8. On the line $x=1$, 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 $45^{\circ}$.
# Solution:

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

From (1) and (3) it follows that $k_{1}=-\frac{1}{2} y_{0}, k_{2}=1+\frac{1}{2} y_{0}$.

From (2) it follows that $k_{1} k_{2}=-\frac{1}{2} y_{0}-\frac{1}{4} y_{0}^{2}=y_{0}, y_{0}^{2}+6 y_{0}=0$. Hence, $\left(y_{0}\right)_{1}=0,\left(y_{0}\right)_{2}=-6$.

Answer: $M_{1}(1 ; 0), M_{2}(1 ;-6)$.