Find a point on the curve

Given:

The curve is $y=x^{2}$

$y=x^{2}$

Differentiating the above w.r.t $x$

$\Rightarrow \frac{d y}{d x}=2 x^{2-1}$

$\Rightarrow \frac{d y}{d x}=2 x \ldots(1)$

Also given the Slope of the tangent is equal to the $x$-coordinate,

$\frac{\mathrm{dy}}{\mathrm{dx}}=x \ldots(2)$

From (1) \& (2), we get,

i.e, $2 x=x$

$\Rightarrow x=0$

Substituting this in $y=x^{2}$, we get,

$y=0^{2}$

$\Rightarrow y=0$'

Thus, the required point is $(0,0)$