373. At what point is the tangent to the curve $y=x^{2}-1$ : a) parallel to the $O x$ axis; b) forms a $45^{\circ}$ angle with the $O x$ axis?
Solution. a) Since the line is parallel to the $O x$ axis, it forms an angle of 0 with it, and its angular coefficient, equal to the tangent of this angle, is zero. It is known that the derivative of the function at a given point is equal to the angular coefficient of the tangent to the curve at this point, i.e., $f^{\prime}\left(x_{0}\right)=$ $=0$.

Let's find the derivative of the function $y=x^{2}-1$; we have $y^{\prime}=\left(x^{2}-1\right)^{\prime}=2 x$. Then $2 x=0$, i.e., $x_{0}=0$, from which $y_{0}=-1$. Thus, the tangent to the given curve is parallel to the $O x$ axis at the point $(0 ;-1)$ (Fig. 103).

b) Since the line forms an angle of $45^{\circ}$ with the $O x$ axis, its angular coefficient, equal to the tangent of this angle, is one. We previously found the derivative of the function $y=x^{2}-1$ at any of its points: $y^{\prime}=2 x$. Let's find the value of the argument at which this derivative equals $1: 2 x_{0}=1$, i.e., $x_{0}=$ $=1 / 2$. Then $y_{0}=-3 / 4$. Thus, the tangent to the given curve forms an angle of $45^{\circ}$ with the $O x$ axis at the point ( $1 / 2 ;-3 / 4$ ) (Fig. 103).