Mathematics - Application of Derivative Question with Solution | TestHub

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