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प्रश्न
Find the points on the curve y2 = 2x3 at which the slope of the tangent is 3 ?
उत्तर
Let (x1, y1) be the required point.
Given :
\[y^2 = 2 x^3 \]
\[\text { Since }\left( x_1 y_1 \right) \text { lies on a curve }, {y_1}^2 = 2 {x_1}^3 . . . . \left( 1 \right)\]
\[ \Rightarrow 2y\frac{dy}{dx} = 6 x^2 \]
\[ \Rightarrow \frac{dy}{dx} = \frac{6 x^2}{2y} = \frac{3 x^2}{y}\]
\[\text { Slope of the tangent at}\left( x, y \right)=\frac{3 {x_1}^2}{y_1}\]
\[\text { Slope of the tangent }=3 [\text { Given }]\]
\[ \therefore \frac{3 {x_1}^2}{y_1} = 3 . . . . \left( 2 \right)\]
\[ \Rightarrow y_1 = {x_1}^2 \]
\[\text { On substituting the value of } y_1 \text { in eq. (1), we get }\]
\[ {x_1}^4 = 2 {x_1}^3 \]
\[ \Rightarrow {x_1}^3 \left( x_1 - 2 \right) = 0\]
\[ \Rightarrow x_1 = 0, 2\]
\[\text { Case }1\]
\[\text { When }x_1 = 0, y_1 = x^2 = 0 . \text { Thus, we get the point }\left( 0, 0 \right). \text { But, it does not satisfy eq }. (2).\]
\[\text { So, we can ignore } (0, 0).\]
\[\text { Case }2\]
\[\text { When } x_1 = 2, y_1 = {x_1}^2 = 4 . \text { Thus, we get the point }\left( 2, 4 \right).\]
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