8. Find the angle between the tangents to the graph of the function $y=x^{2} \sqrt{3} / 24$, passing through the point $M(4 ;-2 \sqrt{3})$.

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# Solution:

$$
y=\frac{x^{2}}{8 \sqrt{3}}, M(4 ;-2 \sqrt{3}) \cdot y=\frac{x_{0}^{2}}{8 \sqrt{3}}+\frac{x_{0}}{4 \sqrt{3}}\left(x-x_{0}\right) ;-2 \sqrt{3}=\frac{x_{0}^{2}}{8 \sqrt{3}}+\frac{x_{0}}{4 \sqrt{3}}\left(4-x_{0}\right) ;
$$

$x_{0}^{2}-8 x_{0}-48=0 ; x_{0}=4 \pm 8 ;\left(x_{0}\right)_{1}=12,\left(x_{0}\right)_{2}=-4$. Equations of the tangents:

1) $y=6 \sqrt{3}+\sqrt{3}(x-12) ; y=\sqrt{3} x-6 \sqrt{3}, \operatorname{tg} \alpha_{1}=\sqrt{3}, \alpha_{1}=60^{\circ}$;
2) $y=\frac{2}{\sqrt{3}}-\frac{1}{\sqrt{3}}(x+4) ; y=-\frac{1}{\sqrt{3}} x-\frac{2}{\sqrt{3}}, \operatorname{tg} \alpha_{2}=-\frac{1}{\sqrt{3}}, \alpha_{2}=-30^{\circ}$.

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

Answer: $90^{\circ}$.