Tangents to the Curve x^3 + y^3 = 3axy at the Origin

Given curve: x^3 + y^3 = 3axy

To find the tangents to this curve at the origin, we need to determine the equations of the lines that pass through the point (0, 0) and are tangents to the curve at that point.

To do this, we will take the derivative of the equation with respect to x, and then find the slope of the tangent line at the origin (0, 0).

1. Taking the derivative:
Differentiating both sides of the equation with respect to x, we get:
3x^2 + 3y^2 * (dy/dx) = 3a(dy/dx) + 3ax(dy/dx)

2. Evaluating at (0, 0):
Substituting x = 0 and y = 0 into the equation, we get:
3(0)^2 + 3(0)^2 * (dy/dx) = 3a(dy/dx) + 3a(0)(dy/dx)
0 = 3a(dy/dx)

From this, we can see that either dy/dx = 0 or a = 0 will give us the tangent lines at the origin.

3. Two possible cases:
a) dy/dx = 0:
If dy/dx = 0, then the slope of the tangent line is 0. This means the tangent line is a horizontal line passing through the origin. The equation of this line is y = 0, which corresponds to option C.

b) a = 0:
If a = 0, then the curve simplifies to x^3 + y^3 = 0. Solving this equation, we find that the only solution is (0, 0). This means the curve is actually a single point at the origin and not a curve. In this case, there are no tangent lines to the curve.

Therefore, the correct answer is option A, which states that the tangents to the curve at the origin are x = 0 and y = 0.