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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 $\mathrm{X}$-axis is $\mathrm{f}(\mathrm{x}, \mathrm{y})=0$, then the point that lies on this curve $f(x, y)=0$ is
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The correct answer is:
$\left(-2, \frac{1}{\sqrt[3]{3}}\right)$
$\because x^3 y^2+\frac{x^2}{y}=5$
Differentiating both sides with respect to ' $x$ ', we get :
$3 x^2 y^2+2 x^3 y \frac{d y}{d x}+\frac{2 x}{y}-\frac{x^2}{y^2} \frac{d y}{d x}=0$
$\begin{aligned} & \Rightarrow\left(2 x^3 y-\frac{x^2}{y^2}\right) \frac{d y}{d x}=-\left(3 x^2 y^2+\frac{2 x}{y}\right) \\ & \Rightarrow \frac{d y}{d x}=\frac{y}{x}\left(\frac{3 x y^3+2}{2 x y^3-1}\right)\end{aligned}$
$\because$ Tangent is parallel to $\mathrm{x}$-axis.
$\begin{aligned} & \therefore \frac{d y}{d x}=0 \Rightarrow \frac{y}{x}\left(\frac{3 x y^3+2}{2 x y^3-1}\right)=0 \\ & \Rightarrow 3 x y^3+2=0\end{aligned}$
The point satisfying above equation, will lies on $f(x, y)=0$
Now, let us check point $(2, \sqrt[3]{3}): 6 \times 3+2 \neq 0$ Point $(\sqrt[3]{2}, 3): 3 \sqrt[3]{2} \cdot 27+2 \neq 0$ Point $\left(-2, \frac{1}{\sqrt[3]{3}}\right): 3(-2) \cdot \frac{1}{3}+2=0$
$\therefore$ Point $\left(-2, \frac{1}{\sqrt[3]{3}}\right)$ lies on the equation $\mathrm{f}(\mathrm{x}, \mathrm{y})=0$
Differentiating both sides with respect to ' $x$ ', we get :
$3 x^2 y^2+2 x^3 y \frac{d y}{d x}+\frac{2 x}{y}-\frac{x^2}{y^2} \frac{d y}{d x}=0$
$\begin{aligned} & \Rightarrow\left(2 x^3 y-\frac{x^2}{y^2}\right) \frac{d y}{d x}=-\left(3 x^2 y^2+\frac{2 x}{y}\right) \\ & \Rightarrow \frac{d y}{d x}=\frac{y}{x}\left(\frac{3 x y^3+2}{2 x y^3-1}\right)\end{aligned}$
$\because$ Tangent is parallel to $\mathrm{x}$-axis.
$\begin{aligned} & \therefore \frac{d y}{d x}=0 \Rightarrow \frac{y}{x}\left(\frac{3 x y^3+2}{2 x y^3-1}\right)=0 \\ & \Rightarrow 3 x y^3+2=0\end{aligned}$
The point satisfying above equation, will lies on $f(x, y)=0$
Now, let us check point $(2, \sqrt[3]{3}): 6 \times 3+2 \neq 0$ Point $(\sqrt[3]{2}, 3): 3 \sqrt[3]{2} \cdot 27+2 \neq 0$ Point $\left(-2, \frac{1}{\sqrt[3]{3}}\right): 3(-2) \cdot \frac{1}{3}+2=0$
$\therefore$ Point $\left(-2, \frac{1}{\sqrt[3]{3}}\right)$ lies on the equation $\mathrm{f}(\mathrm{x}, \mathrm{y})=0$
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