The line which is parallel to X-axis and crosses the curve y= x at...

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 $\left(\mathrmk^{2}, \mathrm{k}\right)$ 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(\mathrmk^{2}, \mathrm{k}\right) \text{ is } \tan \left(\pm \frac{\pi}{4}\right)=\pm$1 $$\begin{array}{l} \Rightarrow \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\left(\mathrmk^{2}, \mathrm{k}\right)}=\pm1 \Rightarrow \left(\frac{1}{2 \sqrt{\mathrm{x}}}\right)_{\left(\mathrmk^{2}, \mathrm{k}\right)}=\pm1 \\ \Rightarrow \mathrm{k}=\pm \frac{1}{2} \end{array}$$