Question

Find the points on the curve y² = 2x³ at which the slope of the tangent is 3 ?

Answer 1

Let (x₁, y₁) be the required point.
Given :

\[y^2 = 2 x^3 \]

\[\text { Since }\left( x_1 y_1 \right) \text { lies on a curve }, {y_1}^2 = 2 {x_1}^3 . . . . \left( 1 \right)\]

\[ \Rightarrow 2y\frac{dy}{dx} = 6 x^2 \]

\[ \Rightarrow \frac{dy}{dx} = \frac{6 x^2}{2y} = \frac{3 x^2}{y}\]

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

\[\text { Slope of the tangent }=3 [\text { Given }]\]

\[ \therefore \frac{3 {x_1}^2}{y_1} = 3 . . . . \left( 2 \right)\]

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

\[\text { On substituting the value of } y_1 \text { in eq. (1), we get }\]

\[ {x_1}^4 = 2 {x_1}^3 \]

\[ \Rightarrow {x_1}^3 \left( x_1 - 2 \right) = 0\]

\[ \Rightarrow x_1 = 0, 2\]

\[\text { Case }1\]

\[\text { When }x_1 = 0, y_1 = x^2 = 0 . \text { Thus, we get the point }\left( 0, 0 \right). \text { But, it does not satisfy eq }. (2).\]

\[\text { So, we can ignore } (0, 0).\]

\[\text { Case }2\]

\[\text { When } x_1 = 2, y_1 = {x_1}^2 = 4 . \text { Thus, we get the point }\left( 2, 4 \right).\]