Problem 41 - College Algebra

Question

Sketch a graph of the hyperbola, labeling vertices and foci. \(64 x^{2}+128 x-9 y^{2}-72 y-656=0\)

Step-by-Step Solution

Answer

Vertices: (2, -4) and (-4, -4); Foci: (-1 + √73, -4) and (-1 - √73, -4).

Step 1: Rewrite in Standard Form

First, we need to rearrange the equation to bring terms with the same variable together: \[ 64x^2 + 128x - 9y^2 - 72y = 656 \]Next, we complete the square for the \(x\) and \(y\) terms separately. For the \(x\) terms: \[ 64(x^2 + 2x) = 64((x+1)^2 - 1) \]So, this becomes \[ 64(x+1)^2 - 64 \]. For the \(y\) terms:\[ -9(y^2 + 8y) = -9((y+4)^2 - 16) \]This becomes \[ -9(y+4)^2 + 144 \]. Thus, we rewrite the equation as:\[ 64(x+1)^2 - 9(y+4)^2 = 656 + 64 - 144 \] \[ 64(x+1)^2 - 9(y+4)^2 = 576 \]

Step 2: Divide by 576

To convert to the standard form of a hyperbola, divide the entire equation by 576:\[ \frac{64(x+1)^2}{576} - \frac{9(y+4)^2}{576} = \frac{576}{576} \]Simplifying each term results in:\[ \frac{(x+1)^2}{9} - \frac{(y+4)^2}{64} = 1 \] This is the standard form of a hyperbola centered at \((-1, -4)\) with horizontal transverse axis.

Step 3: Identify the Vertices

The standard form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \) is used to find the vertices. Here, \(a = 3\) and \(h = -1\), \(k = -4\). The vertices along the horizontal axis are calculated as:\((-1 + 3, -4)\) and \((-1 - 3, -4)\).So, the vertices are \((2, -4)\) and \((-4, -4)\).

Step 4: Find the Foci

The foci of the hyperbola are calculated using \(c^2 = a^2 + b^2\). With \(a^2 = 9\) and \(b^2 = 64\), we find:\[ c^2 = 9 + 64 = 73 \]\(c = \sqrt{73}\). The foci are at:\((-1 + \sqrt{73}, -4)\) and \((-1 - \sqrt{73}, -4)\).

Step 5: Sketch the Hyperbola

Using the center, vertices, and foci, sketch the hyperbola. Draw a rectangle centered at \((-1, -4)\) with width \(2a = 6\) and height \(2b = 16\). Draw the diagonals of the rectangle. The hyperbola approaches these diagonals as asymptotes. Plot the vertices at \((2, -4)\) and \((-4, -4)\) and the foci approximately at \((-1 + \sqrt{73}, -4)\) and \((-1 - \sqrt{73}, -4)\).

Key Concepts

Standard Form

The standard form of a hyperbola is an essential way of expressing its equation, which clearly shows the hyperbola's center and orientation. For hyperbolas, the standard form appears as:

• Horizontal transverse axis: \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\)
• Vertical transverse axis: \(\frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2} = 1\)

Here's a simple breakdown:- \(h\) and \(k\) are the coordinates of the center of the hyperbola.- \(a\) is the distance between the center and each vertex along the transverse axis.- \(b\) is related to the conjugate axis, but we won't focus on that here.For the given problem, the equation was rearranged and simplified to its standard form, \(\frac{(x+1)^2}{9} - \frac{(y+4)^2}{64} = 1\), which reveals that the center is at \((-1, -4)\) and the hyperbola has a horizontal transverse axis.

Vertices

Vertices are crucial points that lie on the hyperbola, specifying where the hyperbola first opens out from the center. These points are located a certain distance from the center, which we calculate using the value \(a\) from the standard form equation.
For a hyperbola in the horizontal form, the vertices are found at:

• \((h+a, k)\)
• \((h-a, k)\)

In our equation, \(a = 3\) (since \(a^2 = 9\)). Thus, the vertices of this hyperbola are located at \((2, -4)\) and \((-4, -4)\). These points highlight the horizontal spread across the transverse axis, confirming the hyperbola's orientation.

Foci

The foci of a hyperbola are points located on its transverse axis, which help define the overall shape and direction. They are crucial because the hyperbola's definition is based on distances to these foci.To find the foci, you use the relationship:- \(c^2 = a^2 + b^2\) where \(c\) is the focal distance.In this situation, \(a^2 = 9\) and \(b^2 = 64\), hence \(c^2 = 73\) and \(c = \sqrt{73}\).The foci are positioned symmetrically around the center at:

• \((h+c, k)\)
• \((h-c, k)\)

For this hyperbola, the foci end up at \((-1 + \sqrt{73}, -4)\) and \((-1 - \sqrt{73}, -4)\), reinforcing its horizontal orientation.

Transverse Axis

The transverse axis is a line that goes through both the center and the vertices of the hyperbola, defining the hyperbola's primary direction. It is integral to understanding hyperbolas because it distinguishes whether the hyperbola is horizontally or vertically oriented based on the positioning of the vertices.In this exercise, because the standard form retains a positive sign on the \((x-h)^2\) portion, we know the transverse axis is horizontal. That means the line extends left and right through the center \((-1, -4)\) and continues through the vertices at \((2, -4)\) and \((-4, -4)\). The transverse axis essentially serves as the backbone of the hyperbola, showing you how far and in what direction it opens out.