8. On the $O y$ axis, find the point $M$ through which two tangents to the graph of the function $y=0.5(x-(1 / 2))^{2}$ pass, the angle between which is $45^{\circ}$.
# Solution:

On the $O y$ axis, find the point $M$ through which two tangents to the graph of the function $y=0.5 \cdot\left(x-\frac{1}{2}\right)^{2}$ pass, with the angle between them being $45^{\circ}$.

Solution (without using derivatives).

$y=0.5 \cdot(x-1 / 2)^{2}, M\left(0 ; y_{0}\right) \cdot$ The equation $0.5 \cdot(x-1 / 2)^{2}=y_{0}+k x$, or $x^{2}-2\left(k+\frac{1}{2}\right) x-2 y_{0}+\frac{1}{4}=0$, has a unique solution if $\frac{D}{4}=k^{2}+k+2 y_{0}=0$.

The two values of $k$ found from this equation must satisfy the conditions $k_{1}+k_{2}=-1$ (1), $k_{1} \cdot k_{2}=2 y_{0}$ (2). From the condition $\alpha_{2}-\alpha_{1}=45^{\circ}, \operatorname{tg}\left(\alpha_{2}-\alpha_{1}\right)=1$ it follows that $\frac{\operatorname{tg} \alpha_{2}-\operatorname{tg} \alpha_{1}}{1+\operatorname{tg} \alpha_{2} \cdot \operatorname{tg} \alpha_{1}}=1$, or $\frac{k_{2}-k_{1}}{1+k_{2} k_{1}}=1$. Hence, $k_{2}-k_{1}=1+k_{2} k_{1}, k_{2}-k_{1}=1+2 y_{0}$, (3). Solving the system $\left\{\begin{array}{c}k_{2}-k_{1}=1+2 y_{0}, \\ k_{1}+k_{2}=-1,\end{array}\right.$ we find $\left\{\begin{array}{c}k_{2}=y_{0}, \\ k_{1}=-y_{0}-1\end{array}\right.$ and $k_{1} k_{2}=-y_{0}^{2}-y_{0}$. Considering (2), we get the condition $2 y_{0}=-y_{0}^{2}-y_{0}$, or $y_{0}^{2}+3 y_{0}=0$. There are two possible solutions. 1)
$y_{0}=0, M(0 ; 0), k_{1}=-1, k_{2}=0$. Equations of the tangents: $y_{1}=-x ; y_{2}=0$.

2) $y_{0}=-3, M(0 ;-3), k_{1}=2, k_{2}=-3$. Equations of the tangents: $y_{1}=2 x-3 ; y_{2}=-3 x-3$.

Answer: $M(0 ; 0)$ or $M(0 ;-3)$.