Question

At what point of the curve $y=x^2$ does the tangent make an angle of ${45}^o$ with the x-axis ?

Answer 1

Slope of the tangent at point $(x,y)$ on the curve $y=x^2$ is

$\frac{dy}{dx}=2x$

The tangent make an angle of ${45}^{\circ }$ with the $x$-axis.

So, slope $=\tan {45}^{\circ }=1$

$\Rightarrow 2x=1$

$x=\frac{1}{2}$

$y=x^2={\left(\frac{1}{2}\right)}^2$

$y=\frac{1}{4}$

So, the point is $(\frac{1}{2},\frac{1}{4})$