8. On the line $y=-13 / 6$ 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).

$$
y=x^{2} / 2, M\left(x_{0} ;-13 / 6\right)
$$

The equation $\frac{1}{2} x^{2}=-\frac{13}{6}+k\left(x-x_{0}\right)$, or $x^{2}-2 k x+2 k x_{0}+\frac{13}{3}=0$, has a unique solution if $\frac{D}{4}=k^{2}-2 k x_{0}-\frac{13}{3}=0$. The two values of $k$ found from this equation must satisfy the conditions $k_{1}+k_{2}=2 x_{0}(1)$, $k_{1} \cdot k_{2}=-13 / 3(3)$.

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, $k_{2}-k_{1}=\sqrt{3}\left(1-\frac{13}{3}\right), k_{2}-k_{1}=-\frac{10}{\sqrt{3}}$

From (1) and (3) it follows that $k_{1}=x_{0}+\frac{5}{\sqrt{3}}, k_{2}=x_{0}-\frac{5}{\sqrt{3}} \cdot$ From (3) it follows that $x_{0}^{2}-\frac{25}{3}=-\frac{13}{3}, x_{0}^{2}=4, x_{0}= \pm 2$. Due to symmetry, both values are valid.

Answer: $M( \pm 2 ;-13 / 6)$.