Problem 27 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the center, vertices, foci, and asymptotes of the hyperbola, and sketch its graph using the asymptotes as an aid. Use graphing utility to verify your graph \(\frac{(x-3)^{2}}{9}-\frac{(y-1)^{2}}{1}=1\)

Step-by-Step Solution

Answer

The center is at (3,1), vertices are at (6,1) and (0,1), foci are at \((3±\sqrt{10}, 1)\), and the asymptotes are \(y = 1 ± 3(x - 3)\).

Step 1 Identify the center

The center, (h,k), is obtained from the denominators of the hyperbola's equation. Thus, the center is at (3,1).

Step 2 Compute a, b and c

a, b, and c are found by examining the denominators and reading off their square roots. a = √9 = 3, b = √1 = 1. Next we compute c, where \(c^2 = a^2 + b^2\), so \(c = \sqrt{3^2 + 1^2} = \sqrt{10}\).

Step 3 Find vertices and foci

The vertices of the hyperbola are \((h±a, k)\), taking the positive root of a, we get the vertices at (6,1) and (0,1). The foci are \((h±c, k)\), taking the positive root of c, we get the foci at \((3±\sqrt{10}, 1)\).

Step 4 Discover the asymptotes

The equations of the asymptotes are \(y = k ± (a/b)(x - h)\). Substituting the values we get, the equations of the asymptotes are \(y = 1 ± 3(x - 3)\).

Step 5 Sketch the graph

Using all the information above, sketch the graph by plotting the center, vertices, and foci, and drawing the asymptotes as dashed lines. The hyperbola approaches these lines as x gets very large or very small.

Key Concepts

Center of Hyperbola

The center of a hyperbola is a crucial point that serves as the middle of its symmetrical paths. For a hyperbola given by the equation \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), the center can be found at the coordinates (h, k).
These coordinates are determined by identifying the values subtracted from x and y.
In the given exercise of \( \frac{(x-3)^2}{9} - \frac{(y-1)^2}{1} = 1 \), we locate the center at (3, 1).
This point is pivotal in positioning the hyperbola on a graph as it acts as the origin for further calculations.
Center points often aid in determining the layout and symmetry of the hyperbola as they are key reference points. Remember: the center is not where the hyperbola crosses the axes; instead, it lies equidistant from each branch.

Vertices of Hyperbola

Vertices are points at which each arm of the hyperbola is closest to one another. For a horizontal hyperbola equation like the one we are analyzing, \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), the vertices are located at \((h \pm a, k)\).
Here, \(a\) is derived from the square root of the numerator beneath \(x^2\) in the equation.
For our example, \(a = \sqrt{9} = 3\), leading to vertices at (6, 1) and (0, 1).
These points mark the start of each arm of the hyperbola, and are essential for drawing an accurate graph.
Plotting these vertices helps us structure the main shape of the hyperbola's curve, as they stretch out along the x-axis in this case.

Foci of Hyperbola

The foci of a hyperbola are two critical points located along the axis of symmetry that define its curvature. For an equation like \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), the foci are at \((h \pm c, k)\).
Before finding the foci, we calculate \(c\) using the equation \(c^2 = a^2 + b^2\).
In our exercise, this results in \(c = \sqrt{9 + 1} = \sqrt{10}\).
Hence, the foci are at \((3 \pm \sqrt{10}, 1)\).
These points are not visible on the graph but importantly contribute to forming the hyperbola's paired curves.
The foci guide the hyperbola's arms toward themselves, providing insight into the path the curves naturally follow around the central axes.

Asymptotes of Hyperbola

Asymptotes are lines that frame the hyperbola, illustrating the direction in which its arms approach but never actually touch. For the hyperbola \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), the asymptote equations are formulated as \(y = k \pm \frac{a}{b}(x - h)\).
These lines serve as invisible boundaries guiding the hyperbola's curvature.

• First, find \(a = \sqrt{9} = 3\) and \(b = \sqrt{1} = 1\).
• Substitute these values along with the center into the asymptote formula.

The resulting equations become \(y = 1 \pm 3(x - 3)\).
When graphing, these lines are typically represented as dashed.
While they don't intersect the hyperbola, they provide a visual framework.
By sketching these asymptotes first, one can align the hyperbola's arms precisely, ensuring an accurate representation of its infinite nature.