Problem 42 - A Graphical Approach to College Algebra

Question

Find the eccentricity \(e\) of each ellipse or hyperbola. $$8 x^{2}-y^{2}=16$$

Step-by-Step Solution

Answer

The eccentricity (e) is 3.

Step 1: Rearrange the Equation

To find the eccentricity, we need the equation in the standard form of a hyperbola. Start by dividing the given equation by 16 to simplify it:\[ \frac{8x^2}{16} - \frac{y^2}{16} = 1 \]Which simplifies to:\[ \frac{x^2}{2} - \frac{y^2}{16} = 1 \]

Step 2: Identify the Standard Form

Identify that the equation \( \frac{x^2}{2} - \frac{y^2}{16} = 1 \) is in the standard form for a hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). Here, \( a^2 = 2 \) and \( b^2 = 16 \).

Step 3: Calculate the Eccentricity

For a hyperbola, the eccentricity \( e \) is given by the formula:\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]Substitute \( a^2 = 2 \) and \( b^2 = 16 \) into the formula:\[ e = \sqrt{1 + \frac{16}{2}} \]This results in:\[ e = \sqrt{1 + 8} = \sqrt{9} \]

Step 4: Simplify the Eccentricity

Simplify the result to find the eccentricity:\[ e = \sqrt{9} = 3 \]

Key Concepts

Standard Form of a Hyperbola

To understand hyperbolas better, we start with the standard form of its equation. For the hyperbola, the standard form is either \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) or \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \). In these equations, \( a^2 \) and \( b^2 \) are positive constants that help determine the hyperbola's shape.

When you see a subtraction in the middle of the terms, it signals a hyperbola, unlike an ellipse which has addition. The parameters \( a \) and \( b \) are crucial as they determine the distances and asymptotes' steepness. In our example, after simplifying the original equation, \( \frac{x^2}{2} - \frac{y^2}{16} = 1 \), we identified that \( a^2 = 2 \) and \( b^2 = 16 \).

This form helps us in comparing any given hyperbola to a standard reference, simplifying future calculations like finding the eccentricity.

Ellipse and Hyperbola Equations

Hyperbolas and ellipses are both curve types but differ in how they stretch on a graph. Their equations, although similar in form, reveal fundamentally different shapes through their signs.

For ellipses, the standard equation is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where the addition results in an oval shape. Hyperbolas, with an equation like \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), using subtraction, create two separate curves or "branches." This minus sign is the defining difference.

Understanding this distinction is essential not only for writing equations but also for graph interpretation. The coefficient arrangement in these formulas tells us about the curve's orientation and dimensions, whether a shape opens sideways, as with horizontal hyperbolas, or up and down, common in vertical hyperbolas.

Hyperbola Properties

Hyperbolas possess unique properties that enhance their understanding in geometry. Let's explore some key features including their asymptotes, axes, and focal points.

• **Asymptotes**: In hyperbolas, these are imaginary lines that the branches approach but never meet. They are diagonally placed and intersect at the center of the hyperbola. The slopes are derived from the formula \( y = \pm \frac{b}{a}x \) when the hyperbola is centered at the origin.
• **Axes**: Like ellipses, hyperbolas have two axes: a transverse axis and a conjugate axis. The transverse axis is the line through the two vertices, while the conjugate axis is perpendicular to it.
• **Foci**: Each hyperbola has two focal points located along the transverse axis. These points are outside the hyperbola, unlike ellipses.

Hyperbolas are uniquely defined by their eccentricity \( e \), which is always greater than 1, indicating the extent of their distortion compared to a circle. Higher values of \( e \) mean a wider spread of the branches, evident in our calculation of \( e = 3 \). These properties assist in visualizing and sketching hyperbolas accurately.