Find the point on the curve


Find the point on the curve $y=x^{2}$ where the slope of the tangent is equal to the $x$-coordinate of the point.


Let the required point be (x1y1).


Point $\left(x_{1}, y_{1}\right)$ lies on a curve.

$\therefore y_{1}=x_{1}^{2} \quad \ldots(1)$


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

Slope of the tangent at $\left(x_{1}, y_{1}\right)=\left(\frac{d y}{d x}\right)_{\left(x_{1}, y_{1}\right)}=2 x_{1}$

Slope of the tangent $=x$ coordinate of the point    $[$ Given $]$

$\therefore 2 x_{1}=x_{1}$

This happens only when $x_{1}=0$.


On putting $x_{1}=0$ in eq. (1), we get


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

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now