8. On the $O y$ axis, find the point $M$ through which two tangents to the graph of the function $y=0.5(x-(\sqrt{3} / 2))^{2}$ pass, the angle between which is $60^{\circ} . \quad$ (12 points)
# Solution:

On the $O y$ axis, find the point $M$ through which two tangents to the graph of the function $y=0.5 \cdot(x-\sqrt{3} / 2)^{2}$ pass, with the angle between them being $60^{\circ}$.

Solution (without using derivatives).

$y=0.5 \cdot(x-\sqrt{3} / 2)^{2}, M\left(0 ; y_{0}\right) \cdot U$ The equation $0.5 \cdot(x-\sqrt{3} / 2)^{2}=y_{0}+k x$, or $x^{2}-2\left(k+\frac{\sqrt{3}}{2}\right) x-2 y_{0}+\frac{3}{4}=0$, has a unique solution if $\frac{D}{4}=k^{2}+\sqrt{3} k+2 y_{0}=0$.

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 $\frac{k_{2}-k_{1}}{1+k_{2} k_{1}}=\sqrt{3}$. Therefore, $k_{2}-k_{1}=\sqrt{3}\left(1+k_{2} k_{1}\right), k_{2}-k_{1}=\sqrt{3}\left(1+2 y_{0}\right)$, (3). Solving the system $\left\{\begin{array}{c}k_{2}-k_{1}=\sqrt{3}\left(1+2 y_{0}\right), \\ k_{1}+k_{2}=-\sqrt{3},\end{array}\right.$ we find $\left\{\begin{array}{c}k_{2}=\sqrt{3} y_{0}, \\ k_{1}=-\sqrt{3}\left(y_{0}+1\right)\end{array}\right.$ and $k_{1} k_{2}=-3 y_{0}^{2}-3 y_{0}$. Considering (2), we get the condition $2 y_{0}=-3 y_{0}^{2}-3 y_{0}$, or $3 y_{0}^{2}+5 y_{0}=0$. There are two possible solutions. 1) $y_{0}=0, M(0 ; 0), k_{1}=-\sqrt{3}, k_{2}=0$. Equations of the tangents: $y_{1}=-\sqrt{3} x ; y_{2}=0$.

2) $y_{0}=-\frac{5}{3}, M\left(0 ;-\frac{5}{3}\right), k_{1}=-\sqrt{3}+\frac{5}{\sqrt{3}}=\frac{2}{\sqrt{3}}, k_{2}=-\frac{5}{\sqrt{3}}$. Equations of the tangents: $y_{1}=\frac{2}{\sqrt{3}} x-\frac{5}{3} ; y_{2}=-\frac{5}{\sqrt{3}} x-\frac{5}{3}$. Answer: $M(0 ; 0)$ or $M(0 ;-5 / 3)$.