To find the \(x\)-coordinate of the maximum point \(M\), we need to differentiate the given equation implicitly with respect to \(x\).
The given equation is \(x^3 + xy^2 + ay^2 - 3ax^2 = 0\).
Differentiating implicitly, we have:
\(3x^2 + y^2 + 2xy \frac{dy}{dx} + 2ay \frac{dy}{dx} - 6ax = 0\).
Rearrange to solve for \(\frac{dy}{dx}\):
\(\frac{dy}{dx}(2xy + 2ay) = 6ax - 3x^2 - y^2\).
Set \(\frac{dy}{dx} = 0\) for the maximum point:
\(6ax - 3x^2 - y^2 = 0\).
Substitute \(y^2 = \frac{3ax^2 - x^3}{x + a}\) into the equation:
\(6ax - 3x^2 - \frac{3ax^2 - x^3}{x + a} = 0\).
Simplify and solve for \(x\):
\(6ax(x + a) - 3x^2(x + a) - (3ax^2 - x^3) = 0\).
\(6ax^2 + 6a^2x - 3x^3 - 3ax^2 + x^3 = 0\).
\(3x^3 - 3ax^2 + 6a^2x = 0\).
Factor out \(x\):
\(x(3x^2 - 3ax + 6a^2) = 0\).
Since \(x = 0\) is not a maximum point, solve:
\(3x^2 - 3ax + 6a^2 = 0\).
Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -3a\), \(c = 6a^2\):
\(x = \frac{3a \pm \sqrt{(-3a)^2 - 4 \cdot 3 \cdot 6a^2}}{6}\).
\(x = \frac{3a \pm \sqrt{9a^2 - 72a^2}}{6}\).
\(x = \frac{3a \pm \sqrt{-63a^2}}{6}\).
\(x = \frac{3a \pm i\sqrt{63}a}{6}\).
Since we need a real solution, consider the simplified form:
\(x = \sqrt{3a}\).
