Problem 42 - 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 equation of the hyperbola is \(\frac{x^2}{324} - \frac{y^2}{36} = -1\).

Step 1: Determine the orientation

The vertices are (0, \pm 6). Since these vertices are aligned vertically (on the y-axis), the hyperbola is a vertical hyperbola. This affects the standard form equation, which is \(\frac{x^2}{b^2} - \frac{y^2}{a^2} = -1\) when centered at the origin \((0,0)\).

Step 2: Identify 'a' from vertices

The distance from the center \((0,0)\) to each vertex \((0, \pm 6)\) is the semi-major axis, \(a = 6\).

Step 3: Use asymptotes to find 'b'

The equations of the asymptotes for this vertical hyperbola are \(y = \pm \frac{a}{b} x\). Given that \(y = \pm \frac{1}{3} x\), we know \(\frac{a}{b} = \frac{1}{3}\). Since \(a = 6\), solve for \(b\): \(\frac{6}{b} = \frac{1}{3}\). Multiplying both sides by \(b\) and by 3 gives \(b = 18\).

Step 4: Write the equation of the hyperbola

Insert \(a = 6\) and \(b = 18\) into the standard form equation: \(\frac{x^2}{b^2} - \frac{y^2}{a^2} = -1\). Therefore, the equation becomes \(\frac{x^2}{324} - \frac{y^2}{36} = -1\).

Key Concepts

Vertices

Understanding vertices is key when dealing with a hyperbola. The vertices of a hyperbola are the points where the hyperbola intersects its axis of symmetry. In simple terms, they are the points closest to the center along the primary direction of the hyperbola. For the given problem, the vertices are at

• (0, 6) and (0, -6)

which indicates a vertical hyperbola.
These points are aligned along the y-axis, confirming that the hyperbola opens upwards and downwards. The distance from the center to these vertices is your 'a' value, representing the semi-major axis of the hyperbola. Therefore, each vertex is 6 units away from the center, leading us to conclude that

This value is essential as it influences the entire equation and shape of the hyperbola.

Asymptotes

Asymptotes are lines that the hyperbola approaches but never actually touches. For a hyperbola, the asymptotes guide the curve as it stretches to infinity.
They help define the tilt of the hyperbola in relation to the coordinate axes. For the given hyperbola, the asymptotes are represented by the equations:

• \(y = \pm \frac{1}{3}x\)

These equations show that as the hyperbola moves away from its vertices, it gets closer and closer to these lines.
Mathematically, for a vertical hyperbola, the relationship between 'a' and 'b' given through the asymptotes is \(y = \pm \frac{a}{b}x\). In our case, \(\frac{a}{b} = \frac{1}{3}\). We already know \(a = 6\), so solving \(\frac{6}{b} = \frac{1}{3}\) leads us to \(b = 18\).
Asymptotes are not just a feature of hyperbolas—they're essential for determining the shape and orientation of the curve.

Standard Form Equation

The standard form equation of a hyperbola gives us a compact way to express its essential geometric properties. For hyperbolas centered at the origin, the standard form is expressed as

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

for vertical hyperbolas. This equation ensures the hyperbola is correctly oriented and scaled.
Plugging in the known values, \(a = 6\) and \(b = 18\), into the standard form gives us:

• \(\frac{x^2}{324} - \frac{y^2}{36} = -1\)

It might be helpful to break down why the minus sign is included. This sign indicates that we subtract \(\frac{y^2}{a^2}\) from \(\frac{x^2}{b^2}\), distinguishing hyperbolas from other conic sections like ellipses.
The values calculated match the problem's conditions for vertices and asymptotes. Hence, the derived equation perfectly describes the hyperbola in question.