At what point on the curve {x^3} - 8{a^2}y = | vedclass

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(C) Given the curve equation: ${x^3} - 8{a^2}y = 0$. Differentiating with respect to $x$: $3{x^2} - 8{a^2} \frac{dy}{dx} = 0$. Thus,the slope of the t

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$(2a, a)$

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(C) Given the curve equation: ${x^3} - 8{a^2}y = 0$. Differentiating with respect to $x$: $3{x^2} - 8{a^2} \frac{dy}{dx} = 0$. Thus,the slope of the tangent is $\frac{dy}{dx} = \frac{3{x^2}}{8{a^2}}$. The slope of the normal is given by $m_n = -\frac{1}{\frac{dy}{dx}} = -\frac{8{a^2}}{3{x^2}}$. We are given that the slope of the normal is $\frac{-2}{3}$. Equating the two: $-\frac{8{a^2}}{3{x^2}} = -\frac{2}{3}$. This simplifies to $\frac{8{a^2}}{{x^2}} = 2$,which means ${x^2} = 4{a^2}$,so $x = 2a$ (considering the positive coordinate). Substituting $x = 2a$ into the original curve equation: ${(2a)^3} - 8{a^2}y = 0$. $8{a^3} - 8{a^2}y = 0$,which implies $8{a^2}(a - y) = 0$,so $y = a$. Therefore,the point is $(2a, a)$.

At what point on the curve ${x^3} - 8{a^2}y = 0$ is the slope of the normal equal to $\frac{-2}{3}$?

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