Question

Find the points on the curve y = x³ where the slope of the tangent is equal to the x-coordinate of the point ?

Answer 1

Let (x₁, y₁) be the required point.

x coordinate of the point is x₁.

\[\text { Since, the point lies on the curve } . \]

\[\text { Hence,} y_1 = {x_1}^3 . . . \left( 1 \right)\]

\[\text { Now }, y = x^3 \]

\[ \Rightarrow \frac{dy}{dx} = 3 x^2 \]

\[\text { Slope of tangent at }\left( x, y \right)= \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) =3 {x_1}^2 \]

\[\text { Given that }\]

\[\text { Slope of tangent at }\left( x_1 , y_1 \right)= x\text {  co-ordinate of the point }\]

\[ \Rightarrow 3 {x_1}^2 = x_1 \]

\[ \Rightarrow x_1 \left( 3 x_1 - 1 \right) = 0\]

\[ \Rightarrow x_1 = 0 \text { or }x_1 = \frac{1}{3}\]

\[ \Rightarrow y_1 = 0^3 \text{or} \  y_1 = \left( \frac{1}{3} \right)^3 (\text { From }(1))\]

\[ \Rightarrow y_1 = 0 \text { or }y_1 = \frac{1}{27}\]

\[\text { So, the points are }\left( x_1 , y_1 \right)=\left( 0, 0 \right),\left( \frac{1}{3}, \frac{1}{27} \right)\]