The two curves `x^(3) - 3xy^(2) + 2 = 0 and 3x^(2) y - y^(3) = 2`

Options

A. touch each other
B. cut at right angles
C. cut at an angle `(pi)/(3)`
D. cut at an angle `(pi)/(4)`

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 at their points of intersection. We will then analyze the slopes to see if they intersect at right angles, or at some other angle.

### Step-by-step Solution:

**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 \( x \):
\[ \frac{d}{dx}(x^3) - \frac{d}{dx}(3xy^2) + \frac{d}{dx}(2) = 0 \]
Using the product rule on \( 3xy^2 \):
\[ 3x^2 - 3\left(y^2 + 2y\frac{dy}{dx}x\right) = 0 \]
Rearranging gives:
\[ 3x^2 - 3y^2 - 6xy\frac{dy}{dx} = 0 \]
Thus,
\[ 3x^2 - 3y^2 = 6xy\frac{dy}{dx} \]
So,
\[ \frac{dy}{dx} = \frac{x^2 - y^2}{2xy} \quad \text{(1)} \]

**Step 2: Differentiate the second curve.**

The second curve is given by:
\[ 3x^2y - y^3 = 2 \]
Differentiating implicitly with respect to \( x \):
\[ \frac{d}{dx}(3x^2y) - \frac{d}{dx}(y^3) = 0 \]
Applying the product rule:
\[ 3(2xy + x^2\frac{dy}{dx}) - 3y^2\frac{dy}{dx} = 0 \]
Rearranging gives:
\[ 6xy + 3x^2\frac{dy}{dx} - 3y^2\frac{dy}{dx} = 0 \]
Thus,
\[ 6xy = (3y^2 - 3x^2)\frac{dy}{dx} \]
So,
\[ \frac{dy}{dx} = \frac{2xy}{y^2 - x^2} \quad \text{(2)} \]

**Step 3: Analyze the slopes at the points of intersection.**

From equations (1) and (2), we have:
\[ \frac{dy}{dx}_1 = \frac{x^2 - y^2}{2xy} \quad \text{and} \quad \frac{dy}{dx}_2 = \frac{2xy}{y^2 - x^2} \]

To check if the curves intersect at right angles, we need to verify if:
\[ \frac{dy}{dx}_1 \cdot \frac{dy}{dx}_2 = -1 \]

Substituting the expressions:
\[ \left(\frac{x^2 - y^2}{2xy}\right) \cdot \left(\frac{2xy}{y^2 - x^2}\right) = \frac{(x^2 - y^2)(2xy)}{2xy(y^2 - x^2)} = \frac{(x^2 - y^2)}{(y^2 - x^2)} = -1 \]

This confirms that the curves intersect at right angles.

### Conclusion:

The two curves intersect at right angles.