380. Form the equations of the tangent and normal to the curve $y=\frac{1}{1+x^{2}}$ at the point $\left(2 ; \frac{1}{5}\right)$
Solution. The equation of the pencil of lines passing through the point $\left(2 ; \frac{1}{5}\right)$ is $y-\frac{1}{5}=k(x-2)$. We find the slope of the tangent:

$$
k_{1}=y_{x=2}^{\prime}=\left(\frac{1}{1+x^{2}}\right)_{x-2}^{\prime}=-\left.\frac{2 x}{\left(1+x^{2}\right)^{2}}\right|_{x=2}=-\frac{4}{25}
$$

Since the normal and the tangent, drawn at the same point of the curve, are perpendicular to each other, the slope of the normal $k_{2}=25 / 4$. Substituting the obtained values of $k_{1}$ and $k_{2}$ into the equation of the pencil of lines, we find the required equations of the tangent and the normal:

equation of the tangent: $y=\frac{-4 x+13}{25}$ or $4 x+25 y-13=0$;

equation of the normal: $y=\frac{125 x-246}{20}$ or $125 x-20 y-246=0$.