Problem 33 - Precalculus with Calculus Previews

Question

In Problems \(21-44,\) find an equation of the hyperbola that satisfies the given conditions. Vertices (±2,0) , asymptotes \(y=\pm \frac{4}{3} x\)

Step-by-Step Solution

Answer

The equation of the hyperbola is \(9x^2 - \frac{27}{16}y^2 = 36\).

Step 1: Understand the Hyperbola Characteristics

The hyperbola has vertices at \((\pm 2, 0)\), which indicates that it is centered at the origin \((0,0)\) and is oriented horizontally. The distance between the center and each vertex is the value of \(a\), so \(a = 2\).

Step 2: Use Asymptote Equation to Find 'b'

The given asymptotes are \(y = \pm \frac{4}{3}x\). For a horizontally oriented hyperbola, the slope of these asymptotes is given by \(\frac{b}{a}\). Therefore, we have \(\frac{b}{2} = \frac{4}{3}\).

Step 3: Solve for b

From the equation \(\frac{b}{2} = \frac{4}{3}\), solve for \(b\).Multiply both sides by 2 to get \(b = \frac{8}{3}\).

Step 4: Write the Standard Equation of the Hyperbola

A horizontally oriented hyperbola centered at the origin has the standard equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). Substitute \(a = 2\) and \(b = \frac{8}{3}\) into this equation to find \(a^2 = 4\) and \(b^2 = \frac{64}{9}\).

Step 5: Formulate the Equation

Substituting the values of \(a^2\) and \(b^2\) into the standard hyperbola equation gives:\[ \frac{x^2}{4} - \frac{y^2}{\frac{64}{9}} = 1 \]

Step 6: Simplify the Equation

Multiply through by 36 to eliminate the fractions:\[ 9x^2 - \frac{9}{64}y^2 = 36 \]Further simplify:\[ 9x^2 - \frac{27}{16}y^2 = 36\].

Key Concepts

Vertices of Hyperbola

Vertices of a hyperbola are crucial in determining its shape and orientation. When we talk about vertices, we are referring to points that lie at the ends of the hyperbola's major axis. It's important to remember that the orientation of the hyperbola depends on the position of these vertices.
For a hyperbola with vertices positioned at \(\pm 2, 0\), the orientation is horizontal. This means the hyperbola opens left and right rather than up and down.

The distance from the center of the hyperbola to each vertex is denoted by the letter \(a\). In our example, the vertices are 2 units away from the origin, so \(a = 2\). This length \(a\) is vital because it helps form the hyperbola's standard equation. From the vertices, we can derive information about the hyperbola's orientation and size, allowing us to construct the entire figure.

Asymptotes of Hyperbola

Asymptotes are the lines that the hyperbola approaches but never actually meets. They essentially serve as guides for the shape of a hyperbola. In the exercise, we are given the asymptotes \(y = \pm \frac{4}{3} x\) which showcase their significance in understanding how the hyperbola behaves at infinity.

In a horizontally-oriented hyperbola, the slopes of the asymptotes are represented by \frac{b}{a}\. Here, the slope is \frac{4}{3}\, allowing us to solve for \(b\). Given our expression \frac{b}{2} = \frac{4}{3}\, solving for \(b\) gives us \(b = \frac{8}{3}\). This length \(b\) corresponds to the semiminor axis distance from the center to an auxiliary point along the hyperbola. Understanding and computing the values of \(m\), \(b\), and \(a\) is key to writing the hyperbola's standard equation.

Standard Equation of Hyperbola

To formulate a hyperbola's equation, we rely on a standard form. For horizontally oriented hyperbolas, we use:

• \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)

This equation showcases the relation between the horizontal difference and vertical separation in a hyperbola. Given that \(a = 2\) and \(b = \frac{8}{3}\), we evaluate \(a^2 = 4\) and \(b^2 = \frac{64}{9}\).

Substituting these squared values back into the standard form, we derive:

• \( \frac{x^2}{4} - \frac{y^2}{\frac{64}{9}} = 1 \)

To simplify, we can multiply through by 36 to clear the fractions, getting \(9x^2 - \frac{27}{16}y^2 = 36\). This form of the equation now expresses the hyperbola clearly. Students should remember that this standard form is crucial in graphing and analyzing hyperbolas.