Problem 36 - Precalculus Mathematics for Calculus

Question

Find an equation for the hyperbola that satisfies the given conditions. Vertices: \((0, \pm 6),\) asymptotes: \(y=\pm \frac{1}{3} x\)

Step-by-Step Solution

Answer

The hyperbola's equation is \( \frac{y^2}{36} - \frac{x^2}{324} = 1 \).

Step 1: Identify the hyperbola type

The problem states that the vertices of the hyperbola are at (0, ±6). Since the vertices are along the y-axis, this is a vertical hyperbola.

Step 2: Standard form of a vertical hyperbola

The standard form of a vertical hyperbola centered at the origin is: \[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \text{, where } a \text{ represents the distance from the center to each vertex along the y-axis.} \]

Step 3: Assign value to \(a\)

Since the vertices are at (0, ±6), \(a = 6\). Thus, \(a^2 = 36\).

Step 4: Use the asymptote formulas

For a vertical hyperbola, the equations of the asymptotes are given by: \( y = \pm \frac{a}{b} x \). Since the asymptotes are given by \( y = \pm \frac{1}{3} x \), equate \( \frac{a}{b} \) to \( \frac{1}{3} \).

Step 5: Solve for \(b\)

Using \( \frac{a}{b} = \frac{1}{3} \) and knowing \(a = 6\), we have \( \frac{6}{b} = \frac{1}{3} \). Solving this equation gives \( b = 18 \). So, \( b^2 = 324 \).

Step 6: Write the equation

Substitute \(a^2\) and \(b^2\) into the standard form to get: \[ \frac{y^2}{36} - \frac{x^2}{324} = 1 \].

Key Concepts

Vertices of a Hyperbola

The vertices of a hyperbola are crucial points that lie on its transverse axis. These are essentially the points that are at the maximum extent away from the center, in the direction defined by the hyperbola's orientation. For a hyperbola centered at the origin, if the vertices are at

• (0, ±a) for a vertical hyperbola, or
• (±a, 0) for a horizontal hyperbola

these coordinates inform us directly about the value of parameter 'a'. This parameter 'a' represents half the distance between the two vertices, crucial for the standard equation of the hyperbola. In our exercise, with vertices at (0, ±6), we deduce it's a vertical hyperbola with a = 6.

Hyperbola Asymptotes

In hyperbolas, asymptotes are straight lines that the curve approaches infinitely but never intersects. Their equations provide insights into the behavior of the hyperbola at large values. The equations for asymptotes of a hyperbola vary depending on its orientation:

• For a vertical hyperbola, the asymptote equations are given by: \[ y = \pm \frac{a}{b} x \]
• For a horizontal hyperbola, they are: \[ y = \pm \frac{b}{a} x \]

In the problem, the asymptotes: \( y = \pm \frac{1}{3} x \) inform us that \( \frac{a}{b} = \frac{1}{3} \). Knowing 'a' helps us determine 'b', which is essential for constructing the hyperbola's equation.

Standard Form of a Hyperbola

The standard form of a hyperbola equation depends on its orientation, either vertical or horizontal. For a vertical hyperbola, the formula is:\[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \] This format highlights the directional nature of the hyperbola with respect to the y-axis. In contrast, for a horizontal orientation,\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] would be used, stressing the primary direction along the x-axis. Each part of the equation plays a role in defining the hyperbola's shape and orientation. In our exercise, since the hyperbola is vertical, its equation follows the first formula.

Vertical Hyperbola

A vertical hyperbola implies that its vertices are on the y-axis, making the y-axis its transverse axis. This orientation influences both its shape and equation structure, given by:\[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \] In the problem context, given vertices suggest a vertical orientation. The distance between these vertices directly informs the 'a' value in the standard form of the equation. The relation \( \frac{a}{b} \) derived from the asymptotes gives us the 'b' value, completing the equation's setup. Hence, for our vertical hyperbola, parameters lead us to construct\[ \frac{y^2}{36} - \frac{x^2}{324} = 1 \]