Question

If the tangent to the curve $x^3y^2 + \dfrac{x^2}{y} = 5$ is parallel to X-axis, then a point on its locus is:

• A. $(2, \sqrt[3]{3})$
• B. $(\sqrt[3]{2}, 3)$
• C. $(-2, \dfrac{1}{\sqrt[3]{3}})$
• D. $(-\sqrt[3]{2}, \dfrac{1}{\sqrt[3]{3}})$

Hint:

Horizontal Tangents:

• $dy/dx = 0$ implies RHS = 0 in implicit diff.
• Always reduce equations before checking options.

Answer 1

The Correct Option is C

Differentiate implicitly: \[ \frac{d}{dx}(x^3 y^2) + \frac{d}{dx}\left(\frac{x^2}{y}\right) = 0 \Rightarrow 3x^2 y^2 + 2x^3 y \cdot y' + 2x y^{-1} - x^2 y^{-2} y' = 0 \] Group $y'$ terms: \[ \left(2x^3 y - \frac{x^2}{y^2}\right)y' = -\left(3x^2 y^2 + \frac{2x}{y}\right) \] Set $y' = 0$: \[ 3x^2 y^2 + \frac{2x}{y} = 0 \Rightarrow 3x y^3 + 2 = 0 \Rightarrow xy^3 = -\frac{2}{3} \] Check option (3): $x = -2$, $y = \frac{1}{\sqrt[3]{3}}$ $\Rightarrow xy^3 = -\frac{2}{3}$