Problem 24 - College Algebra with Corequisite Support

Question

For the following exercises, 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

Standard form: \(\frac{(y+\frac{1}{2})^{2}}{9} - \frac{(x-4)^{2}}{16} = 1\). Vertices: \((4, \frac{5}{2})\), \((4, -\frac{7}{2})\). Foci: \((4, \frac{9}{2})\), \((4, -\frac{11}{2})\). Asymptotes: \(y + \frac{1}{2} = ±\frac{3}{4}(x - 4)\).

Step 1: Rearrange the Equation

Start by rearranging the terms of the given hyperbola equation \(-9x^{2} + 72x + 16y^{2} + 16y + 4 = 0\) to group the x and y terms together: \(-9x^{2} + 72x + 16y^{2} + 16y = -4\).

Step 2: Complete the Square

To make the equation into a standard form, we need to complete the square for both the x and y terms. First, factor out the coefficients of \(x^2\) and \(y^2\). For x, factor \(-9\) and complete the square: \(-9(x^{2} - 8x + 16) = -9((x-4)^{2})\). For y, factor \(16\) and complete the square: \(16(y^{2} + y + \frac{1}{4}) = 16((y+\frac{1}{2})^{2})\).

Step 3: Construct the Standard Form

Using the completed squares, the equation results in \(-9(x-4)^{2} + 16(y+\frac{1}{2})^{2} = -4 + 144 - 4\). Simplify and rearrange to achieve the standard form: \(\frac{(y+\frac{1}{2})^{2}}{9} - \frac{(x-4)^{2}}{16} = 1\). This represents a hyperbola centered at \((4, -\frac{1}{2})\) with it's transverse axis along the y-axis.

Step 4: Identify the Vertices

For a hyperbola given by \(\frac{(y-k)^{2}}{a^{2}} - \frac{(x-h)^{2}}{b^{2}} = 1\), the vertices are at \((h, k±a)\). Here, \(a^{2} = 9\), so \(a = 3\). The vertices are \((4, -\frac{1}{2} ± 3)\) i.e., \((4, \frac{5}{2})\) and \((4, -\frac{7}{2})\).

Step 5: Determine the Foci

The foci of a hyperbola are \((h, k±c)\) where \(c\) is found using \(c^{2} = a^{2} + b^{2}\). With \(b^2 = 16\), compute \(c\): \(c = \sqrt{9 + 16} = \sqrt{25} = 5\). Therefore, the foci are \((4, -\frac{1}{2} ± 5)\), thus \((4, \frac{9}{2})\) and \((4, -\frac{11}{2})\).

Step 6: Write the Asymptotes

For a hyperbola oriented vertically like \(\frac{(y-k)^{2}}{a^{2}} - \frac{(x-h)^{2}}{b^{2}} = 1\), the asymptotes have equations \(y - k = ±\frac{a}{b}(x - h)\). Substituting in the known values, \(y + \frac{1}{2} = ±\frac{3}{4}(x - 4)\) provides the asymptotes.

Key Concepts

Standard Form of Hyperbola

When working with hyperbolas, it is essential to first express its equation in the standard form. The standard form for a hyperbola can be expressed in two different ways depending on its orientation:

• If the transverse axis is horizontal: \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \)
• If the transverse axis is vertical: \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \)

Here, \(h\) and \(k\) represent the coordinates of the center of the hyperbola, while \(a\) and \(b\) are constants that determine the shape of the hyperbola.
By completing the square for grouped \(x\) and \(y\) terms, we can convert the general form of a hyperbola equation into its standard form. This transformation reveals the center of the hyperbola and identifies which axis is transverse.

Vertices of Hyperbola

In a hyperbola, the vertices are critical points that lie on the transverse axis. For the standard form of a hyperbola, identifying vertices depends on whether the hyperbola is oriented horizontally or vertically.

• If vertical: The vertices come from \((h, k \pm a)\)
• If horizontal: The vertices are located at \((h \pm a, k)\)

The value \(a\) is key here as it determines the distance from the center to each vertex. \(a\) is extracted from the term under the square of the variable linked to the transverse axis. In our example, the hyperbola centered at \((4, -\frac{1}{2})\) has its transverse axis along the y-axis, giving us vertices at \((4, \frac{5}{2})\) and \((4, -\frac{7}{2})\).

Foci of Hyperbola

The foci (or focus points) of a hyperbola are points located along the transverse axis, further from the center than the vertices. The distance of these points from the center is determined by another parameter, \(c\), where \(c > a\).
To find \(c\), we use the equation \(c^2 = a^2 + b^2\). This tells us how far along the transverse axis the foci are positioned.

• For vertically oriented hyperbolas: The foci are at \((h, k \pm c)\)
• For horizontally oriented: The foci appear at \((h \pm c, k)\)

From the given example, with center \((4, -\frac{1}{2})\), and using \(a = 3\) and \(b = 4\), we find \(c = 5\). Thus, the foci are located at \((4, \frac{9}{2})\) and \((4, -\frac{11}{2})\).

Equations of Asymptotes

A hyperbola's asymptotes offer a guide to its overall shape, slanting towards but never touching these lines as they extend infinitely. The asymptotes reflect the slopes formed between the center and the vertices in an extended plane.

• For hyperbolas with vertical transverse axis, the equations are given by: \( y - k = \pm \frac{a}{b}(x - h) \)
• If the transverse axis is horizontal: \( y - k = \pm \frac{b}{a}(x - h) \)

In the exercised example, with the hyperbola oriented vertically, it's shown that the equations of the asymptotes are \(y + \frac{1}{2} = \pm\frac{3}{4}(x - 4)\). This reflects the slope determined by \(\frac{a}{b}\), hinting at the relative steepness as the hyperbola opens.