8. On the line $x-y=5$ find the point through which two perpendicular to each other tangents to the graph of the function $y=x^{2} / 8$ pass. Write the equations of these tangents.
# Solution:

Let $y=a x^{2}, A\left(x_{1} ; y_{1}\right), B\left(x_{2} ; y_{2}\right)$ be the points of tangency, $C\left(x_{0} ; y_{0}\right)$ be the point of intersection of the tangents.

Equations of the tangents

$$
\begin{aligned}
& y=a x_{1}^{2}+2 a\left(x-x_{1}\right), \text { or } y=2 a x_{1} x-a x_{1}^{2} \\
& y=a x_{2}^{2}+2 a\left(x-x_{2}\right), \text { or } y=2 a x_{2} x-a x_{2}^{2}
\end{aligned}
$$

From the system of equations $\left\{\begin{array}{l}y_{0}=2 a x_{1} x_{0}-a x_{1}^{2}, \\ y_{0}=2 a x_{2} x_{0}-a x_{2}^{2}\end{array}\right.$ we find $x_{0}=\frac{x_{1}+x_{2}}{2}, y_{0}=a x_{1} x_{2}$.

By the condition of perpendicularity of the tangents $2 a x_{1} \cdot 2 a x_{2}=-1$, from which we get $y_{0}=-\frac{1}{4 a}$. From the equation of the given line we find $x_{0}$.

To find $x_{1}$ and $x_{2}$, we use the quadratic equation $x^{2}-2 x_{0} x-\frac{1}{4 a^{2}}=0$. Hence, $x_{1,2}=x_{0} \pm \sqrt{x_{0}^{2}+\frac{1}{4 a^{2}}}$.

Answer: $a=\frac{1}{8}, y_{0}=-2, x_{0}=5+y_{0}=3 . \quad x_{1,2}=3 \pm \sqrt{9+16}=3 \pm 5 ; x_{1}=8, x_{2}=-2$.