8. Find the angle between the tangents to the graph of the function $y=x^{2} \sqrt{3} / 6$, passing through the point $M(1 ;-\sqrt{3} / 2)$.
Solution:

$$
y=\frac{x^{2}}{2 \sqrt{3}}, M\left(1 ;-\frac{\sqrt{3}}{2}\right) \cdot y=\frac{x_{0}^{2}}{2 \sqrt{3}}+\frac{x_{0}}{\sqrt{3}}\left(x-x_{0}\right) ;-\frac{\sqrt{3}}{2}=\frac{x_{0}^{2}}{2 \sqrt{3}}+\frac{x_{0}}{\sqrt{3}}\left(1-x_{0}\right) ; x_{0}^{2}-2 x_{0}-3=0
$$

$; x_{0}=1 \pm 2 ;\left(x_{0}\right)_{1}=3,\left(x_{0}\right)_{2}=-1$. Equations of the tangents:

1) $y=\frac{3 \sqrt{3}}{2}+\sqrt{3}(x-3) ; y=\sqrt{3} x-\frac{3 \sqrt{3}}{2}, \operatorname{tg} \alpha_{1}=\sqrt{3}, \alpha_{1}=60^{\circ}$;
2) $y=\frac{1}{2 \sqrt{3}}-\frac{1}{\sqrt{3}}(x+1) ; y=-\frac{1}{\sqrt{3}} x-\frac{1}{2 \sqrt{3}}, \operatorname{tg} \alpha_{2}=-\frac{1}{\sqrt{3}}, \alpha_{2}=-30^{\circ}$.

Angle between the tangents: $\varphi=60^{\circ}-\left(-30^{\circ}\right)=90^{\circ}$.

Answer: $90^{\circ}$.