Problem 34 - Algebra and Trigonometry

Question

use the center, vertices, and asymptotes to graph each hyperbola. Locate the foci and find the equations of the asymptotes. $$ \frac{(x+2)^{2}}{9}-\frac{(y-1)^{2}}{25}=1 $$

Step-by-Step Solution

Answer

The center of the hyperbola is at (-2, 1), the vertices are at (-5, 1) and (1, 1), and the foci are approximately at (-7.83, 1) and (3.83, 1). The equations of the asymptotes are \(y= \frac{5}{3}x+4\) and \(y=-\frac{5}{3}x-2\).

Step 1: Identify the Center, a, and b

From the given equation \(\frac{(x+2)^2}{9} - \frac{(y-1)^2}{25} = 1\), the center of the hyperbola (h, k) can be identified as (-2, 1). The values a and b are given by the square roots of the denominators of the fractions, so a=3 and b=5.

Step 2: Plot the Center and Vertices

Plot the center point (-2, 1) on the graph. Because this is a horizontal hyperbola (the x term is positive), the vertices are a units to the left and right of the center. So, plot the points (-2±3, 1), which are (-5, 1) and (1, 1).

Step 3: Sketch the Asymptotes

The asymptotes of the hyperbola are the lines that pass through the center with slope ±b/a. In this case, the slopes are ±5/3. Draw the lines y=1±5/3*(x+2), remembering that these are not part of the hyperbola but help guide in sketching the hyperbola.

Step 4: Locate the Foci

The foci of a hyperbola are c units from the center, where \(c=\sqrt{a^2+b^2}\). Calculate \(c=\sqrt{9+25}= \sqrt{34}\), approximately 5.83. The foci will be (-2±5.83, 1), or approximately (-7.83, 1) and (3.83, 1).

Step 5: Find the Equations of the Asymptotes

The equation of the asymptotes for a horizontal hyperbola is given by \(y=k \pm \frac{b}{a}(x-h)\). Substituting the given values, the equations of the asymptotes are \(y=1 \pm \frac{5}{3}(x+2)\), which simplifies to \(y= \frac{5}{3}x+4\) and \(y=-\frac{5}{3}x-2\).

Key Concepts

Asymptotes of a Hyperbola

Asymptotes are imaginary lines that a hyperbola approaches but never touches, guiding the shape of the hyperbola's arms as they extend to infinity.
When graphing a hyperbola, the asymptotes form an 'X' whose center point is the center of the hyperbola. The slopes of these lines depend on the coefficients of the hyperbolic equation.

In our example, for the equation \(\frac{(x+2)^{2}}{9}-\frac{(y-1)^{2}}{25}=1\), the slopes of the asymptotes are given by \(\pm\frac{b}{a}\) where 'a' and 'b' are the square roots of the denominators from the hyperbola's equation. The '±' indicates that there are two asymptotes, one with a positive slope and one with a negative slope, intersecting at the hyperbola's center.

Understanding where to position the asymptotes is essential to accurately draw the hyperbola. Remember, asymptotes don't intersect the hyperbola; they act like boundary lines dictating where the curve exists.

Locating Foci of a Hyperbola

The foci (singular: focus) of a hyperbola are two fixed points located inside each curving branch of the hyperbola. To locate the foci, you need to find the distance from the center to each focus, which is the value 'c'.

In mathematical terms, you calculate 'c' using the equation \(c=\sqrt{a^2+b^2}\), where 'a' is the distance from the center to the vertices, and 'b' is the distance associated with the conjugate axis.
For our example, \(c=\sqrt{9+25}=\sqrt{34}\), which is roughly 5.83. This tells us the foci, (-2±5.83, 1), lie approximately 5.83 units away from the center at (-2, 1) along the horizontal axis, because this hyperbola is oriented horizontally.

Visually locating the foci on a graph helps to ensure the shape of the hyperbola is accurate, thus becoming a crucial element of the graphing process.

Finding Equations of Asymptotes

To find the equations of the asymptotes for a hyperbola, you use the center (h, k) along with 'a' and 'b'.
The generic form of an asymptote’s equation for a horizontal hyperbola is \(y=k \pm \frac{b}{a}(x-h)\).

Inserting the values from our hyperbola, with the center at (-2, 1), 'a' being 3, and 'b' being 5, we have \(y=1 \pm \frac{5}{3}(x+2)\). This simplifies to the two linear equations \(y= \frac{5}{3}x+4\) and \(y=-\frac{5}{3}x-2\), representing the asymptotes of our hyperbola.

These equations outline the slopes and positions of the lines that the hyperbola will never cross but will infinitely approach. Recognizing and calculating the equations of asymptotes is vital for graphing hyperbolas, as it sets the framework for the hyperbola’s branches.