5. Determine the angle between the tangents to the parabola $y^{2}=4 x$ at the points of its intersection $\mathrm{s}$

(8) with the line $2 x+y-12=0$.
Solution. The points of intersection of the parabola and the line are obtained from the equations:

$$
y^{2}=4 x \quad \text { and } \quad y=-2 x+12
$$

From here we have

$$
\begin{aligned}
(-2 x+12)^{2}= & 4 x \\
x^{2}-13 x+36= & 0 \\
x_{1}=9, & x_{2}=4 \\
S_{1}(9,-6), & S_{2}(4,4)
\end{aligned}
$$

The equations of the tangents to the parabola are:

$$
\begin{array}{lll}
t_{1} & \cdots & y \cdot(-6)=2 \cdot(x+9) \quad \Rightarrow \quad y=-\frac{1}{3} x-3 \quad \Rightarrow \quad k_{1}=-\frac{1}{3} \\
t_{2} & \cdots & y \cdot 4=2 \cdot(x+4) \quad \Rightarrow \quad y=\frac{1}{2} x+2 \quad \Rightarrow \quad k_{2}=\frac{1}{2}
\end{array}
$$

The angle $\alpha$ between the tangents is determined by the relation

$$
\operatorname{tg} \alpha=\left|\frac{k_{2}-k_{1}}{1+k_{1} k_{2}}\right|=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \cdot \frac{1}{3}}=1
$$

so the desired angle is $\alpha=45^{\circ}$.

(full solution 8 points)

(for correct slope coefficients of the tangents $k_{1}, k_{2}, 4$ points)