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Question
The point on the curve 9y2 = x3, where the normal to the curve makes equal intercepts with the axes is
(a) \[\left( 4, \frac{8}{3} \right)\]
(b) \[\left( - 4, \frac{8}{3} \right)\]
(c) \[\left( 4, - \frac{8}{3} \right)\]
(d) none of these
Solution
(a) \[\left( 4, \frac{8}{3} \right)\] and (c) \[\left( 4, - \frac{8}{3} \right)\]
Let (x1, y1) be the required point.
\[\text { Since, } \left( x_1 , y_1 \right) \text { lies on the given curve} \]
\[ \therefore 9 {y_1}^2 = {x_1}^3 . . . \left( 1 \right)\]
\[\text { Now }, 9 y^2 = x^3 \]
\[18y \frac{dy}{dx} = 3 x^2 \]
\[ \Rightarrow \frac{dy}{dx} = \frac{3 x^2}{18y} = \frac{x^2}{6y}\]
\[\text { Slope of the tangent } = \left( \frac{dy}{dx} \right)_\left( x_1 , y_1 \right) =\frac{{x_1}^2}{6 y_1}\]
\[\text { Slope of the normal } =\frac{- 1}{\frac{{x_1}^2}{6 y_1}}=\frac{- 6 y_1}{{x_1}^2}\]
\[\text { It is given that the normal makes equal intercepts with the axes }.\]
\[\therefore \text { Slope of the normal } = \pm1\]
\[\text { Now }, \]
\[\frac{- 6 y_1}{{x_1}^2} = \pm 1\]
\[ \Rightarrow \frac{- 6 y_1}{{x_1}^2} = 1 or \frac{- 6 y_1}{{x_1}^2}=-1\]
\[ \Rightarrow y_1 = \frac{- {x_1}^2}{6} \ or \ y_1 = \frac{{x_1}^2}{6} . . . \left( 2 \right)\]
\[\text { Case 1: When }y_1 = \frac{- {x_1}^2}{6}\]
\[\text { From (1), we have}\]
\[9\left( \frac{{x_1}^4}{36} \right) = {x_1}^3 \]
\[ \Rightarrow {x_1}^4 = 4 {x_1}^3 \]
\[ \Rightarrow {x_1}^4 - 4 {x_1}^3 = 0\]
\[ \Rightarrow {x_1}^3 \left( x_1 - 4 \right) = 0\]
\[ \Rightarrow x_1 = 0, 4\]
\[\text { Putting } x_1 = 0 \text { in } \left( 1 \right), \text { we get }, \]
\[9 {y_1}^2 = 0\]
\[ \Rightarrow y_1 = 0\]
\[\text { Putting } x_1 = 4 \text { in } \left( 1 \right), \text { we get }, \]
\[9 {y_1}^2 = 4^3 \]
\[ \Rightarrow y_1 = \pm \frac{8}{3}\]
\[\text { But, the line making the equal intercepts with the coordinate axes can not pass through the origin } . \]
\[\text { So, the points are } \left( 4, \pm \frac{8}{3} \right) \]
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