Problem 26 - Thomas Calculus

Question

Exercises \(25-28\) give the eccentricities and the vertices or foci of hyperbolas centered at the origin of the \(x y\) -plane. In each case, find the hyperbola's standard-form equation in Cartesian coordinates. $$\begin{array}{l}{\text { Eccentricity: } 2} \\ {\text { Vertices: }( \pm 2,0)}\end{array}$$

Step-by-Step Solution

Answer

The equation of the hyperbola is \(\frac{x^2}{4} - \frac{y^2}{12} = 1\).

Step 1: Understand Eccentricity and Relationship

The eccentricity of a hyperbola (\(e\)) is given by the formula:\[e = \frac{c}{a}\]where \(c\) is the distance from the center to each focus, and \(a\) is the distance from the center to each vertex. We are given that \(e = 2\) and vertices at \((\pm 2, 0)\), indicating that \(a = 2\).

Step 2: Calculate the Distance to each Focus, \(c\)

Using the relationship \(e = \frac{c}{a}\), substitute \(e = 2\) and \(a = 2\):\[2 = \frac{c}{2}\]Multiply through by 2 to solve for \(c\):\[c = 4\].

Step 3: Calculate \(b\) Using \(a^2 + b^2 = c^2\)

For hyperbolas, \(b\) can be found using the equation:\[c^2 = a^2 + b^2\]Substitute \(a = 2\) and \(c = 4\):\[4^2 = 2^2 + b^2\]\[16 = 4 + b^2\]Subtract 4 from both sides:\[b^2 = 12\].

Step 4: Write the Standard Form Equation

The equation for a hyperbola centered at the origin and oriented horizontally is:\[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\]Substitute \(a^2 = 4\) and \(b^2 = 12\):\[\frac{x^2}{4} - \frac{y^2}{12} = 1\].

Key Concepts

Eccentricity

Eccentricity is a key characteristic of conic sections, including hyperbolas. It measures how "stretched out" a conic section is compared to a circle. Specifically for hyperbolas,eccentricity (\( e \) ) is given by the formula:

• \( e = \frac{c}{a} \)

where \( c \) is the distance from the center to each focus, and \( a \) is the distance from the center to each vertex.
The hyperbola's eccentricity tells us how much the shape deviates from being circular. If \( e = 1 \), the conic is a parabola. For hyperbolas, \( e > 1 \). For instance, in our exercise, the hyperbola has an eccentricity of 2, indicating it's quite stretched, with foci well-separated from the vertices.
This determines how "open" the hyperbola is. The larger \( e \), the more elongated the shape is.

Vertices

Vertices are crucial points on a hyperbola. These are the closest points on the hyperbola to the center. In the coordinate plane, they define how wide or narrow the hyperbola is horizontally or vertically. For the given exercise, the vertices are at \( (+2, 0) \) and \( (-2, 0) \), thus indicating that the hyperbola opens horizontally.

• The distance from the center to each vertex is represented by \( a \).

This information is vital for forming the equation. Since \( a = 2 \) in the exercise, it is used to calculate the value necessary for the equation parameters.

Cartesian Coordinates

Cartesian coordinates provide a grid-like system to locate points in the plane using two numbers: \( x \) for the horizontal position and \( y \) for the vertical position. Hyperbolas centered at the origin ((0,0)) use these coordinates for plotting. Knowing this, it becomes possible to place the geometric shape accurately within this plane.
Understanding how to work within Cartesian coordinates allows us to derive formulas, analyze displacement, and construct graphical representations clearly. In our exercise, this system helps express the hyperbola's orientation (horizontally), and the positioning of vertices and foci accurately.

Standard-Form Equation

The standard-form equation of a hyperbola provides a structured look at its geometric nature. This form helps in discerning key properties, like direction and shape, without delving into a graph.

• For a horizontally oriented hyperbola, the standard-form equation is:

\[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\]Here, \( a^2 \) is beneath the \( x^2 \) term as our hyperbola is horizontal. From the solution, we calculated \( a^2 = 4 \) and \( b^2 = 12 \).
Substituting these values gives us the precise formula: \[\frac{x^2}{4} - \frac{y^2}{12} = 1\]This equation succinctly captures the hyperbola's shape and orientation while being easily interpreted for graphing or further analysis.