Problem 55 - Precalculus: Mathematics for Calculus

Question

Complete the square to determine whether the graph of the equation is an ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an cllipse, 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 hyperbola centered at (4, 0). Vertices at (6, 0) and (2, 0); foci at (4 ± √5, 0).

Step 1: Expand and Simplify the Equation

Start by expanding and simplifying the given equation. The expression is initially given as: \[x^{2}+16=4(y^{2}+2x)\]First, distribute the 4 on the right-hand side:\[x^2 + 16 = 4y^2 + 8x\]Rearrange the terms to bring all terms to one side:\[x^2 - 8x + 16 - 4y^2 = 0\]

Step 2: Complete the Square for the x Terms

Focus on the quadratic terms involving \(x\): \[x^2 - 8x\]To complete the square, take half of the coefficient of \(x\) (which is -8), square it, and add it inside the completion:\[x^2 - 8x = (x - 4)^2 - 16\]The equation now becomes:\[(x - 4)^2 - 16 - 4y^2 + 16 = 0\]This simplifies further to:\[(x - 4)^2 = 4y^2\]

Step 3: Rewrite the Equation in Standard Form

Divide both sides by 4 to express the equation in terms of \(x\) and \(y\):\[\frac{(x - 4)^2}{4} = y^2\]This is the standard form of a hyperbola centered at \((h,k)\) with horizontal transverse symmetry.

Step 4: Identify the Conic Section

Since the equation is rewritten as:\[\frac{(x - 4)^2}{4} - \frac{y^2}{1} = 0\]This indicates a hyperbola centered at \((4, 0)\). Thus, the conic section is a hyperbola.

Step 5: Determine the Components of the Hyperbola

The center of the hyperbola is at \((4, 0)\). The transverse axis is horizontal due to \((x - h)^2\) being the positive term and has vertices at \((4 \pm 2, 0)\) which are \((6, 0)\) and \((2, 0)\).Calculate the foci using the formula \(c^2 = a^2 + b^2\). Here, \(a = 2\) and \(b = 1\):\[c = \sqrt{4 + 1} = \sqrt{5}\]The foci are at \((4 \pm \sqrt{5}, 0)\). The asymptotes are given by the slopes \(\pm \frac{b}{a}\). Hence the equations are:\[y = \pm \frac{1}{2}(x - 4)\]

Step 6: Sketch the Hyperbola

To sketch, plot the center \((4, 0)\), vertices \((6,0)\), \((2,0)\), and foci \((4 \pm \sqrt{5}, 0)\). Draw the transverse axis from the vertices and the asymptotes through the center, establishing the general shape of the hyperbola.

Key Concepts

Completing the Square

Completing the square is an essential algebraic technique often used to simplify quadratic expressions and play a crucial role in identifying conic sections. In this context, it involves adjusting a quadratic equation so it becomes a perfect square trinomial. This makes it easier to rewrite the equation in a form that reveals the type of conic section it represents.

• First, look at the quadratic terms in the equation, and gather terms accordingly.
• Calculate half of the coefficient of the linear term, square it, and add or subtract as needed.
• This allows you to express the quadratic component as a squared term plus or minus a constant.

In our problem:

• The expression for completing the square is: \(x^2 - 8x\).
• Half of \(-8\) is \(-4\), and squaring it gives \(16\).
• Add and subtract this value, leading to \((x-4)^2 - 16\).

Properly completing the square facilitates transforming the equation into a recognizable conic section form.

Hyperbola

A hyperbola is one of the four types of conic sections, each formed by the intersection of a plane with a double-napped cone. Unlike ellipses and parabolas, a hyperbola has two distinct pieces called lobes. It can be visualized as the set of all points where the difference of the distances to two fixed points (foci) is constant. Key characteristics of a hyperbola include:

• It has two branches that open either horizontally or vertically.
• The transverse axis runs through the foci and defines the direction in which the hyperbola opens.
• Its center is the midpoint of the line segment joining the foci.

Understanding its geometric properties is vital:

• Recognize the form of its standard equation: \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) for horizontally oriented hyperbolas.
• Analyze the asymptotes, which intersect at the hyperbola's center and guide its overall shape.

Graphing Conics

Graphing conic sections like hyperbolas involves several steps and considerations. This not only helps in visualization but also enhances understanding of their geometric properties. Here are the basic steps to graph a hyperbola:

• Identify the center: From the given equation, determine the coordinates \((h, k)\) of the center.
• Locate the vertices: Use the term associated with the axis of opening to find the distance to the vertices from the center.
• Determine the foci: Calculate using the relation \(c^2 = a^2 + b^2\).
• Draw asymptotes: These are highly important as they define the hyperbola's shape. The slope is given by \(\pm \frac{b}{a}\).

Start by plotting these elements on a graph:

• Draw the center point, vertices, and foci.
• Sketch the asymptotes as straight lines through the center.
• Draw the branches of the hyperbola, ensuring they approach the asymptotes.

This methodical approach ensures a precise graph of the hyperbola.

Equation of Hyperbola

The equation of a hyperbola reflects its geometric properties and provides valuable insights into its shape and orientation. The standard form equation tells us a lot about the hyperbola's attributes, including:

• Vertices: The terms \(a^2\) and \(b^2\) help locate the vertices along the transverse axis.
• Asymptotes: Key components dictated by the equation's form and slopes \(\pm \frac{b}{a}\).
• Foci: Calculated using the formula \(c^2 = a^2 + b^2\), critical for understanding the hyperbola's geometry.

Given the equation from the exercise: \(\frac{(x-4)^2}{4} = y^2\), we notice:

• The standard form is \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), indicating a horizontally opening hyperbola.
• Its center \((4,0)\) and vertices at \((6,0)\) and \((2,0)\).
• The foci are calculated to be \((4 \pm \sqrt{5}, 0)\).

Interpreting the equation of a hyperbola provides a strong foundation for sketching and understanding the conic's shape.