Problem 31 - Intermediate Algebra

Question

We have graphed various equations of the form \(A x^{2}+\) \(B y^{2}=C\), where \(C\) is a nonzero constant. Now graph each of the following. (a) \(x^{2}+y^{2}=0\) (b) \(2 x^{2}+3 y^{2}=0\) (c) \(x^{2}-y^{2}=0\) (d) \(4 y^{2}-x^{2}=0\)

Step-by-Step Solution

Answer

(a) & (b): Point at origin. (c): Lines \(y = x\) and \(y = -x\). (d): Lines \(y = \frac{x}{2}\) and \(y = -\frac{x}{2}\).

Step 1: Understanding the Problem Structure

Equations of the form \(Ax^2 + By^2 = C\) represent conic sections. The term \(C\) determines the nature of these conics. However, when \(C = 0\), special cases arise that we need to explore individually.

Step 2: Analyze the Equation \(x^2 + y^2 = 0\)

For the equation \(x^2 + y^2 = 0\), both terms are perfect squares and they add to zero. Since squares of real numbers are non-negative, the only solution is \(x = 0\) and \(y = 0\). Thus, the graph is a single point at the origin.

Step 3: Analyze the Equation \(2x^2 + 3y^2 = 0\)

Here, \(2x^2 + 3y^2 = 0\) implies \(2x^2\) and \(3y^2\) must both be zero to satisfy the equation. This leads to \(x = 0\) and \(y = 0\). The graph is a single point at the origin, similar to (a).

Step 4: Analyze the Equation \(x^2 - y^2 = 0\)

The equation \(x^2 - y^2 = 0\) can be factored as \((x - y)(x + y) = 0\). This implies two solutions: \(x = y\) and \(x = -y\). The graph is two lines intersecting at the origin, \(y = x\) and \(y = -x\).

Step 5: Analyze the Equation \(4y^2 - x^2 = 0\)

Similarly, \(4y^2 - x^2 = 0\) can be factored as \((2y - x)(2y + x) = 0\), giving solutions \(2y = x\) and \(2y = -x\), or \(y = \frac{x}{2}\) and \(y = -\frac{x}{2}\). These represent two lines through the origin.

Key Concepts

Equation Graphing

Graphing equations involves plotting the solutions of the equations on a coordinate plane. For conic sections, these equations often have the form \(Ax^2 + By^2 = C\). Here, \(A\) and \(B\) are coefficients, and \(C\) is a constant that influences the shape of the graphed conic. However, when \(C = 0\), the equations take on special characteristics.

To begin graphing, identify the type of conic represented by the equation. When \(C = 0\), observe how terms balance each other to solve for potential solutions.

• For terms like \(x^2 + y^2 = 0\), both terms must equate to zero simultaneously. The graph is a single point: the origin.
• Equations like \(x^2 - y^2 = 0\) factor into linear components, creating lines as solutions.

Remember, the key is understanding the relationships between terms and how they contribute to the graph's shape.

Special Cases of Conics

Special cases of conic sections occur when the equation \(Ax^2 + By^2 = C\) has \(C=0\). In these cases, the expected shape of the conic alters, leading to different outcomes:

- **Point Conic**: This occurs when an equation such as \(x^2 + y^2 = 0\) produces no real space other than a point. The point is located at the origin \((0,0)\) because any non-zero value for either variable results in an impossible condition: two positive squares adding up to zero.
- **Line Conics**: Equations like \(x^2 - y^2 = 0\) or \(4y^2 - x^2 = 0\) may break down into factors that suggest lines. Here, the solution is not a curve but one or two intersecting lines. This transformation happens because the negative and positive coefficients create a balance that resolves into linear factors.

These cases are essential in understanding the diversity of conic sections and how they behave under different conditions.

Linear Equations in Graphing

In specific scenarios of conic equations where \(C=0\), the equation can transform into linear components, leading to a graphical representation involving lines. Unlike curved conics, these cases simplify the process by employing basic linear graphing methods.

For instance, the equation \(x^2 - y^2 = 0\) can be factored as \((x - y)(x + y) = 0\). Each factor represents a line: \(y = x\) and \(y = -x\), which intersect through the origin. Such graphing results from equations that can be split into linear terms.

• Identify possible factorizations of the equation.
• Consider each factor as a potential line on the graph.
• Verify intersection points between these lines, often the origin for \(C=0\) equations.

By understanding these procedures, graphing becomes manageable, turning initially complex-looking equations into straightforward linear representations.