Problem 5 - Thomas Calculus

Question

Use the discriminant \(B^{2}-4 A C\) to decide whether the equations represent parabolas, ellipses, or hyperbolas. \(x^{2}+2 x y+y^{2}+2 x-y+2=0\)

Step-by-Step Solution

Answer

The equation represents a parabola since the discriminant is zero.

Step 1: Identify Equation Structure

The given equation can be written in the form of a general second-degree equation: \[ A x^2 + B xy + C y^2 + Dx + Ey + F = 0 \]Comparing it with the given equation \( x^2 + 2xy + y^2 + 2x - y + 2 = 0 \), we find:- \( A = 1 \)- \( B = 2 \)- \( C = 1 \)- \( D = 2 \)- \( E = -1 \)- \( F = 2 \)

Step 2: Calculate the Discriminant

The discriminant for a second-degree equation is given by:\[ \Delta = B^2 - 4AC \]Substitute the values of \( A = 1 \), \( B = 2 \), and \( C = 1 \) into the discriminant formula:\[ \Delta = 2^2 - 4 \times 1 \times 1 = 4 - 4 = 0 \]

Step 3: Interpret the Discriminant

The value of the discriminant helps classify the conic section:- If \( \Delta > 0 \), the conic is a hyperbola.- If \( \Delta = 0 \), the conic is a parabola.- If \( \Delta < 0 \), the conic is an ellipse.Since \( \Delta = 0 \) for this equation, the conic section represented is a **parabola**.

Key Concepts

Understanding the Discriminant

The discriminant is a crucial tool in determining the nature of conic sections. It's a part of the quadratic formula but is generalized for conic sections through the equation: \[ \Delta = B^2 - 4AC \] where \( A \), \( B \), and \( C \) come from the standard form of a second-degree equation: \[ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \].The discriminant tells us about the roots of the equation, which determines the type of conic section:

• If \( \Delta > 0 \), the conic is a hyperbola.
• If \( \Delta = 0 \), the conic is a parabola.
• If \( \Delta < 0 \), the conic is an ellipse.

In this exercise, once we calculated \( \Delta = 0 \), we understood that the equation represents a parabola, because the discriminant is exactly zero.

What is a Parabola?

A parabola is a symmetrical open plane curve that you might have seen in the path of a thrown ball. It's one of the conic sections, which include circles, ellipses, and hyperbolas. When graphed, a parabola has a vertex, which is its highest or lowest point.Parabolas can open upwards, downwards, left, or right, depending on the equation. In a coordinate system, a standard parabola that opens upwards has the equation \( y = ax^2 + bx + c \), where "a" determines how "wide" or "narrow" the parabola is. The discriminant helps identify when an equation without an explicit "xy" term simplifies to a parabola.In the context of this exercise, since \( \Delta = 0 \), the given equation represents a parabola. This aligns with the fact that no matter the orientation, every parabola's discriminant will be zero in a second-degree conic equation.

Exploring Second-Degree Equations

Second-degree equations, often called quadratic equations, form the basis for understanding conic sections. These equations generally look like: \[ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \].Each term provides information about the shape formed:

• \( Ax^2 \) and \( Cy^2 \) define the curve's inward (ellipse/parabola) or outward (hyperbola) arc.
• "Bxy" shows a degree of rotation; with non-zero "B", the graph can tilt.
• "D" and "E" terms handle shifting along x and y axes.
• "F" provides orientation and distance from the origin.

In practical applications, identifying these parts helps decipher the equation's real-world meaning. In our exercise, we broke down the coefficients, found \( \Delta = 0 \), and concluded that the equation was a parabola. Understanding how each component contributes to the whole enriches comprehension of the conic sections' properties.