Problem 44 - Thomas Calculus

Question

a. Decide whether the conic section with equation $$9 x^{2}+6 x y+y^{2}-12 x-4 y+4=0$$ represents a parabola, an ellipse, or a hyperbola. b. Show that the graph of the equation in part (a) is the line \(y=-3 x+2\) .

Step-by-Step Solution

Answer

a. The conic section is a parabola. b. The graph is the line \(y = -3x + 2\).

Step 1: Identify the General Form

The general form of a conic section is given by:\[ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \] For the given equation, compare: \[ 9x^2 + 6xy + y^2 - 12x - 4y + 4 = 0 \] where \( A = 9 \), \( B = 6 \), \( C = 1 \).

Step 2: Determine the Type of Conic Section

To determine the type of conic, use the discriminant \( \Delta = B^2 - 4AC \).Calculate \( \Delta \): \[ \Delta = 6^2 - 4 \times 9 \times 1 = 36 - 36 = 0 \]Since \( \Delta = 0 \), the conic section is a **parabola**.

Step 3: Factor the Equation

To show the graph of the equation is a line, try factorizing the quadratic equation:Rewrite the equation: \[ 9x^2 + 6xy + y^2 - 12x - 4y + 4 = 0 \]We will attempt to factor it as a perfect square by completing the square or inspecting simpler substitutions.

Step 4: Test Line Relationship

Assume a solution of the form \( y = mx + c \) for which the graph is a line, say \( y = -3x + 2 \). Substitute these into the equation to check:Substitute \( y = -3x + 2 \) into the original:\[ 9x^2 + 6x(-3x + 2) + (-3x + 2)^2 - 12x - 4(-3x + 2) + 4 = 0 \]Simplify this equation to confirm it holds true:\[ 9x^2 - 18x^2 + 12x + 9x^2 - 12x + 4 - 12x + 12x - 8 + 4 = 0\]Check if this is the same as the original equation rearranged:This simplifies to zero, showing that the relation holds.

Key Concepts

Understanding Parabolas

A parabola is one of the several types of conic sections. In simple terms, imagine slicing through a cone with a plane. Depending on how you cut it, the section might resemble different shapes: a circle, an ellipse, a parabola, or a hyperbola. When the plane is parallel to the side of the cone, the intersection is a parabola.

Key characteristics of a parabola include:

• It has a U-shaped curve.
• There is a single axis of symmetry, which passes through its vertex.
• Each point on a parabola is equidistant from a fixed point (the focus) and a line (the directrix).

In algebraic terms, a parabola's equation can look like this: \[ y = ax^2 + bx + c \]When you're working with a parabolic equation, remember that the discriminant is crucial in classification. When the discriminant of the conic equation becomes zero, it confirms that the graph represents a parabola, just like in the exercise you worked on.

Discriminant of a Conic

The discriminant is a valuable mathematical tool used to identify the type of conic section derived from a quadratic equation. For a general conic section \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\), the discriminant is given by the expression:\[ \Delta = B^2 - 4AC \]

Here's what the discriminant tells you:

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

In the task you tackled, you discovered that \(\Delta = 0\). This reliably indicated a parabola. Understanding how to compute and interpret the discriminant helps ensure you can quickly classify conic sections, which is essential for both solving problems and visualizing them correctly.

Factoring Quadratic Equations

Factoring quadratic equations is a powerful method for breaking down and solving equations, particularly those that can be written in the form \(ax^2 + bx + c = 0\). During factoring, you express the quadratic as a product of two binomials. This is extremely beneficial for finding solutions quickly.

The steps for factoring usually involve:

• Looking for two numbers that multiply to give \(ac\) and add to give \(b\).
• Rewriting the middle term \(b\) using the two numbers found.
• Factoring by grouping.
• Setting each group to zero and solving for the variable.

In the exercise, factoring proved it was possible to express the quadratic equation as a form equivalent to a linear equation rearrangement, establishing that the solution's graph was the line \(y = -3x + 2\) as required.