The point on the curve $y^2 = x,$ where the tangent makes an angle of $\frac{\pi }{4}$ with x-axis is :

$\left(\frac{1}{4}, \frac{1}{2}\right)$

The correct answer is Option (2) → $\left(\frac{1}{4}, \frac{1}{2}\right)$

$y^2=x$

so $2y\frac{dy}{dx}=1⇒\frac{dy}{dx}=\frac{1}{2y}$  (tangent makes $\frac{π}{4}$ with x-axis)

so $\frac{dy}{dx}=\tan\frac{π}{4}=\frac{1}{2y}$

so $y=\frac{1}{2}$

$⇒x=\frac{1}{4}$

Point $\left(\frac{1}{4}, \frac{1}{2}\right)$