
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_247a158c2fc9e6452974g-21.jpg?height=919&width=770&top_left_y=457&top_left_x=648)

From the equation of the parabola, we get $2p = 16$, hence $\frac{p}{2} = 4$.

The focus of the parabola is the point $(4,0)$. The directrix is the line $x = -4$.

The center of the circle is at point $S(4,0)$, and the radius of the circle is the distance from $S$ to the directrix, $r = 8$. Therefore, the equation of the circle is $(x-4)^2 + y^2 = 64$.

(2 points)

The intersection of the parabola and the circle is the solution to the system of equations $(x-4)^2 + y^2 = 64, y^2 = 16x$.

\[
\begin{aligned}
(x-4)^2 + 16x & = 64 \\
(x+4)^2 & = 64
\end{aligned}
\]

$x + 4 = \pm 8$, i.e., $x = 4$ or $x = -12$.

The solution is $x = 4$ ( $-12$ is not possible because then $y^2 < 0$ ).

Then $y^2 = 16 \cdot 4, y = \pm 8$, and the intersection points are $T_1 = (4,8)$ and $T_2 = (4,-8)$.

Due to symmetry, the desired angle will not depend on the choice of the point.

The tangent to the parabola at $T_1 = (4,8)$ is $y \cdot 8 = 8(x + 4)$, or $y = x + 4$.

The tangent to the circle at $T_1 = (4,8)$ is the line $y = 8$.

Then the angle between the given curves or the angle between the obtained tangents is equal to the angle $\varphi$ that the line $y = x + 4$ makes with the positive part of the $x$-axis.

Since $\operatorname{tg} \varphi = k = 1$, it follows that $\varphi = 45^\circ$.

(2 points)