At which point(s) on the curve y^3 + 3x^2 = 1 | vedclass

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

Vedclass · Accepted Answer

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

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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 $\frac{dy}{dx}$: $\frac{dy}{dx} (3y^2 - 12) = -6x \implies \frac{dy}{dx} = -\frac{6x}{3y^2 - 12} = -\frac{2x}{y^2 - 4}$. For the tangent to be vertical,the slope $\frac{dy}{dx}$ must be undefined,which occurs when the denominator is zero: $y^2 - 4 = 0 \implies y^2 = 4 \implies y = \pm 2$. Case $1$: If $y = 2$,substitute into the original equation: $(2)^3 + 3x^2 = 12(2) \implies 8 + 3x^2 = 24 \implies 3x^2 = 16 \implies x^2 = \frac{16}{3} \implies x = \pm \frac{4}{\sqrt{3}}$. Case $2$: If $y = -2$,substitute into the original equation: $(-2)^3 + 3x^2 = 12(-2) \implies -8 + 3x^2 = -24 \implies 3x^2 = -16$. Since $x^2$ cannot be negative for real $x$,there are no real points for $y = -2$. Thus,the points are $\left( \pm \frac{4}{\sqrt{3}}, 2 \right)$.

At which point$(s)$ on the curve $y^3 + 3x^2 = 12y$ is the tangent vertical?

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