The point on the curve

Given that $\mathrm{y}^{2}=\mathrm{x}$

The tangent makes $45^{\circ}$ angle with $\mathrm{x}$-axis.

So, slope of tangent $=\tan 45^{\circ}=1$

$\because$ the point lies on the curve

$\therefore$ Slope of the curve at that point must be 1

$2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=1$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2 \mathrm{y}}$

$\Rightarrow \frac{1}{2 \mathrm{y}}=1$

$\Rightarrow \mathrm{y}=\frac{1}{2}$

And $\mathrm{x}=\frac{1}{4}$

So, the correct option is B.