V is the set of points on the curve y^3 - 3xy | vedclass

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Question

(C) Given the curve equation: $y^3 - 3xy + 2 = 0$. Differentiating both sides with respect to $x$: $3y^2 \frac{dy}{dx} - 3(y + x \frac{dy}{dx}) = 0$.

Vedclass · Accepted Answer

$\{(1, 1)\}$

Vedclass · Answer

(C) Given the curve equation: $y^3 - 3xy + 2 = 0$. Differentiating both sides with respect to $x$: $3y^2 \frac{dy}{dx} - 3(y + x \frac{dy}{dx}) = 0$. Dividing by $3$: $y^2 \frac{dy}{dx} - y - x \frac{dy}{dx} = 0$. $\frac{dy}{dx}(y^2 - x) = y$. Thus,$\frac{dy}{dx} = \frac{y}{y^2 - x}$. For the tangent to be vertical,the denominator must be zero and the numerator must be non-zero: $y^2 - x = 0 \Rightarrow x = y^2$ and $y 
eq 0$. Substituting $x = y^2$ into the original curve equation: $y^3 - 3(y^2)y + 2 = 0$. $y^3 - 3y^3 + 2 = 0$. $-2y^3 + 2 = 0 \Rightarrow y^3 = 1 \Rightarrow y = 1$. Since $y = 1$,then $x = (1)^2 = 1$. The point is $(1, 1)$. Thus,$V = \{(1, 1)\}$.

$V$ is the set of points on the curve $y^3 - 3xy + 2 = 0$ where the tangent is vertical. Then $V = $

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