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:

\[y = x^2 \]

\[\text { Point} \left( x_1 , y_1 \right) \text { lies on a curve } . \]

\[ \therefore y_1 = {x_1}^2 . . . \left( 1 \right)\]

\[\text { Now,} \]

\[y = x^2 \Rightarrow \frac{dy}{dx} = 2x\]

\[\text { Slope of the tangent at }\left( x_1 , y_1 \right)= \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) = 2 x_1 \]

\[\text { Slope of the tangent =x coordinate of the point [Given] }\]

\[ \therefore 2 x_1 = x_1 \]

\[\text { This happens only when } x_1 = 0.\]

\[\text{ On putting } x_1 = 0 \text { in eq }. \left( 1 \right), \text { we get }\]

\[ y_1 = {x_1}^2 = 0^2 = 0\]

\[\text { Thus, the required point is }\left( 0, 0 \right).\]