Find the point on the curve

Question:

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

Solution:

Let the required point be (x1y1).
Given:

$y=x^{2}$

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

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

Now,

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

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

$y_{1}=x_{1}^{2}=0^{2}=0$

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

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