The point on the curve $y^2=x$ at which the tangent makes an angle of $45^{\circ}$ with the $x$ axis

Question

The point on the curve $y^2=x$ at which the tangent makes an angle of $45^{\circ}$ with the $x$ axis, is

Professor PC Thomas Classes · Answer

Slope $m = 	an 45^\circ = 1$. Differentiating $y^2 = x$ yields $2y y' = 1 \Rightarrow y = 1/2$. Substituting back gives $x = 1/4$.