At which point is the tangent to the curve y^ | vedclass

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(D) Given the curve equation: $y^3 + 3x^2 = 12y$.Differentiating both sides with respect to $x$:$3y^2 \frac{dy}{dx} + 6x = 12 \frac{dy}{dx}$.Rearrangi

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$\left( \pm \frac{4}{\sqrt{3}}, 2 \right)$

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(D) Given the curve equation: $y^3 + 3x^2 = 12y$.Differentiating both sides with respect to $x$:$3y^2 \frac{dy}{dx} + 6x = 12 \frac{dy}{dx}$.Rearranging to solve for $\frac{dy}{dx}$:$6x = (12 - 3y^2) \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{6x}{12 - 3y^2} = \frac{2x}{4 - y^2}$.$A$ tangent is vertical when the slope $\frac{dy}{dx}$ is undefined,which occurs when the denominator is zero:$4 - y^2 = 0 \implies y^2 = 4 \implies y = \pm 2$.Case $1$: If $y = 2$,then $2^3 + 3x^2 = 12(2) \implies 8 + 3x^2 = 24 \implies 3x^2 = 16 \implies x = \pm \frac{4}{\sqrt{3}}$.Case $2$: If $y = -2$,then $(-2)^3 + 3x^2 = 12(-2) \implies -8 + 3x^2 = -24 \implies 3x^2 = -16$,which has no real solution for $x$.Thus,the points where the tangent is vertical are $\left( \pm \frac{4}{\sqrt{3}}, 2 \right)$.

At which point is the tangent to the curve $y^3 + 3x^2 = 12y$ vertical?

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