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

$$
y=x^{2} / 2, \quad M\left(\sqrt{3} / 2 ; y_{0}\right) . \text { The equation } \frac{1}{2} x^{2}=y_{0}+k\left(x-\frac{\sqrt{3}}{2}\right), \text { or } x^{2}-2 k x+k \sqrt{3}-2 y_{0}=0,
$$

has a unique solution if $\frac{D}{4}=k^{2}-k \sqrt{3}+2 y_{0}=0$.

The two values of $k$ found from this equation must satisfy the conditions $k_{1}+k_{2}=\sqrt{3}(1), k_{1} \cdot k_{2}=2 y_{0}(2)$.

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 $\frac{k_{2}-k_{1}}{1+k_{2} k_{1}}=\sqrt{3}$. Hence, $k_{2}-k_{1}=\sqrt{3}\left(1+2 y_{0}\right)(3)$.

From (1) and (3) it follows that $k_{1}=-\sqrt{3} y_{0}, k_{2}=\sqrt{3}+\sqrt{3} y_{0}$. From (2) it follows that $k_{1} k_{2}=-3 y_{0}-3 y_{0}^{2}=2 y_{0}$,
$3 y_{0}^{2}+5 y_{0}=0$. Hence, $\left(y_{0}\right)_{1}=0,\left(y_{0}\right)_{2}=-5 / 3$.
Answer: $M_{1}(\sqrt{3} / 2 ; 0), M_{2}(\sqrt{3} / 2 ;-5 / 3)$.