Problem 15 - 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. \(\frac{(x-1)^{2}}{9}-\frac{(y-2)^{2}}{16}=1\)

Step-by-Step Solution

Answer

Vertices: (4, 2), (-2, 2); Foci: (6, 2), (-4, 2); Asymptotes: \(y = 2 \pm \frac{4}{3}(x-1)\).

Step 1: Identify the Standard Form

The given equation \(\frac{(x-1)^2}{9} - \frac{(y-2)^2}{16} = 1\) is already in the standard form of a hyperbola, which is \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\). Here, \(h = 1\), \(k = 2\), \(a^2 = 9\), and \(b^2 = 16\).

Step 2: Determine the Vertices

For a hyperbola in the form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), the vertices are located \(a\) units left and right from the center \((h, k)\). Since \(a = 3\), the vertices are at \((h \pm a, k)\), which gives the vertices \((1+3, 2) = (4, 2)\) and \((1-3, 2) = (-2, 2)\).

Step 3: Find the Foci

For hyperbolas, the distance \(c\) from the center to each focus is given by \(c = \sqrt{a^2 + b^2}\). Here, \(c = \sqrt{9 + 16} = \sqrt{25} = 5\). The foci are at \((h \pm c, k)\), so the foci are \((1+5, 2) = (6, 2)\) and \((1-5, 2) = (-4, 2)\).

Step 4: Find the Equations of the Asymptotes

For the given form, the equations of the asymptotes are \(y = k \pm \frac{b}{a}(x-h)\). Substituting the values, we get \(y = 2 \pm \frac{4}{3}(x-1)\). So, the equations are \(y = 2 + \frac{4}{3}(x-1)\) and \(y = 2 - \frac{4}{3}(x-1)\).

Key Concepts

Vertices of Hyperbola

Hyperbolas are fascinating geometric shapes with unique properties. One important feature of a hyperbola is its vertices. These are the points where the hyperbola is closest to its center. Consider the given hyperbola equation in standard form: \(\frac{(x-1)^2}{9} - \frac{(y-2)^2}{16} = 1\). This equation reveals the center of the hyperbola at \((h, k) = (1,2)\).
In this form, the vertices are located "\(a\)" units left and right from the center, which corresponds to the values \(a^2 = 9\), giving \(a = 3\). Thus, the vertices are at:

• \((1+3, 2) = (4, 2)\)
• \((1-3, 2) = (-2, 2)\)

Understanding vertices helps in visualizing the width of the hyperbola along the x-axis, highlighting how these points mark significant transitions in the curve's direction.

Foci of Hyperbola

Focusing on the foci of a hyperbola helps illuminate some of its intriguing properties. The foci are critical points that significantly influence the hyperbola's shape and definition.
For the hyperbola \(\frac{(x-1)^2}{9} - \frac{(y-2)^2}{16} = 1\), remember the rule: the distance from the center to each focus is given by \(c = \sqrt{a^2 + b^2}\).
Calculating, we find \(c = \sqrt{9 + 16} = 5\).
The foci are then positioned "\(c\)" units to the left and right of the center, specifically at:

• \((1+5, 2) = (6, 2)\)
• \((1-5, 2) = (-4, 2)\)

These foci are essential for understanding how hyperbolas maintain their geometric integrity. It is as if two invisible strings, each connected to a focus, guide the formation of the hyperbola's branches.

Asymptotes of Hyperbola

The asymptotes of a hyperbola play a crucial role in guiding its overall shape and direction. These lines are the paths that the hyperbola's branches approach yet never touch as they extend towards infinity.
For the hyperbola given by \(\frac{(x-1)^2}{9} - \frac{(y-2)^2}{16} = 1\), the asymptotes can be calculated using its center \((1, 2)\) and the formula for the slope: \(\frac{b}{a}\), where \(b = 4\) and \(a = 3\).
The equations of the asymptotes are:

• \(y = 2 + \frac{4}{3}(x-1)\)
• \(y = 2 - \frac{4}{3}(x-1)\)

These linear equations define the asymptotic paths that help predict the trajectory of each hyperbola branch. Asymptotes are particularly useful when sketching a hyperbola to ensure that its shape is accurately represented and in understanding the dynamic behavior of the hyperbola as \(x\) moves towards infinity.