Problem 24 - College Algebra

Question

Write the equation for the hyperbola in standard form if it is not already, and identify the vertices and foci, and write equations of asymptotes. \(-9 x^{2}+72 x+16 y^{2}+16 y+4=0\)

Step-by-Step Solution

Answer

The hyperbola's equation is \(\frac{(x-4)^2}{16} - \frac{(y+\frac{1}{2})^2}{9} = 1\), with vertices \((0, -\frac{1}{2}), (8, -\frac{1}{2})\), foci \((-1, -\frac{1}{2}), (9, -\frac{1}{2})\), and asymptotes \(y=\frac{3}{4}x-3.25\) and \(y=-\frac{3}{4}x+2.25\).

Step 1: Rearrange the Equation

Reorganize the given equation by grouping the x-terms and y-terms together:\[-9x^2 + 72x + 16y^2 + 16y + 4 = 0\] becomes \[-9(x^2 - 8x) + 16(y^2 + y) = -4.\]

Step 2: Complete the Squares

Complete the square for both the x-terms and the y-terms.For the x-terms:\[x^2 - 8x\] becomes \[(x - 4)^2 - 16.\] Thus, \[-9((x - 4)^2 - 16) = -9(x - 4)^2 + 144.\]For the y-terms:\[y^2 + y\] becomes \[(y + \frac{1}{2})^2 - \frac{1}{4}.\] Thus, \[16((y + \frac{1}{2})^2 - \frac{1}{4}) = 16(y + \frac{1}{2})^2 - 4.\]

Step 3: Substitute and Simplify

Substitute the completed squares back into the equation:\[-9(x - 4)^2 + 144 + 16(y + \frac{1}{2})^2 - 4 = -4.\]Combine the constants:\[ -9(x - 4)^2 + 16(y + \frac{1}{2})^2 = -144.\]

Step 4: Divide by the Constant

To express the equation in standard form, divide through by -144:\[\frac{(x - 4)^2}{16} - \frac{(y + \frac{1}{2})^2}{9} = 1.\]This is the equation of a hyperbola in standard form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\).

Step 5: Identify Vertices and Foci

The equation \(\frac{(x - 4)^2}{16} - \frac{(y + \frac{1}{2})^2}{9} = 1\) implies vertices at \((h\pm a, k)\) and foci at \((h\pm c, k)\).Vertices: \((4 \pm 4, -\frac{1}{2})\) or \((0, -\frac{1}{2})\) and \((8, -\frac{1}{2})\).Calculate c: \[c^2 = a^2 + b^2 = 16 + 9 = 25\] so \(c=5\).Foci: \((4 \pm 5, -\frac{1}{2})\) or \((-1, -\frac{1}{2})\) and \((9, -\frac{1}{2})\).

Step 6: Determine Equations of Asymptotes

The asymptote equations for hyperbolas in the form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) are given by:\[y - k = \pm \frac{b}{a} (x - h)\].Substituting our values:\[y + \frac{1}{2} = \pm \frac{3}{4} (x - 4)\].Thus, the equations are \(y = \frac{3}{4}x - 3.25\) and \(y = -\frac{3}{4}x + 2.25\).

Key Concepts

Standard Form Equation

Hyperbolas are fascinating shapes and understanding their standard form equation is an essential aspect of mastering this mathematical concept. A hyperbola in its standard form can typically be expressed as:
\[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\]Here,

• \((h, k)\) represents the center of the hyperbola,
• \(a\) is the distance from the center to each vertex along the x-axis,
• \(b\) is the distance from the center to the vertices along the y-axis.

To simplify a given equation into this standard form, we typically:

• Rearrange terms to group \'x\' and \'y\' together.
• Complete the square for both x and y terms.
• Divide through by constants to get "1" on the right side.

For instance, the equation \[-9 x^{2}+72 x+16 y^{2}+16 y+4=0\] is transformed through these steps into \[\frac{(x-4)^2}{16} - \frac{(y+\frac{1}{2})^2}{9} = 1.\]This step is critical in not only understanding the properties of the hyperbola but also in identifying its components.

Vertices and Foci

To truly grasp the concept of a hyperbola, it's crucial to pinpoint its vertices and foci. These points help define the hyperbola's distinct shape.

• **Vertices**: Under the standard form equation \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), the vertices are located at \((h \pm a, k)\). Using our example, we find the vertices at \((0, -\frac{1}{2})\) and \((8, -\frac{1}{2})\) by adding and subtracting the value of \(a = 4\) from the x-coordinate of the center.
• **Foci**: The foci are crucial points inside each branch of the hyperbola, found by calculating \(c\) where \(c^2 = a^2 + b^2\). For our hyperbola, \(c = 5\), leading to foci at \((-1, -\frac{1}{2})\) and \((9, -\frac{1}{2})\).

The positioning of vertices and foci helps in visualizing the hyperbola's span and its opening direction. Accurate calculation of these points is essential for sketching a precise graph.

Asymptotes

Asymptotes are straight lines that closely approach the hyperbola but never touch it, defining its direction as it extends. Understanding the asymptote equations is pivotal in graphing the hyperbola accurately. For the standard hyperbola equation \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), the equations of asymptotes are:

\[y - k = \pm \frac{b}{a} (x - h).\]By substituting the respective values of our given hyperbola:

• \(b = 3\) and \(a = 4\) provide the slope \(\frac{3}{4}\).
• The center point \(h = 4\) and \(k = -\frac{1}{2}\) are used to find the intercepts.

The resulting equations are \(y = \frac{3}{4}x - 3.25\) and \(y = -\frac{3}{4}x + 2.25\). These asymptotes are ideal guides for extending the branches of a hyperbola, offering boundaries within which the curve fits. Understanding the characteristics of asymptotes allows for a more dynamic and thorough visualization of hyperbolic graphs.