The point on the curve 9y^2 = x^3 where the n | vedclass

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(B) Given the curve $9y^2 = x^3$. Differentiating with respect to $x$, we get $18y \frac{dy}{dx} = 3x^2$, so $\frac{dy}{dx} = \frac{x^2}{6y}$.The slop

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$(4, 8/3)$

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(B) Given the curve $9y^2 = x^3$. Differentiating with respect to $x$, we get $18y \frac{dy}{dx} = 3x^2$, so $\frac{dy}{dx} = \frac{x^2}{6y}$.The slope of the normal at point $(x_1, y_1)$ is $m = -\frac{1}{dy/dx} = -\frac{6y_1}{x_1^2}$.Since the normal makes equal intercepts with the axes, its slope must be $\pm 1$. Given the curve, the normal must have a slope of $-1$ to make equal intercepts (i.e., $y - y_1 = -1(x - x_1) \implies x + y = x_1 + y_1$).Setting $-\frac{6y_1}{x_1^2} = -1$, we get $x_1^2 = 6y_1$.Substitute $y_1 = \frac{x_1^2}{6}$ into the curve equation $9y_1^2 = x_1^3$:$9(\frac{x_1^2}{6})^2 = x_1^3 \implies 9(\frac{x_1^4}{36}) = x_1^3 \implies \frac{x_1^4}{4} = x_1^3$.Since $x_1 
eq 0$, $x_1 = 4$. Then $y_1^2 = \frac{4^3}{9} = \frac{64}{9}$, so $y_1 = \pm 8/3$.Checking the slope condition: if $x_1 = 4, y_1 = 8/3$, slope $m = -\frac{6(8/3)}{16} = -1$. Thus, the point is $(4, 8/3)$.

The point on the curve $9y^2 = x^3$ where the normal to the curve makes equal intercepts with the coordinate axes is

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