Problem 27 - Algebra and Trigonometry

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 with no graph, forming two intersecting lines.

Step 1: Simplify the Equation

Begin by expanding and rewriting the equation: \[ x^2 + 16 = 4(y^2 + 2x) \] Expanding the right-hand side gives: \[ x^2 + 16 = 4y^2 + 8x \]. Rearrange terms to:\[ x^2 - 8x + 16 = 4y^2 \].

Step 2: Complete the Square (X-terms)

To complete the square for the \( x \)-terms, begin with the expression:\[ x^2 - 8x \].Take half of the coefficient of \( x \), square it, and add inside the expression: \[ rac{-8}{2} = -4 \], and \((-4)^2 = 16\).Thus, the equation becomes:\[ (x - 4)^2 = 4y^2 \].

Step 3: Analyze the Equation Form

The equation now is:\[ (x - 4)^2 = 4y^2 \].This equation can be rewritten as:\[ rac{(x - 4)^2}{1} - rac{y^2}{rac{1}{4}} = 0 \].This is of the form \( A(x - h)^2 - B(y - k)^2 = 0 \), representing a degenerate conic, which is essentially two intersecting lines.

Key Concepts

Complete the Square

Completing the square is a handy algebraic technique used to simplify quadratic equations. It helps rewrite an expression from the standard quadratic form into a perfect square trinomial. This is especially useful when dealing with determining the nature of conic sections. In our exercise, we started with the equation:\[ x^2 + 16 = 4(y^2 + 2x) \]To complete the square for the \(x\)-terms, focus on \(x^2 - 8x\). To complete it:

• Take half of the coefficient of \(x\), which is \(-4\).
• Square this number to get \(16\).
• Add and subtract \(16\) within the equation to turn it into a square trinomial.

The expression now becomes \((x - 4)^2\). This method reveals more about the equation's form, aiding in identifying the conic section it represents.

Ellipse

An ellipse is one of the notable types of conic sections, characterized by its stretched circular shape. It's important to recognize whether an equation can represent an ellipse. Typically, an ellipse has:\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]In this format, \(h, k\) represent the center of the ellipse, and \(a, b\) are the lengths from the center to the vertices along the respective axes. However, in the given exercise, this equation aimed to determine whether we have an ellipse. Upon completing the square, the equation reflects a different conic type instead of an ellipse. Thus, understanding the criteria for an ellipse is crucial in such analyses, though it does not apply to this specific equation.

Parabola

Parabolas are a common form of conic sections, graphically represented as a U-shaped curve. The standard form of a parabola's equation is:\[ (x-h)^2 = 4p(y-k) \]or vice versa for those opening sideways. In this format:

• The vertex is \((h, k)\).
• The focus is at distance \(|p|\) away from the vertex.
• The directrix is a line \(|p|\) units away from the vertex, opposite the focus.

In the exercise, you may initially consider this form for the equation. However, after completing the square process, there is an absence of variable components that isolate parabola characteristics.Hence, determining the conic type moves past this consideration.

Hyperbola

Hyperbolas appear as two mirror-image curves, similar to open-ended parabolas. The typical form of a hyperbola can be:\[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]or the subtraction term can precede depending on the orientation. Key elements of a hyperbola include:

• The center \((h, k)\).
• The vertices at distance \(a\) from the center.
• The foci are \(c\) units from the center, determined via \(c^2 = a^2 + b^2\).
• Asymptotes are lines that the curves approach, given by slopes \(\pm\frac{b}{a}\).

In the exercise's solution, the rearranged completed-square equation \((x - 4)^2 = 4y^2\) suggests it as a degenerate conic, not a traditional hyperbola. However, understanding hyperbolas is essential, as they always start by analyzing if an equation resembles this form.