414. At which points of the parabola $y=\sqrt{2} x^{2}$ is the radius of curvature equal to one?
Solution. We find the derivatives $y^{\prime}=2 \sqrt{2} x, y^{\prime \prime}=2 \sqrt{2}$ and by formula (1) the radius of curvature of the parabola at any point with abscissa $x$:

$$
R(x)=\frac{\left(1+8 x^{2}\right)^{\frac{3}{2}}}{2 \sqrt{2}}
$$

Setting $R(x)=1$, we get the abscissas of the desired points $2 \sqrt{ } \overline{2}=\left(1+8 x^{2}\right)^{\frac{2}{2}} ;(2 \sqrt{2})^{\frac{2}{3}}=1+8 x^{2} ; 8 x^{2}=1 ; x= \pm \frac{1}{2 \sqrt{2}}$.