8. On the line $x=\sqrt{3}$, find the point $M$ through which two tangents to the graph of the function $y=x^{2} / 4$ pass, such that the angle between them is $60^{\circ}$.
Solution (without using derivatives).

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

From (1) and (3) it follows that $k_{1}=-\frac{\sqrt{3}}{2} y_{0}, k_{2}=\sqrt{3}+\frac{\sqrt{3}}{2} y_{0}$. From (2) it follows that $k_{1} k_{2}=-\frac{3}{2} y_{0}-\frac{3}{4} y_{0}^{2}=y_{0}$, $3 y_{0}^{2}+10 y_{0}=0$. Hence, $\left(y_{0}\right)_{1}=0,\left(y_{0}\right)_{2}=-10 / 3$.

Answer: $M_{1}(\sqrt{3} ; 0), M_{2}(\sqrt{3} ;-10 / 3)$.