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

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(B) Given the curve $y^2 = \frac{x^3}{9}$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = \frac{3x^2}{9} = \frac{x^2}{3}$,so $\frac{dy

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

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(B) Given the curve $y^2 = \frac{x^3}{9}$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = \frac{3x^2}{9} = \frac{x^2}{3}$,so $\frac{dy}{dx} = \frac{x^2}{6y}$.The slope of the normal at point $(x_1, y_1)$ is $m_n = -\frac{1}{dy/dx} = -\frac{6y_1}{x_1^2}$.The equation of the normal is $y - y_1 = -\frac{6y_1}{x_1^2}(x - x_1)$.Since the normal makes equal intercepts with the axes,its slope must be $\pm 1$. Given the geometry,the slope is $-1$.So,$-\frac{6y_1}{x_1^2} = -1 \implies x_1^2 = 6y_1$.Substitute $y_1^2 = \frac{x_1^3}{9}$ into the equation: $y_1 = \frac{x_1^2}{6} \implies y_1^2 = \frac{x_1^4}{36}$.Equating the two expressions for $y_1^2$: $\frac{x_1^3}{9} = \frac{x_1^4}{36} \implies 4x_1^3 = x_1^4 \implies x_1 = 4$ (since $x_1 
eq 0$).If $x_1 = 4$,then $y_1^2 = \frac{64}{9} \implies y_1 = \pm \frac{8}{3}$.Thus,the points are $(4, \pm \frac{8}{3})$.

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

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