378. At what angle does the parabola $y=x^{2} / 2$ intersect the line $3 x-2 y-2=0$?
Solution. This question can be rephrased as: at what angle do the tangents to the curve $y=x^{2} / 2$, drawn at the points of intersection with the line $3 x-2 y-2=0$, and the line itself intersect?

We find the points of intersection of the parabola $y=x^{2} / 2$ and the line $3 x-2 y-2=0$ by solving the system of equations $3 x-2 y-2=0, y=x^{2} / 2$. Substituting $y=x^{2} / 2$ into the first equation, we get $3 x-x^{2}-2=0$ or $x^{2}-3 x+2=0$, i.e., $x_{A}=1, x_{B}=2$. From this, we obtain $y_{A}=1 / 2, y_{B}=2$ (Fig. 106).

We have found two points of intersection of the parabola $y=x^{2} / 2$ and the line $3 x-2 y-2=0: A(1 ; 1 / 2)$ and $B(2 ; 2)$. We find the derivatives of the function $y=x^{2} / 2$ at these points:

$$
y^{\prime}=\left(x^{2} / 2\right)^{\prime}=x ; y^{\prime}(1)=1 ; y^{\prime}(2)=2
$$

Thus, the slope of the tangent $A F$, drawn to the curve $y=x^{2} / 2$ at the point $A(1 ; 1 / 2)$, is 1, and the slope of the tangent $B N$ at the point $B(2 ; 2)$ is 2. We find the angles between these tangents and the line $3 x-2 y-2=0$. The equation of the line $3 x-2 y-2=0$ can be written as $y=\frac{3}{2} x-1$. Thus, $k_{A B}=\frac{3}{2}$; $k_{A F}=1, k_{B N}=2$.

From analytic geometry, the formula for finding the tangent of the angle between two lines given their slopes is known:

$$
\operatorname{tg} \varphi_{1}=\frac{k_{A B}-k_{A F}}{1+k_{A B} k_{A F}}=\frac{\frac{3}{2}-1}{1+\frac{3}{2} \cdot 1}=\frac{1}{5}
$$

$$
\operatorname{tg} \varphi_{2}=\frac{k_{B N}-k_{A B}}{1+k_{B N} k_{A B}}=\frac{2-\frac{3}{2}}{1+2 \cdot \frac{3}{2}}=\frac{1}{8}
$$

Thus, $\varphi_{1}=\operatorname{arctg}(1 / 5)$ is the angle formed by the line $A B$ with the tangent $A F$, and $\varphi_{2}=\operatorname{arctg}(1 / 8)$ is the angle formed by this line with the tangent $B N$.