8. On the line $y=-15 / 2$ find the point $M$, through which two tangents to the graph of the function $y=x^{2} / 2$ pass, the angle between which is $30^{\circ}$.
# Solution:

(without using derivatives)

$y=x^{2} / 2, M\left(x_{0} ;-15 / 2\right)$. The equation

$\frac{1}{2} x^{2}=-\frac{15}{2}+k\left(x-x_{0}\right)$, or

$x^{2}-2 k x+2 k x_{0}+15=0$, has a unique solution if $\frac{D}{4}=k^{2}-2 k x_{0}-15=0$.

The solutions of this equation for $k$ satisfy the conditions $k_{1}+k_{2}=2 x_{0}(1)$, $k_{1} \cdot k_{2}=-15(3)$.

From the condition $\alpha_{2}-\alpha_{1}=30^{\circ}, \operatorname{tg}\left(\alpha_{2}-\alpha_{1}\right)=1 / \sqrt{3}$, it follows that $\frac{\operatorname{tg} \alpha_{2}-\operatorname{tg} \alpha_{1}}{1+\operatorname{tg} \alpha_{2} \cdot \operatorname{tg} \alpha_{1}}=\frac{1}{\sqrt{3}}$, or $\frac{k_{2}-k_{1}}{1+k_{2} k_{1}}=\frac{1}{\sqrt{3}}$. $k_{2}-k_{1}=-\frac{14}{\sqrt{3}}$ $(3)$.

From (1) and (3) it follows that $k_{1}=x_{0}+\frac{7}{\sqrt{3}}, k_{2}=x_{0}-\frac{7}{\sqrt{3}}$. From (3) it follows that $x_{0}^{2}-\frac{49}{3}=-15, x_{0}^{2}=\frac{4}{3}, x_{0}= \pm \frac{2}{\sqrt{3}}$. Both values are valid due to symmetry.

Answer: $M( \pm 2 / \sqrt{3} ;-15 / 2)$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9cd2bbfc846d490b2ee8g-04.jpg?height=1268&width=877&top_left_y=1131&top_left_x=861)
the solutions must be symmetric.