BITSAT 2015 Paper

Given equation of a line parallel to X-axis isy = 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 ($k^2$, k) It is given that the line y = k intersect the curve y = $\sqrt{x}$ at an angle of π/4. This means that the slope of the tangent to y = $\sqrt{x}$ at $(k^2,k)$ is tan $\left( \pm \frac{\pi}{4} \right)$ = ± 1 ⇒ $\left( \frac{dy}{dx} \right)_{(k^2,k)}$ = ± 1 ⇒ $\left( \frac{1}{2\sqrt{x}} \right)_{(k^2,k)}$ = ± 1 ⇒ k ± $\frac{1}{2}$