Find the x-coordinate of the point on the cur | vedclass

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(D) Let the point on the curve be $(x_1, y_1)$.Given the curve $ay^2 = x^3$.Differentiating with respect to $x$,we get $2ay \frac{dy}{dx} = 3x^2$,so $

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$4a/9$

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(D) Let the point on the curve be $(x_1, y_1)$.Given the curve $ay^2 = x^3$.Differentiating with respect to $x$,we get $2ay \frac{dy}{dx} = 3x^2$,so $\frac{dy}{dx} = \frac{3x^2}{2ay}$.The slope of the normal at $(x_1, y_1)$ is $m = -\frac{1}{(dy/dx)_{(x_1, y_1)}} = -\frac{2ay_1}{3x_1^2}$.Since the normal makes equal intercepts on the axes,its slope must be $\pm 1$. Given the geometry of the curve,we set the slope to $-1$ for equal intercepts (i.e.,$x$-intercept = $y$-intercept).Thus,$-\frac{2ay_1}{3x_1^2} = -1$,which implies $2ay_1 = 3x_1^2$.From the curve equation,$ay_1^2 = x_1^3$. Squaring the first relation: $4a^2y_1^2 = 9x_1^4$.Substituting $ay_1^2 = x_1^3$,we get $4a(x_1^3) = 9x_1^4$.Since $x_1 
eq 0$,we have $4a = 9x_1$,so $x_1 = \frac{4a}{9}$.

Find the $x$-coordinate of the point on the curve $ay^2 = x^3$ where the normal makes equal intercepts on the axes.

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