Find the points on the curve 9y^2 = x^3 where | vedclass

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(A) Let the point on the curve $9y^2 = x^3$ be $P(x_1, y_1)$.Differentiating with respect to $x$,we get $18y \frac{dy}{dx} = 3x^2$,so $\frac{dy}{dx} =

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

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(A) Let the point on the curve $9y^2 = x^3$ be $P(x_1, y_1)$.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 tangent at $P(x_1, y_1)$ is $m_t = \frac{x_1^2}{6y_1}$.The slope of the normal at $P(x_1, y_1)$ is $m_n = -\frac{1}{m_t} = -\frac{6y_1}{x_1^2}$.Since the normal makes equal intercepts with the axes,its slope must be $\pm 1$.Case $1$: $m_n = -1 \implies -\frac{6y_1}{x_1^2} = -1 \implies x_1^2 = 6y_1$.Since $P$ lies on the curve,$9y_1^2 = x_1^3$. Substituting $x_1^2 = 6y_1$,we get $9y_1^2 = x_1(6y_1) \implies 9y_1^2 = 6x_1y_1$. Assuming $y_1 
eq 0$,$9y_1 = 6x_1 \implies y_1 = \frac{2}{3}x_1$.Substituting into $x_1^2 = 6y_1$,we get $x_1^2 = 6(\frac{2}{3}x_1) = 4x_1 \implies x_1^2 - 4x_1 = 0 \implies x_1 = 0$ or $x_1 = 4$.If $x_1 = 4$,$y_1 = \frac{2}{3}(4) = \frac{8}{3}$. Point is $(4, 8/3)$.Case $2$: $m_n = 1 \implies -\frac{6y_1}{x_1^2} = 1 \implies x_1^2 = -6y_1$.Substituting into $9y_1^2 = x_1^3$,we get $9y_1^2 = x_1(-6y_1) \implies 9y_1 = -6x_1 \implies y_1 = -\frac{2}{3}x_1$.Substituting into $x_1^2 = -6y_1$,we get $x_1^2 = -6(-\frac{2}{3}x_1) = 4x_1 \implies x_1 = 4$ or $x_1 = 0$.If $x_1 = 4$,$y_1 = -\frac{2}{3}(4) = -\frac{8}{3}$. Point is $(4, -8/3)$.Thus,the points are $(4, 8/3)$ and $(4, -8/3)$.

Find the points on the curve $9y^2 = x^3$ where the normal to the curve makes equal intercepts with the axes.

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