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प्रश्न
Find the points on the curve y = x3 where the slope of the tangent is equal to the x-coordinate of the point ?
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उत्तर
Let (x1, y1) be the required point.
x coordinate of the point is x1.
\[\text { Since, the point lies on the curve } . \]
\[\text { Hence,} y_1 = {x_1}^3 . . . \left( 1 \right)\]
\[\text { Now }, y = x^3 \]
\[ \Rightarrow \frac{dy}{dx} = 3 x^2 \]
\[\text { Slope of tangent at }\left( x, y \right)= \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) =3 {x_1}^2 \]
\[\text { Given that }\]
\[\text { Slope of tangent at }\left( x_1 , y_1 \right)= x\text { co-ordinate of the point }\]
\[ \Rightarrow 3 {x_1}^2 = x_1 \]
\[ \Rightarrow x_1 \left( 3 x_1 - 1 \right) = 0\]
\[ \Rightarrow x_1 = 0 \text { or }x_1 = \frac{1}{3}\]
\[ \Rightarrow y_1 = 0^3 \text{or} \ y_1 = \left( \frac{1}{3} \right)^3 (\text { From }(1))\]
\[ \Rightarrow y_1 = 0 \text { or }y_1 = \frac{1}{27}\]
\[\text { So, the points are }\left( x_1 , y_1 \right)=\left( 0, 0 \right),\left( \frac{1}{3}, \frac{1}{27} \right)\]
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