To find where the tangent is parallel to the \(x\)-axis, we need \(\frac{dy}{dx} = 0\).
Differentiate the given equation implicitly:
\(\frac{d}{dx}(2x^2y - xy^2) = \frac{d}{dx}(a^3)\).
Using the product rule, we have:
\(4xy + 2x^2\frac{dy}{dx} - (y^2 + 2xy\frac{dy}{dx}) = 0\).
Simplify to get:
\(4xy + 2x^2\frac{dy}{dx} - y^2 - 2xy\frac{dy}{dx} = 0\).
Combine terms:
\((2x^2 - 2xy)\frac{dy}{dx} = y^2 - 4xy\).
Set \(\frac{dy}{dx} = 0\) to find:
\(y^2 - 4xy = 0\).
Factor to get:
\(y(y - 4x) = 0\).
So, \(y = 0\) or \(y = 4x\).
Reject \(y = 0\) because it does not satisfy the original equation unless \(a = 0\), which is not allowed as \(a\) is positive.
Substitute \(y = 4x\) into the original equation:
\(2x^2(4x) - x(4x)^2 = a^3\).
Simplify to get:
\(8x^3 - 16x^3 = a^3\).
\(-8x^3 = a^3\).
\(x^3 = -\frac{a^3}{8}\).
\(x = -\frac{a}{2}\).
Substitute \(x = -\frac{a}{2}\) back into \(y = 4x\):
\(y = 4\left(-\frac{a}{2}\right) = -2a\).
Thus, the \(y\)-coordinate of the point is \(y = -2a\).
