Exploring the Curve 2x2 3xy y2 3: No Stationary Points and Parallel Lines

Understanding the behavior of a curve is crucial in mathematics, especially when dealing with conic sections such as the equation 2x2 3xy y2 3. This article will delve into two aspects of this curve: the absence of stationary points and its representation as a pair of parallel straight lines.

1. Showing the Curve Has No Stationary Points

Stationary points, also known as critical points, are crucial in determining the local extrema of a function. To find these points, we differentiate the given equation implicitly with respect to x and set the derivative to zero.

Step 1: Implicit Differentiation

We start with the given equation:

(2x^2 3xy y^2 - 3 0)

Implicitly differentiating both sides with respect to x:

(frac{d}{dx}(2x^2) frac{d}{dx}(3xy) frac{d}{dx}(y^2) frac{d}{dx}(3))

This leads to:

(4x 3left(y xfrac{dy}{dx}right) 2yfrac{dy}{dx} 0)

Simplifying further:

(4x 3y 3xfrac{dy}{dx} 2yfrac{dy}{dx} 0)

Factoring out (frac{dy}{dx}):

((3x 2y)frac{dy}{dx} -4x - 3y)

Solving for (frac{dy}{dx}):

(frac{dy}{dx} frac{-4x - 3y}{3x 2y})

Step 2: Finding Stationary Points

Setting (frac{dy}{dx} 0):

(4x 3y 0)

Solving for y:

(y -frac{4}{3}x)

Substituting y -frac{4}{3}x back into the original equation:

(2x^2 3xleft(-frac{4}{3}xright) left(-frac{4}{3}xright)^2 3)

This simplifies to:

(2x^2 - 4x^2 frac{16}{9}x^2 3)

Further simplifying:

(left(2 - 4 frac{16}{9}right)x^2 3)

(left(frac{18}{9} - frac{36}{9} frac{16}{9}right)x^2 3)

(left(-frac{2}{9}right)x^2 3)

This implies:

(x^2 -frac{27}{2})

Since (x^2) cannot be negative, there are no real solutions for x. Therefore, the curve has no stationary points.

2. Showing that the Curve Represents a Pair of Parallel Straight Lines

Another key aspect of the curve is to determine if it represents a pair of parallel straight lines. We can check this by analyzing the quadratic form of the equation (2x^2 3xy y^2 - 3 0).

Step 1: Analyzing the Conic Section

The general form of a conic section is:

(Ax^2 Bxy Cy^2 Dx Ey F 0)

For the given equation:

(A 2), (B 3), (C 1), (D 0), (E 0), (F -3)

The discriminant (D) is given by:

(D B^2 - 4AC)

Substituting the values:

(D 9 - 4 cdot 2 cdot 1 9 - 8 1)

Since (D > 0), the conic section is either a hyperbola or a pair of intersecting lines. To determine if it is a pair of parallel lines, we need to check if the quadratic part can be factored into two linear factors that are multiples of each other.

Step 2: Using the Quadratic Formula

The given equation is:

(y^2 3xy 2x^2 - 3 0)

This is a quadratic in y. Using the quadratic formula (y frac{-b pm sqrt{b^2 - 4ac}}{2a}), we get:

(y frac{-3x pm sqrt{(3x)^2 - 4 cdot 2 cdot 1}}{2 cdot 1})

(y frac{-3x pm sqrt{9x^2 - 8x^2 12}}{2})

(y frac{-3x pm sqrt{x^2 12}}{2})

This results in two distinct lines, indicating that the curve represents a pair of parallel lines.

Conclusion

Through detailed analysis, we have shown that the curve (2x^2 3xy y^2 3) has no stationary points because substituting the condition leads to a contradiction, specifically a negative (x^2).

Furthermore, the curve represents a pair of parallel straight lines due to the positive discriminant and the fact that the quadratic part can be factored into two linear factors that are multiples of each other.