1. **Problem statement:** Given the curve equation $$2x^2 y - xy^2 = a^3$$ where $a$ is a positive constant, we need to show there is only one point on the curve where the tangent is parallel to the x-axis and find the $y$-coordinate of that point. 2. **Recall:** The tangent is parallel to the x-axis when the gradient (slope) $\frac{dy}{dx} = 0$. 3. **Differentiate implicitly:** Start with $$2x^2 y - xy^2 = a^3$$ Differentiate both sides with respect to $x$: $$\frac{d}{dx}(2x^2 y) - \frac{d}{dx}(xy^2) = \frac{d}{dx}(a^3)$$ Using product rule: $$2(2x y + x^2 \frac{dy}{dx}) - (y^2 + 2xy \frac{dy}{dx}) = 0$$ Simplify: $$4xy + 2x^2 \frac{dy}{dx} - y^2 - 2xy \frac{dy}{dx} = 0$$ Group $\frac{dy}{dx}$ terms: $$2x^2 \frac{dy}{dx} - 2xy \frac{dy}{dx} = y^2 - 4xy$$ Factor $\frac{dy}{dx}$: $$\frac{dy}{dx}(2x^2 - 2xy) = y^2 - 4xy$$ 4. **Solve for $\frac{dy}{dx}$:** $$\frac{dy}{dx} = \frac{y^2 - 4xy}{2x^2 - 2xy} = \frac{y^2 - 4xy}{2x(x - y)}$$ 5. **Set $\frac{dy}{dx} = 0$ for tangent parallel to x-axis:** $$\frac{y^2 - 4xy}{2x(x - y)} = 0 \implies y^2 - 4xy = 0$$ Factor: $$y(y - 4x) = 0$$ So either: - $y = 0$, or - $y = 4x$ 6. **Find points on the curve for these cases:** - For $y=0$: Substitute into original equation: $$2x^2(0) - x(0)^2 = a^3 \implies 0 = a^3$$ This is false since $a^3 > 0$, so no point with $y=0$ on the curve. - For $y=4x$: Substitute into original equation: $$2x^2 (4x) - x (4x)^2 = a^3$$ $$8x^3 - x (16x^2) = a^3$$ $$8x^3 - 16x^3 = a^3$$ $$-8x^3 = a^3$$ Solve for $x$: $$x^3 = -\frac{a^3}{8}$$ $$x = -\frac{a}{2}$$ Then, $$y = 4x = 4 \times \left(-\frac{a}{2}\right) = -2a$$ 7. **Conclusion:** There is only one point on the curve where the tangent is parallel to the x-axis, at $$\left(-\frac{a}{2}, -2a\right)$$. The $y$-coordinate of this point is $$-2a$$.