Question

Find the point on the curve y = x² where the slope of the tangent is equal to the x-coordinate of the point.

Answer 1

Let the required point be (x₁, y₁).
Given:
$$\begin{gathered} y=x^2 \\ \mathrm{Point}\left(x_1,y_1\right)\mathrm{lies}\mathrm{on}\mathrm{a}\mathrm{curve}. \\ \therefore y_{\mathit{1}}={x_1}^2...\left(1\right) \\ \mathrm{Now}, \\ y=x^2\Rightarrow \frac{dy}{dx}=2x \\ \text{Slope of the tangent at}\left(x_1,y_1\right)\text{=}{\left(\frac{dy}{dx}\right)}_{\left(x_1,y_1\right)}=2x_1 \\ \mathrm{S}\text{lope of the tangent =}x\mathrm{coordinate}\mathrm{of}\mathrm{the}\mathrm{point}\left[\mathrm{Given}\right] \\ \therefore 2x_1=x_1 \\ \text{This happens only when}{\mathit{\text{x}}}_1\text{= 0.} \\ \mathrm{On}\mathrm{putting}{x}_1=0\mathrm{in}\mathrm{eq}.\left(1\right),\mathrm{we}\mathrm{get} \\ {y}_1={x_1}^2=0^2=0 \\ \text{Thus, the required point is}\left(0,0\right)\text{.} \end{gathered}$$