Problem 24 - 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 \(\frac{x^{2}}{9}-\frac{y^{2}}{1}=1\)

Step-by-Step Solution

Answer

The center of the hyperbola is at the origin (0, 0). The vertices are at (3, 0) and (-3, 0). The foci are at (\(\sqrt{10}\), 0) and (-\(\sqrt{10}\), 0). The equations of the asymptotes are y = 3x and y = -3x.

Step 1: Identify a, b, and c from the given equation

The given equation for the hyperbola is \(\frac{x²}{9} - \(\frac{y²}{1} = 1\). The values under the x² and y² are a² and b², respectively. So a² = 9, hence a = 3 and b² = 1, hence b = 1. We can calculate c by using Pythagorean theorem for hyperbolas, which is, \(c² = a² + b²\). So, \(c² = 3² + 1²\) = 10, hence c = \(\sqrt{10}\)

Step 2: Determine the center, vertices, and foci

For hyperbolas, the center is always at the origin (0,0) when the equation is not shifted. Therefore, the center in our case is at (0,0). The vertices are then at the distance ±a units along the x-axis from the center. Hence the vertices are (±3, 0). The foci are ±c units on the x-axis from the center, hence foci are at (±\(\sqrt{10}\), 0)

Step 3: Determine the asymptotes

The equations for the asymptotes in the standard form for hyperbola are \(\frac{y}{b} = ± \frac{x}{a}\). Hence, substituting values of a and b here, we get y = ±3x as equations of the asymptotes.

Step 4: Sketch the graph

Having determined the center, vertices, foci, and asymptotes, one can now draw the hyperbola. Draw a pair of perpendicular lines representing the x and y axes, mark the center (0,0), the vertices and foci at respective distances, draw the asymptotes, and finally draw the hyperbola making sure it approaches the asymptotes as it extends.

Key Concepts

Graphing a Hyperbola

Graphing a hyperbola might seem tricky at first, but it becomes manageable by following the right steps. A hyperbola consists of two separate curves called branches. These branches mirror each other across the center of the hyperbola and in this case are aligned along the x-axis.

To start, identify the center. For the equation \(\frac{x^{2}}{9}-\frac{y^{2}}{1}=1\), the center is at (0,0). This is always the same when no shifts are present in the equation.

Next, locate the vertices. These are points where the hyperbola crosses the x-axis, positioned at the distance of \(\pm a\) from the center. In this equation, \(a=3\), so the vertices are at (3,0) and (-3,0).

The drawing begins by sketching the asymptotes, helping guide the shape of the hyperbola, and then drawing the curves that approach but never touch these lines.

Equation of Asymptotes

Asymptotes are invisible lines that the hyperbola approaches but never intersects. They are crucial for accurately plotting the graph. Asymptotes provide a framework for the hyperbola's shape. In the equation \(\frac{x^{2}}{9}-\frac{y^{2}}{1}=1\), the asymptotes can be found using the standard formula for hyperbolas: \(\frac{y}{b} = \pm \frac{x}{a}\).

Here, \(a = 3\) and \(b = 1\). Substituting these values into the formula results in the equations \(y = \pm 3x\).

These equations mean there are two lines: one with a positive slope and another with a negative slope. They cross through the origin, forming a cross-like guide that dictates how the branches of the hyperbola expand.

Vertices and Foci Calculation

Finding the vertices and foci helps define the key features of the hyperbola. The vertices are straightforward, as they are determined by the value of \(a\). As previously determined, \(a = 3\), so the vertices are located at (3,0) and (-3,0). These represent the closest points of each hyperbola branch to the center.

The foci are more challenging as they involve a calculation using \(c\), defined by the equation \(c^2 = a^2 + b^2\). Substituting in our values, \(c^2 = 3^2 + 1^2 = 10\), resulting in \(c = \sqrt{10}\).

This positions the foci at (\(\sqrt{10}\),0) and (-\(\sqrt{10}\),0), indicating where each hyperbola branch points towards.

Algebraic Graph Solutions

Utilizing algebraic methods ensures precise graphing of the hyperbola. An algebraic solution incorporates all the components calculated earlier.

Start by defining the center, which helps with understanding the symmetry. Then calculate the vertices and foci to pinpoint crucial points along the hyperbola.

• The vertices provide the width of the branches.
• The foci indicate their direction and extent.

Finally, incorporate the asymptotes. These guide how the hyperbola bends and where it broadens, ensuring the graph's curvature accurately represents the algebraic properties of the expression.

Carefully sketching each part together provides a comprehensive view, aiding in the understanding and visualization of the hyperbola's elegant structure.