8. On the line $y=-5 / 3$ find the point $M$, through which two tangents to the graph of the function $y=x^{2} / 2$ pass, the angle between which is $60^{\circ}$.
Solution (without using derivatives).

$$
y=x^{2} / 2, \quad M\left(x_{0} ;-5 / 3\right)
$$

The equation $\frac{1}{2} x^{2}=-\frac{5}{3}+k\left(x-x_{0}\right)$, or $x^{2}-2 k x+2 k x_{0}+\frac{10}{3}=0$, has a unique solution if $\frac{D}{4}=k^{2}-2 k x_{0}-\frac{10}{3}=0$. The two values of $k$ found from this equation must satisfy the conditions $k_{1}+k_{2}=2 x_{0}$ (1), $k_{1} \cdot k_{2}=-10 / 3$ (3).

From the condition $\alpha_{2}-\alpha_{1}=60^{\circ}, \operatorname{tg}\left(\alpha_{2}-\alpha_{1}\right)=\sqrt{3}$, it follows that $\frac{\operatorname{tg} \alpha_{2}-\operatorname{tg} \alpha_{1}}{1+\operatorname{tg} \alpha_{2} \cdot \operatorname{tg} \alpha_{1}}=\sqrt{3}$, or $\frac{k_{2}-k_{1}}{1+k_{2} k_{1}}=\sqrt{3}$.

From this, $k_{2}-k_{1}=\sqrt{3}\left(1-\frac{10}{3}\right)$,

$$
k_{2}-k_{1}=-\frac{7}{\sqrt{3}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_9cd2bbfc846d490b2ee8g-11.jpg?height=946&width=854&top_left_y=1463&top_left_x=1092)

From (1) and (3) it follows that $k_{1}=x_{0}+\frac{7}{2 \sqrt{3}}, k_{2}=x_{0}-\frac{7}{2 \sqrt{3}}$. From (3) it follows that $x_{0}^{2}-\frac{49}{12}=-\frac{10}{3}, x_{0}^{2}=\frac{3}{4}, x_{0}= \pm \frac{\sqrt{3}}{2}$. Due to symmetry, both values are valid.

Answer: $M( \pm \sqrt{3} / 2 ;-5 / 3)$.