Problem 29 - Algebra and Trigonometry

Question

\(23-34\) 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 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. $$ 16 x^{2}-9 y^{2}-96 x+288=0 $$

Step-by-Step Solution

Answer

The equation represents a hyperbola centered at (3, 0) with vertices (6, 0), (0, 0), and foci (8, 0), (-2, 0).

Step 1: Move constant term to the right

First, we isolate the constant term by moving it to the opposite side of the equation.\[ 16x^2 - 9y^2 - 96x + 288 = 0 \] becomes \[ 16x^2 - 9y^2 - 96x = -288 \]

Step 2: Group x and y terms

Separate the equation into groups containing only the x-terms and the y-terms.\[ 16(x^2 - 6x) - 9y^2 = -288 \]

Step 3: Complete the square for x-terms

We complete the square for the x-terms inside the parenthesis. Take half the coefficient of x, square it, and add it inside the parenthesis. Remember to balance the equation by adding the same term outside the parenthesis, considering the coefficient '16'.Half of -6 is -3, and (-3)^2 is 9.\[ 16(x^2 - 6x + 9) - 16(9) - 9y^2 = -288 \]\[ 16(x-3)^2 - 9y^2 = -288 + 144 \]\[ 16(x-3)^2 - 9y^2 = -144 \]

Step 4: Rearrange like hyperbola standard form

Divide the entire equation by -144 to get it into a recognizable conic form.\[ \frac{16(x-3)^2}{144} - \frac{9y^2}{144} = 1 \]Simplify:\[ \frac{(x-3)^2}{9} - \frac{y^2}{16} = 1 \]

Step 5: Identify the conic section

The equation fits the standard form of a hyperbola: \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]. This confirms it is a hyperbola centered at (h, k). Here, the center is \((3, 0)\).

Step 6: Calculate vertices and foci of hyperbola

For a hyperbola of the form \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \], the vertices are \( (h \,\pm\, a, k) \), and the foci are \( (h \,\pm\, c, k) \) with \( c = \sqrt{a^2 + b^2} \).Since \( a^2 = 9 \Rightarrow a = 3 \) and \( b^2 = 16 \Rightarrow b = 4 \),\( c = \sqrt{9 + 16} = 5 \).Thus, vertices are \((3 \pm 3, 0) = (6, 0) \) and \( (0, 0) \),and foci are \( (3 \pm 5, 0) = (8, 0) \) and \( (-2, 0) \).

Step 7: Determine asymptotes of hyperbola

The asymptotes for this hyperbola are given by the equations:\( y = \pm \frac{b}{a}(x-h) + k = \pm \frac{4}{3}(x-3) \).This simplifies to \( y = \pm \frac{4}{3}(x-3) \).

Step 8: Sketch the hyperbola

Using the vertices, foci, center, and asymptotes, sketch the hyperbola:- Center at (3, 0).- Vertices at (6, 0) and (0, 0).- Foci at (8, 0) and (-2, 0).- Draw asymptotes through the center with slopes \( \pm \frac{4}{3} \).

Key Concepts

Completing the Square

Completing the square is a fundamental technique used in algebra to transform quadratic equations into a form that is easier to analyze. In this process, you manipulate a quadratic equation such that a binomial square is achieved. This method is particularly useful for identifying the properties of conic sections such as circles, ellipses, parabolas, and hyperbolas.
For the given problem, the equation includes both \( x \) and \( y \) terms, and our goal is to complete the square for the \( x \) components. This involves the following key steps:

• Group the \( x \) and \( y \) terms separately.
• Add and subtract the appropriate constant inside the \( x \) term group. This involves dividing the coefficient of \( x \) by two, squaring it, and adjusting the equation accordingly.
• Consider the influence of any coefficient outside the squared term, such as the factor of 16 in our problem.

By carrying out these steps, we streamline the equation into a form that reveals its nature as a hyperbola. This makes subsequent identification of features like vertices and asymptotes more manageable.

Hyperbola

A hyperbola is a type of conic section that is defined as the set of points, the difference of whose distances from two fixed foci remains constant. Unlike other conic sections, a hyperbola consists of two separate curves called branches.
The standard form of a hyperbola's equation can be expressed as \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\]This form signals the graph of a hyperbola with a horizontal transverse axis.

• Here, \(h\) and \(k\) represent the center of the hyperbola.
• \(a\) and \(b\) are the distances that relate to the vertices and asymptotes.

In practice, recognizing the specific features of the hyperbola in the enhanced equation allows us to delve deeper into its geometric properties, like finding its vertices, foci, and defining its asymptotes. The recognition of the hyperbola's structure is crucial for graphing and understanding its behavior.

Vertices and Foci

In hyperbolas, the vertices and foci are crucial characteristics for understanding the shape and positioning of the branches. These points determine how stretched or compressed the hyperbola appears.
For a hyperbola of the form \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1,\] the vertices are located at \((h \pm a, k)\), and the foci are positioned at \((h \pm c, k)\), where \(c\) is calculated using \(c = \sqrt{a^2 + b^2}\).

• The vertices are points where each branch of the hyperbola is nearest to the center.
• The foci are inside each branch and lie further away on the transverse axis than the vertices.

In the given problem, \(a^2 = 9\) and \(b^2 = 16\), giving \(a = 3\) and \(b = 4\). This results in vertices at \((6, 0)\) and \((0, 0)\), and foci calculated using \(c = 5\), placing them at \((8, 0)\) and \((-2, 0)\). Understanding these aspects helps in accurately sketching and interpreting the hyperbola.

Asymptotes

Asymptotes provide a framework for hyperbolas, guiding the direction of their branches. These lines show the slope that the hyperbola approaches but never quite reaches as it extends to infinity.
For the standard hyperbola form \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1,\]asymptotes are given by the equations:\[y = \pm \frac{b}{a}(x-h) + k.\]

• The slopes of the asymptotes, \(\pm \frac{b}{a}\), describe how steep or shallow the branches spread away from the center.
• Their intersection point is at the hyperbola's center, \((h, k)\).

In the case of the exercise, the asymptotes are calculated as \(y = \pm \frac{4}{3}(x-3)\), making them pass through \((3, 0)\) with slopes \(\pm \frac{4}{3}\). Incorporating asymptotes helps in sketching a more precise graph, showing the accurate direction and extent of the hyperbola's branches.