The points on the curve 9y^{2} = x^{3},where | vedclass

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(D) The equation of the given curve is $9y^{2} = x^{3}$.Differentiating with respect to $x$,we get:$9(2y) \frac{dy}{dx} = 3x^{2} \Rightarrow \frac{dy}

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$\left( 4, \pm \frac{8}{3} \right)$

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(D) The equation of the given curve is $9y^{2} = x^{3}$.Differentiating with respect to $x$,we get:$9(2y) \frac{dy}{dx} = 3x^{2} \Rightarrow \frac{dy}{dx} = \frac{x^{2}}{6y}$.The slope of the normal to the curve at point $(x_{1}, y_{1})$ is $-\frac{1}{\left( \frac{dy}{dx} ight)_{(x_{1}, y_{1})}} = -\frac{6y_{1}}{x_{1}^{2}}$.The equation of the normal at $(x_{1}, y_{1})$ is $y - y_{1} = -\frac{6y_{1}}{x_{1}^{2}}(x - x_{1})$.Rearranging into intercept form $\frac{x}{a} + \frac{y}{b} = 1$,we get:$x_{1}^{2}y - x_{1}^{2}y_{1} = -6y_{1}x + 6x_{1}y_{1} \Rightarrow 6y_{1}x + x_{1}^{2}y = 6x_{1}y_{1} + x_{1}^{2}y_{1} = y_{1}(6x_{1} + x_{1}^{2})$.Dividing by $y_{1}(6x_{1} + x_{1}^{2})$,we get the intercepts as $a = \frac{y_{1}(6x_{1} + x_{1}^{2})}{6y_{1}} = \frac{6x_{1} + x_{1}^{2}}{6}$ and $b = \frac{y_{1}(6x_{1} + x_{1}^{2})}{x_{1}^{2}}$.Since the intercepts are equal,$a = b$ or $a = -b$. For the normal to make equal intercepts,the slope of the normal must be $-1$.Thus,$-\frac{6y_{1}}{x_{1}^{2}} = -1 \Rightarrow x_{1}^{2} = 6y_{1}$.Substituting $y_{1} = \frac{x_{1}^{2}}{6}$ into the curve equation $9y_{1}^{2} = x_{1}^{3}$:$9 \left( \frac{x_{1}^{2}}{6} ight)^{2} = x_{1}^{3} \Rightarrow 9 \cdot \frac{x_{1}^{4}}{36} = x_{1}^{3} \Rightarrow \frac{x_{1}^{4}}{4} = x_{1}^{3}$.Since $x_{1} 
eq 0$,we have $x_{1} = 4$.Then $y_{1}^{2} = \frac{4^{3}}{9} = \frac{64}{9} \Rightarrow y_{1} = \pm \frac{8}{3}$.Thus,the points are $\left( 4, \pm \frac{8}{3} ight)$.

The points on the curve $9y^{2} = x^{3}$,where the normal to the curve makes equal intercepts with the axes are

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