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

The required point be $(x,y)$

The tangent makes an angle of ${45}^o$ with the $x-$axis.

$\therefore$ Slop of the tangent $=\tan {45}^o=1$

Since, the point lies on the curve.

Now, $y=x^2$

Differentiate w.r.t $x,$

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

Slop of the tangent $={\left(\frac{dy}{dx}\right)}_{(x,y)}=2x$

$\Rightarrow$ $2x=1$ [ Given ]

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

Now,

$\Rightarrow$ $y=x^2={\left(\frac{1}{2}\right)}^2=\frac{1}{4}$

$\therefore$ $(x,y)=(\frac{1}{2},\frac{1}{4})$