If the locus of the points on the curve x^3 y | vedclass

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(C) Given the curve equation: $x^3 y^2 + \frac{x^2}{y} = 5$. Differentiating both sides with respect to $x$: $3x^2 y^2 + 2x^3 y \frac{dy}{dx} + \frac{

Vedclass · Accepted Answer

$\left(-2, \frac{1}{\sqrt[3]{3}}\right)$

Vedclass · Answer

(C) Given the curve equation: $x^3 y^2 + \frac{x^2}{y} = 5$. Differentiating both sides with respect to $x$: $3x^2 y^2 + 2x^3 y \frac{dy}{dx} + \frac{2x}{y} - \frac{x^2}{y^2} \frac{dy}{dx} = 0$. Rearranging to solve for $\frac{dy}{dx}$: $\left(2x^3 y - \frac{x^2}{y^2}\right) \frac{dy}{dx} = -\left(3x^2 y^2 + \frac{2x}{y}\right)$. $\frac{dy}{dx} = -\frac{3x^2 y^2 + \frac{2x}{y}}{2x^3 y - \frac{x^2}{y^2}} = -\frac{3x^2 y^3 + 2x}{y(2x^3 y^2 - x^2)} = -\frac{x(3xy^3 + 2)}{y(2x^3 y^2 - x^2)}$. For the tangent to be parallel to the $X$-axis,$\frac{dy}{dx} = 0$,which implies $3xy^3 + 2 = 0$. This is the equation $f(x, y) = 0$. Testing the given options: For $\left(-2, \frac{1}{\sqrt[3]{3}}\right)$: $3(-2)\left(\frac{1}{\sqrt[3]{3}}\right)^3 + 2 = 3(-2)\left(\frac{1}{3}\right) + 2 = -2 + 2 = 0$. Thus,the point $\left(-2, \frac{1}{\sqrt[3]{3}}\right)$ satisfies the equation.

If the locus of the points on the curve $x^3 y^2+\frac{x^2}{y}=5$ at which the tangent is parallel to the $X$-axis is $f(x, y)=0$,then the point that lies on this curve $f(x, y)=0$ is

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