801. Write the equation of the tangent and normal to the parabola $f(x)=x^{2}-2 x+5$ at the point where the abscissa is $x=2$.
Solution. We will use formulas (1) and (2).

Given $x_{0}=2$, substituting this value into the given parabola equation, we find $y_{0}: y_{0}=2^{2}-2 \cdot 2+5, y_{0}=5$.

Therefore, the point of tangency is $M_{0}(2 ; 5)$.

Now let's find the slope of the tangent to the given curve at the point $x=2$:

$$
f^{\prime}(x)=2 x-2
$$

hence $f^{\prime}(2)=2$.

The equation of the tangent $M_{0} T$:

$$
y-5=2(x-2) \text{ or } 2 x-y+1=0
$$

The equation of the normal $M_{0} N$ (Fig. 62):

$$
y-5=-\frac{1}{2}(x-2) \text{ or } x+2 y-12=0
$$