At which points 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 $x$: $3y^2 \frac{dy}{dx} + 6x = 12 \frac{dy}{dx}$. 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 find $\frac{dy}{dx}$: $\frac{dy}{dx} (3y^2 - 12) = -6x$ $\frac{dy}{dx} = \frac{-6x}{3y^2 - 12} = \frac{6x}{12 - 3y^2}$. For the tangent to be parallel to the $y$-axis,the slope $\frac{dy}{dx}$ must be undefined,which means the denominator must be zero: $12 - 3y^2 = 0 \implies y^2 = 4 \implies y = \pm 2$. Case $1$: If $y = 2$,then $y^3 + 3x^2 = 12y \implies (2)^3 + 3x^2 = 12(2) \implies 8 + 3x^2 = 24 \implies 3x^2 = 16 \implies x = \pm \frac{4}{\sqrt{3}}$. Case $2$: If $y = -2$,then $y^3 + 3x^2 = 12y \implies (-2)^3 + 3x^2 = 12(-2) \implies -8 + 3x^2 = -24 \implies 3x^2 = -16$,which has no real solution for $x$. Thus,the points are $\left( \pm \frac{4}{\sqrt{3}}, 2 \right)$.

At which points is the tangent to the curve $y^3 + 3x^2 = 12y$ parallel to the $y$-axis?

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