The shortest distance between the parabolas 2y² = 2x -1 and 2x² = 2y - 1 is

The given parabola $2{\mathrm{y}}^2=2\mathrm{x}-1\text{ and }2{\mathrm{x}}^2=2\mathrm{y}-1$ are symmetrical about the line y = x.
Also, the shortest distance occurs along the common normal which is perpendicular to the line y = x.
Differentiating ${\text{2y}}^2=2\mathrm{x}-1 \mathrm{W}.\mathrm{r}.\mathrm{t} \mathrm{x}, \mathrm{weget} 2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}=1 \mathrm{or} \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2\mathrm{y}}=1 \mathrm{or} \mathrm{y}=\frac{1}{2}$
The points are as shown in the figure. Then the shortest distance is $\mathrm{d}=\sqrt{\frac{1}{16}+\frac{1}{16}}=\frac{1}{2\sqrt{2}}$