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 find the slopes of the tangents to these curves and analyze their product.

### Step 1: Differentiate the first curve
The first curve is given by:
\[ x^3 - 3xy^2 + 2 = 0 \]
We differentiate this implicitly with respect to \( x \):
\[ \frac{d}{dx}(x^3) - \frac{d}{dx}(3xy^2) + \frac{d}{dx}(2) = 0 \]
Calculating the derivatives:
\[ 3x^2 - 3(y^2 + 2xy\frac{dy}{dx}) = 0 \]
Rearranging gives:
\[ 3x^2 - 3y^2 - 6xy\frac{dy}{dx} = 0 \]
Now, isolate \( \frac{dy}{dx} \):
\[ 6xy\frac{dy}{dx} = 3x^2 - 3y^2 \]
\[ \frac{dy}{dx} = \frac{3(x^2 - y^2)}{6xy} = \frac{x^2 - y^2}{2xy} \]
Let this slope be \( m_1 \):
\[ m_1 = \frac{x^2 - y^2}{2xy} \]

### Step 2: Differentiate the second curve
The second curve is given by:
\[ 3x^2y - y^3 = 2 \]
Differentiating this implicitly with respect to \( x \):
\[ \frac{d}{dx}(3x^2y) - \frac{d}{dx}(y^3) = \frac{d}{dx}(2) \]
Calculating the derivatives:
\[ 3(2xy + x^2\frac{dy}{dx}) - 3y^2\frac{dy}{dx} = 0 \]
Rearranging gives:
\[ 6xy + 3x^2\frac{dy}{dx} - 3y^2\frac{dy}{dx} = 0 \]
Now, isolate \( \frac{dy}{dx} \):
\[ (3x^2 - 3y^2)\frac{dy}{dx} = -6xy \]
\[ \frac{dy}{dx} = \frac{-6xy}{3(x^2 - y^2)} = \frac{-2xy}{x^2 - y^2} \]
Let this slope be \( m_2 \):
\[ m_2 = \frac{-2xy}{x^2 - y^2} \]

### Step 3: Analyze the product of slopes
Now we will find the product \( m_1 \cdot m_2 \):
\[ m_1 \cdot m_2 = \left(\frac{x^2 - y^2}{2xy}\right) \cdot \left(\frac{-2xy}{x^2 - y^2}\right) \]
Simplifying this:
\[ m_1 \cdot m_2 = \frac{(x^2 - y^2)(-2xy)}{2xy(x^2 - y^2)} = -1 \]

### Conclusion
Since the product of the slopes \( m_1 \cdot m_2 = -1 \), this indicates that the two curves are perpendicular to each other.