Problem 42 - A Graphical Approach to Precalculus with Limits

Question

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

Step-by-Step Solution

Answer

The eccentricity \(e\) of the hyperbola is 3.

Step 1: Recognize the Conic Section

The equation given is \(8x^2 - y^2 = 16\). This has the form \(Ax^2 - By^2 = C\), where \(A\) and \(B\) are positive constants. This is the standard form of a hyperbola due to the subtraction sign between \(x^2\) and \(y^2\) terms.

Step 2: Convert to Standard Form of Hyperbola

The standard form of a hyperbola is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). Start by dividing the entire equation by 16 to get \(\frac{8x^2}{16} - \frac{y^2}{16} = 1\), which simplifies to \(\frac{x^2}{2} - \frac{y^2}{16} = 1\). This gives \(a^2 = 2\) and \(b^2 = 16\).

Step 3: Determine Values of a, b, and c for the Hyperbola

From \(a^2 = 2\), we find \(a = \sqrt{2}\). From \(b^2 = 16\), we find \(b = 4\). In a hyperbola, \(c\) is found using the relationship \(c^2 = a^2 + b^2\). Substitute the known values to find \(c^2 = 2 + 16 = 18\), thus \(c = \sqrt{18} = 3\sqrt{2}\).

Step 4: Find the Eccentricity of the Hyperbola

For hyperbolas, the eccentricity \(e\) is given by \(e = \frac{c}{a}\). Substitute the values \(c = 3\sqrt{2}\) and \(a = \sqrt{2}\) to find \(e = \frac{3\sqrt{2}}{\sqrt{2}} = 3\).

Key Concepts

Eccentricity

In the world of conic sections, eccentricity is a measure that determines how much a curve deviates from being circular. It plays a crucial role in defining the shape of ellipses and hyperbolas. Eccentricity is represented by the symbol \( e \), and its value varies depending on the type of conic section you are dealing with.

• For an ellipse, the eccentricity is between 0 and 1, which indicates how "stretched" it is compared to a perfect circle.
• For a hyperbola, like in our exercise, the eccentricity is greater than 1, showing how the two branches are open and extend outwards.
• A perfect circle, on the other hand, has an eccentricity of 0.

Understanding eccentricity is important because it helps us predict the overall shape and geometry of the curve we are dealing with. In our exercise, we computed the eccentricity of a hyperbola and found it to be \( e = 3 \), which confirms that the hyperbola is indeed a conic section with branches opening outward.

Hyperbola

A hyperbola is one of the fascinating conic sections you will encounter in mathematics. It is characterized by two distinct branches that mirror each other, forming a symmetrical shape about a central axis. Hyperbolas are defined by an equation that typically has a subtraction sign between its squared terms. In our case, the equation is \( 8x^2 - y^2 = 16 \).

• This subtraction is key, distinguishing it from an ellipse, which uses addition.
• A hyperbola's defining feature is its two disconnected curves, which resemble open-ended U-shapes or bowls facing away from each other.

By understanding and recognizing the equation form of a hyperbola, we can better solve problems and calculate characteristics such as the eccentricity. Hyperbolas appear in several real-world applications, such as in navigation systems and the paths of celestial bodies.

Standard Form of Hyperbola

The standard form of a hyperbola equation makes it easier to handle and interpret. It is written as:\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]This standard form involves:

• \( a^2 \) and \( b^2 \), which are constants derived from the hyperbola’s equation.
• "\( - \)" sign that separates the two fractions, indicating that the graph will be a hyperbola.

Let's see how this applies to our problem: the equation \( 8x^2 - y^2 = 16 \) was converted to the standard form by dividing through by 16, simplifying it to \( \frac{x^2}{2} - \frac{y^2}{16} = 1 \). From here, we identified that \( a^2 = 2 \) and \( b^2 = 16 \), which helped us calculate the values needed to determine other properties like eccentricity and the hyperbola's branches. Understanding how to manipulate and utilize the standard form is an essential skill when dealing with hyperbolas.