Problem 27 - Precalculus: Mathematical 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. $$x^{2}+16=4\left(y^{2}+2 x\right)$$

Step-by-Step Solution

Answer

The equation represents a degenerate conic, specifically intersecting lines.

Step 1: Rearrange the Equation

Start by rewriting the given equation as follows:\[x^2 + 16 = 4(y^2 + 2x)\]Expanding the right side, we have:\[x^2 + 16 = 4y^2 + 8x\]Next, rearrange all terms to one side of the equation:\[x^2 - 8x - 4y^2 + 16 = 0\]

Step 2: Complete the Square for x

To complete the square for the terms involving \(x\), take the expression:\[x^2 - 8x\]To complete the square, take half of the coefficient of \(x\) which is \(8/2 = 4\), and square it to get \(16\). Add and subtract 16:\[(x^2 - 8x + 16) - 16 = (x-4)^2 - 16\]

Step 3: Substitute and Simplify

Substitute the completed square form for \(x\) into the equation and rearrange:\[(x-4)^2 - 16 - 4y^2 + 16 = 0\]Simplify to get:\[(x-4)^2 - 4y^2 = 0\]

Step 4: Identify the Conic Type

The equation \((x-4)^2 - 4y^2 = 0\) can be rewritten as:\[(x-4)^2 = 4y^2\]This can further be rearranged to give:\[\frac{(x-4)^2}{4} - \frac{y^2}{1} = 0\]Since there are no distinct non-zero positive or negative terms for \(x\) and \(y\), this represents a degenerate conic instead of a hyperbola, ellipse, or parabola.

Step 5: Conclusion: Nature of the Equation

Given the equation represents a degenerate conic, it does not have a standard graph associated with conics like a parabola, ellipse, or hyperbola. Degenerate conics are typically a point, a line, or a pair of intersecting lines. Here, the equation simplifies to intersecting lines at \(x - 4 = \pm 2y\).

Key Concepts

Completing the Square

Completing the square is a handy tool used in algebra to transform a quadratic equation into a perfect square trinomial. This technique is essential, especially when dealing with conic sections, as it helps to identify the type of conic and to rewrite the equation in a more recognizable form.
To complete the square for a quadratic expression, follow these simple steps:

• Identify the coefficient of the linear term, usually in the form of ax² + bx + c.
• Take half of the coefficient of the x-term, then square it.
• Add and subtract this squared number inside the equation to form a complete square.
• Factor the resulting perfect square trinomial.

The process aims to create an expression like \((x-h)^2\) so that it can be directly assessed in terms of distance from a particular value. In the original exercise, the equation \(x^2 - 8x\) got transformed into \((x-4)^2 - 16\) by completing the square. This facilitated the further analysis of the equation into its simplest form.

Degenerate Conics

Degenerate conics occur when a conic section does not form a typical figure like an ellipse, parabola, or hyperbola. Typically, a degenerate conic appears as points, a single line, or a pair of lines. This situation arises from specific configurations of an equation and its underlying geometric structure.
In the exercise, we encountered a degenerate hyperbola. When we simplified the equation \((x-4)^2 - 4y^2 = 0\), it resulted in intersecting lines. Degenerate conics can sometimes be confusing since their graphical representations differ from the standard curves of conics. However, recognizing them is crucial in conic section analysis. In this case, knowing that \((x-4)^2 = 4y^2\) does not meet typical values for an ellipse, parabola, or hyperbola leads to identifying the simplified form of a degenerate hyperbola.

Hyperbola Identification

A hyperbola is one of the classic conic sections formed when a plane intersects both halves of a double cone. Hyperbolas have two disconnected curves known as branches, and they are defined by their centers, vertices, foci, and asymptotes. Generally represented by equations like \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), hyperbolas have distinctive features that help differentiate them from other conics.
In our original problem, the transformed equation initially suggested a hyperbola because of its form \((x-4)^2 - 4y^2 = 0\). However, further simplification showed a contrast, reducing it to \[(x-4)^2 = 4y^2\], indicating that the terms didn't define a valid hyperbola. Instead, it became evident that the equation represented a degenerate conic. Recognizing the distinction between a genuine hyperbola and its degenerate forms requires observing the specific structures apparent within the final simplified equation.