Problem 33 - Precalculus Mathematics for Calculus

Question

Complete the square to determine whether the equation represents an ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an ellipse, find the center, foci, vertices, and lengths of the major and minor axes. If it is a parabola, find the vertex, focus, and directrix. If it is a hyperbola, find the center, foci, vertices, and asymptotes. Then sketch the graph of the equation. If the equation has no graph, explain why. $$3 x^{2}+4 y^{2}-6 x-24 y+39=0$$

Step-by-Step Solution

Answer

The equation represents a degenerate conic, a single point at (1, 3). There is no typical graph.

Step 1: Group and rearrange terms

First, we rearrange the given equation so that all terms with \(x\) and \(y\) are grouped together. The equation is already grouped: \(3x^2 - 6x + 4y^2 - 24y + 39 = 0\).

Step 2: Move constant term to the right

Next, move the constant term to the other side of the equation: \(3x^2 - 6x + 4y^2 - 24y = -39\).

Step 3: Complete the square for x terms

For the \(x\) terms \(3x^2 - 6x\), factor out the 3: \(3(x^2 - 2x)\). To complete the square, take half of \(-2\), square it to get 1, and add/subtract inside the parentheses: \(3(x^2 - 2x + 1 - 1)\) or \(3((x-1)^2 - 1)\).

Step 4: Complete the square for y terms

For the \(y\) terms \(4y^2 - 24y\), factor out the 4: \(4(y^2 - 6y)\). Take half of \(-6\), square it to get 9, and add/subtract: \(4(y^2 - 6y + 9 - 9)\) or \(4((y-3)^2 - 9)\).

Step 5: Substitute back into the equation

The equation becomes \(3((x-1)^2 - 1) + 4((y-3)^2 - 9) = -39\). Simplify to \(3(x-1)^2 - 3 + 4(y-3)^2 - 36 = -39\) or \(3(x-1)^2 + 4(y-3)^2 = 0\).

Step 6: Analyze the conic section

The equation \(3(x-1)^2 + 4(y-3)^2 = 0\) indicates a degenerate conic because the left side only equals zero at the point \((x-1)^2 = 0\) and \((y-3)^2 = 0\), which only occurs when \(x=1\) and \(y=3\). This equation doesn't form a typical ellipse, parabola, or hyperbola.

Step 7: Conclusion

Since the expression only equals zero at a single point (1, 3), it is a degenerate conic and represents just that point.

Key Concepts

Completing the Square

"Completing the square" is a mathematical technique used to transform quadratic equations into a form that is much easier to work with. This method can be particularly useful when solving problems related to conic sections, as it allows us to rewrite equations so that we can more readily identify their properties. Here's a breakdown of the process:

• Identify the quadratic terms you need to complete first, such as terms involving \(x\) or \(y\).
• Factor out any common coefficients from these terms. For instance, in the equation \(3x^2 - 6x\), you would factor out a 3 to get \(3(x^2 - 2x)\).
• To complete the square, take half of the linear coefficient, square it, and both add and subtract this square term inside the parenthesis. For \(x^2 - 2x\), this would give you \(x^2 - 2x + 1 - 1\).
• Rewrite the expression as a perfect square such as \((x-1)^2\) minus what you had to subtract and factor back the main coefficient to obtain the final squared term.

This turns a possibly confusing quadratic expression into something more apparent, revealing more about the conic section it represents.

Conic Sections

Conic sections are the curves formed by the intersection of a plane and a cone. Depending on the angle and location of the intersection, different shapes such as ellipses, parabolas, and hyperbolas are formed.

• Ellipse: Formed when the plane cuts all the way through the cone; it looks like a stretched circle and has two foci.


• Parabola: Created when the plane is parallel to the edge of the cone; this U-shaped curve has a single focus and directrix.


• Hyperbola: Occurs when the plane intersects both halves of the double cone; it features two foci and two branches.

Conic sections can often be put into different standard forms using algebraic techniques such as completing the square. However, when these shapes shrink down to one point or a line, we get what is known as a degenerate conic. Degenerate conics like the one described in the original problem, where \(3(x-1)^2 + 4(y-3)^2 = 0\), indicate that instead of extending over an area, the shapes have reduced to a singular point (1,3).

Coordinate Geometry

Coordinate geometry, also called analytic geometry, is the study of geometric properties and relationships using algebraic calculus in a coordinate plane. This approach turns geometric situations into algebraic equations, which can be manipulated to extract useful information.When analyzing conic sections through coordinate geometry:

• You begin by representing shapes using equations. For example, circles use \(x^2 + y^2 = r^2\) for radius \(r\).
• The use of equations allows for transformations using techniques like completing the square to identify shapes, such as ellipses or parabolas.
• Coordinate geometry also provides tools for finding specific parts of the shape easily, such as the center, foci, or lengths of axes for ellipses, or vertices and directrix for parabolas.

Practically, this makes it easier to analyze curves and assess properties like symmetry or graph interactions by translating between the algebraic and geometric viewpoints. In the context of the problem, coordinate geometry helps confirm that \(3(x-1)^2 + 4(y-3)^2 = 0\) resonates only at the point \(1, 3\). This transition from algebra to geometry reveals the nature of the degenerate conic present.