The two curves `x^3-3xy^2+2=0` and `3x^2y-y^3=2`, are:

Options

A. Parallel
B. Perpendicular
C. No common point of intersection
D. None of these

Answer

To determine the relationship between the two curves given by the equations \(x^3 - 3xy^2 + 2 = 0\) and \(3x^2y - y^3 = 2\), we will differentiate both equations to find the slopes of the tangents at their points of intersection.

### Step 1: Differentiate the first curve
The first curve is given by:
\[ f(x, y) = x^3 - 3xy^2 + 2 = 0 \]
To find the derivative \(\frac{dy}{dx}\), we differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}(x^3) - \frac{d}{dx}(3xy^2) + \frac{d}{dx}(2) = 0 \]
Using the product rule on \(3xy^2\):
\[ 3x^2 - 3\left(y^2 + 2y\frac{dy}{dx}x\right) = 0 \]
This simplifies to:
\[ 3x^2 - 3y^2 - 6xy\frac{dy}{dx} = 0 \]
Rearranging gives:
\[ 6xy\frac{dy}{dx} = 3x^2 - 3y^2 \]
Thus, we can express \(\frac{dy}{dx}\) as:
\[ \frac{dy}{dx} = \frac{x^2 - y^2}{2xy} \quad \text{(Equation 1)} \]

### Step 2: Differentiate the second curve
The second curve is given by:
\[ g(x, y) = 3x^2y - y^3 - 2 = 0 \]
Differentiating both sides with respect to \(x\):
\[ \frac{d}{dx}(3x^2y) - \frac{d}{dx}(y^3) - \frac{d}{dx}(2) = 0 \]
Using the product rule on \(3x^2y\) and the chain rule on \(y^3\):
\[ 6xy + 3x^2\frac{dy}{dx} - 3y^2\frac{dy}{dx} = 0 \]
Rearranging gives:
\[ 3x^2\frac{dy}{dx} - 3y^2\frac{dy}{dx} = -6xy \]
Factoring out \(\frac{dy}{dx}\):
\[ \frac{dy}{dx}(3x^2 - 3y^2) = -6xy \]
Thus, we can express \(\frac{dy}{dx}\) as:
\[ \frac{dy}{dx} = \frac{-6xy}{3(x^2 - y^2)} = \frac{-2xy}{x^2 - y^2} \quad \text{(Equation 2)} \]

### Step 3: Determine the relationship between the curves
Now, we have the slopes of the tangents at the points of intersection:
- From the first curve: \(m_1 = \frac{x^2 - y^2}{2xy}\)
- From the second curve: \(m_2 = \frac{-2xy}{x^2 - y^2}\)

To check if the curves are perpendicular, we multiply the slopes:
\[ m_1 \cdot m_2 = \left(\frac{x^2 - y^2}{2xy}\right) \cdot \left(\frac{-2xy}{x^2 - y^2}\right) \]
This simplifies to:
\[ m_1 \cdot m_2 = \frac{-(x^2 - y^2)(2xy)}{2xy(x^2 - y^2)} = -1 \]

### Conclusion
Since the product of the slopes \(m_1\) and \(m_2\) equals \(-1\), the two curves are perpendicular to each other.