The line which is parallel to X-axis and crosses the curve $y=\sqrt{x}$ at an angle of $45^{\circ}$, is

Given equation of a line parallel to X-axis is $\mathrm{y}=\mathrm{k}$ Given equation of the curve is $y=\sqrt{x}$, On solving equation of line with the equation of curve, we get $x=k^{2}$ Thus the intersecting point is $\left(\mathrm{k}^{2}, \mathrm{k}\right)$ It is given that the line $\mathrm{y}=\mathrm{k}$ intersect the curve $y=\sqrt{x}$ at an angle of $\pi / 4$. This means that the slope of the tangent to $$ \mathrm{y}=\sqrt{\mathrm{x}} \text { at }\left(\mathrm{k}^{2}, \mathrm{k}\right) \text { is } \tan \left(\pm \frac{\pi}{4}\right)=\pm 1 $$ $$ \begin{array}{l} \Rightarrow\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\left(\mathrm{k}^{2}, \mathrm{k}\right)}=\pm 1 \Rightarrow\left(\frac{1}{2 \sqrt{\mathrm{x}}}\right)_{\left(\mathrm{k}^{2}, \mathrm{k}\right)}=\pm 1 \\ \Rightarrow \mathrm{k}=\pm \frac{1}{2} \end{array} $$