Problem 38 - Thomas Calculus

Question

Exercises \(35-38\) give foci and corresponding directrices of hyperbolas centered at the origin of the \(x y\) -plane. In each case, find the hyperbola's eccentricity. Then find the hyperbola's standard-form equation. $$ \begin{array}{l}{\text { Focus: }(-6,0)} \\ {\text { Directrix: } x=-2}\end{array} $$

Step-by-Step Solution

Answer

Eccentricity is 2; equation: \(\frac{x^2}{9} - \frac{y^2}{27} = 1\).

Step 1: Identify the Components of the Hyperbola

We are given the focus at \((-6, 0)\) and the directrix at \(x = -2\). Knowing the hyperbola is centered at the origin and symmetrical about the x-axis, we need to find the eccentricity \(e\) in order to write the standard-form equation.

Step 2: Understand the Relationship Between Focus, Directrix, and Eccentricity

The eccentricity \(e\) of a hyperbola is calculated using the formula \(e = \frac{c}{a}\), where \(c\) is the distance from the center to a focus, and \(a\) is the distance from the center to the directrix. The formula linking the directrix \(x = -2\) and focus \((-6, 0)\) helps in obtaining \(e\) by \(e = \frac{|a - (-a)|}{|a|} = \frac{|-6 - (-2)|}{|-2|} = \frac{4a}{a} = 2\).

Step 3: Calculate the Eccentricity

Using the formula \(e = \frac{c}{a}\) and the given details, note that the distance from the center to the focus \(c = |-6| = 6\) and the distance to the directrix \(|-2| = 2\). Thus, \(e = 2\).

Step 4: Determine the Equation Parameters \(a, b, c\)

For a hyperbola \(x^2/a^2 - y^2/b^2 = 1\), we know \(e = \frac{c}{a}\) and \(c = \sqrt{a^2 + b^2}\). Given \(e = 2\), we solve \(2 = \frac{c}{a}\), \(c = 2a\). Since \(c = 6\), we solve for \(a\): \(2a = 6\) implies \(a = 3\). To find \(b\), use \(c^2 = a^2 + b^2\): \(36 = 9 + b^2\), thus \(b^2 = 27\).

Step 5: Write the Standard-Form Equation

With \(a = 3\), \(b^2 = 27\), the hyperbola's formula is \(\frac{x^2}{9} - \frac{y^2}{27} = 1\), matching the standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\).

Key Concepts

Hyperbola Standard Equation

The hyperbola standard equation is a mathematical expression that describes the shape and orientation of a hyperbola on a coordinate system. It is important for making calculations involving hyperbolas easier. A hyperbola has two branches that reflect across an axis, and its equation captures this structure.

When dealing with a hyperbola centered at the origin, the standard form of the equation is:

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

The letters \(a\) and \(b\) in the equation represent the distances from the center to vertices and co-vertices, respectively. In our specific case, with a focus at \((-6, 0)\) and the directrix \(x = -2\), the hyperbola is horizontally oriented with its standard equation being \(\frac{x^2}{9} - \frac{y^2}{27} = 1\). This equation is derived after calculating the necessary parameters, allowing us to capture the characteristics of this specific hyperbola.

Focus and Directrix Relationship

The relationship between a hyperbola's focus and directrix is critical to understanding its shape. The focus is a point, while the directrix is a line. The hyperbola is defined such that the difference in distances from any point on the hyperbola to the foci and the distances to the directrix is constant. They are connected by the eccentricity \(e\), which is a measure of how stretched the hyperbola is.

The eccentricity can be calculated using the formula \(e = \frac{c}{a}\), where:

• \(c\) is the distance from the center of the hyperbola to the focus.
• \(a\) is the distance from the center to the directrix.

For the given hyperbola centered at the origin, with a focus at \((-6,0)\) and a directrix \(x = -2\), the eccentricity is calculated as \(e = \frac{6}{3} = 2\). This implies the hyperbola is quite stretched, as \(e = 2\) is greater than 1, a hallmark of hyperbolas.

Centered Hyperbola

A centered hyperbola is a hyperbola that has its geometrical center, sometimes called the 'origin' of the hyperbola, located right at the origin of the Cartesian plane. This means it is symmetrically aligned along the coordinate axes, with the center at \( (0, 0) \).

When discussing a centered hyperbola, the key elements include its foci, vertices, and co-vertices all aligned symmetrically based on the x and y axes. The presence of a directrix parallel to one of the axes is also a feature of centered hyperbolas. Such hyperbolas simplify the calculation of various properties due to their symmetrical nature.

In our example, the hyperbola is centered at the origin with its foci on the x-axis at \((-6, 0)\) and a directrix at \(x = -2\). The symmetry allows us to confidently focus on distances purely along the x-axis when computing for parameters like the eccentricity and the standard equation. Handling centered hyperbolas becomes straightforward due to their simplified position on the coordinate plane.