Problem 35 - 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. $$ \begin{array}{l}{\text { Focus: }(4,0)} \\ {\text { Directrix: } x=2}\end{array} $$

Step-by-Step Solution

Answer

The hyperbola's eccentricity is 2, and the equation is \(\frac{x^2}{4} - \frac{y^2}{12} = 1\).

Step 1: Understand the given information

The focus of the hyperbola is given as \((4,0)\) and the directrix as \(x=2\). A hyperbola is defined as the set of all points \((x, y)\) such that the absolute difference of the distances to the two foci is constant. The eccentricity \(e\) is a measure of how much the conic section deviates from being circular.

Step 2: Understand what eccentricity represents

The eccentricity \(e\) of a hyperbola is greater than 1 and is expressed as the ratio \(e = \frac{c}{a}\), where \(c\) is the distance from the center to the focus, and \(a\) is the distance from the center to a vertex of the hyperbola.

Step 3: Use given focus and directrix to determine \(e\)

The distance from the center \((0,0)\) to the focus is \(c = 4\). The distance from a point on the hyperbola to the corresponding focus over the distance from that point to the directrix is the eccentricity, so for this hyperbola, the eccentricity is \(e = \frac{4}{2} = 2\).

Step 4: Establish the equation of the hyperbola

For a hyperbola centered at the origin with horizontal transverse axis, the standard form is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). Since \(c = 4\) and \(e = \frac{c}{a} = 2\), it follows that \(c = 2a\), so \(a = 2\), and \(c = 4\).

Step 5: Calculate \(b^2\)

Using the relationship \(c^2 = a^2 + b^2\), we know \(c = 4\) and \(a = 2\). Substituting these values gives us \(4^2 = 2^2 + b^2\), leading to \(16 = 4 + b^2\). Solving for \(b^2\) yields \(b^2 = 12\).

Step 6: Write the standard-form equation

Substitute \(a^2\) and \(b^2\) into the hyperbola equation: \(\frac{x^2}{4} - \frac{y^2}{12} = 1\). This equation represents the hyperbola with the given focus and directrix.

Key Concepts

Eccentricity

Eccentricity is a fundamental concept when studying hyperbolas and other conic sections. It measures the deviation of a conic from being a perfect circle. For hyperbolas, the eccentricity, denoted as \(e\), is always greater than 1. This means a hyperbola is more elongated or stretched compared to circles and ellipses.
To calculate the eccentricity of a hyperbola, we use the formula:

• \(e = \frac{c}{a}\)

where \(c\) is the distance from the center of the hyperbola to its focus, and \(a\) is the distance from the center to a vertex.
In the given exercise, the focus is located at \((4,0)\) and the directrix at \(x=2\). By examining these points, we can find that \(c = 4\) and \(a = 2\). Thus, the eccentricity is calculated as follows:

• \(e = \frac{4}{2} = 2\)

This result means our hyperbola is considerably more elongated compared to an ellipse, as it has an eccentricity of 2.

Standard-form Equation

The standard-form equation represents the general form an equation of a hyperbola takes when it is centered at the origin. For hyperbolas, there are two common configurations:

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

The orientation of a hyperbola is determined by its transverse axis, the axis that contains the foci and vertices. In this exercise, due to having the focus at \((4,0)\), we use the horizontally oriented form.
Knowing that \(a = 2\), we calculate \(b^2\) using the relation \(c^2 = a^2 + b^2\). Here, \(c = 4\) and \(a = 2\), hence:

• \(4^2 = 2^2 + b^2\rightarrow 16 = 4 + b^2\rightarrow b^2 = 12\)

Now, we can substitute \(a^2\) and \(b^2\) into the standard-form equation:

• \(\frac{x^2}{4} - \frac{y^2}{12} = 1\)

This equation describes the shape, size, and orientation of the hyperbola based on provided parameters.

Directrix

The directrix of a hyperbola is a significant line that helps define the curve. It is located on the outside of the branches of the hyperbola and remains linked to the eccentricity's definition.
For a hyperbola centered at the origin, having horizontal orientation, a directrix line denotes significant proportionality. The role of the directrix helps uphold the geometric property that each point on the hyperbola maintains a consistent ratio of distances to a focus and the directrix.
Typically, if the equation of the directrix is \(x = k\) or \(y = k\), the hyperbola adheres to its unique alignment rules.
In the given exercise, the directrix is specified as \(x = 2\). The relation of the directrix to the hyperbola helps in identifying the localized parameters and allows for eccentricity verification. It complements the attributes of the focus for a stable hyperbola construction.

Focus

The focus (or foci, for plural in hyperbolas) is a vital point that significantly shapes the structure of a hyperbola. It is one of the two fixed points whose geometric relationship with points on the hyperbola establishes the curve's form.
For the hyperbola in question, we have a focus located at \((4, 0)\), playing a primary role in formulating both the geometry and equation of the curve. The focus's location helps in calculating eccentricity since it affects the values of \(c\) and, by extension, \(a\) when deducing the hyperbola's specific measurements.
The constant difference in distances from any point on the hyperbola to each of the foci is the defining property. This unique feature implies that any point on a hyperbola retains equivalent eccentric measures to the directrix, underpinning the breadth and design of the hyperbola.