Problem 44 - A Graphical Approach to College Algebra

Question

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

Step-by-Step Solution

Answer

The eccentricity \(e\) is \(\frac{\sqrt{2}}{2}\).

Step 1: Identify the type of conic

The given equation is \(x^2 + 2y^2 = 8\). By comparing it with the standard forms, we identify it as an ellipse because the coefficients of \(x^2\) and \(y^2\) are different and both positive.

Step 2: Write the standard form of ellipse

The standard form of an ellipse aligned with the coordinate axes is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). We need to transform the given equation into this form.

Step 3: Divide by the constant term

Divide the whole equation by 8 to simplify: \(\frac{x^2}{8} + \frac{2y^2}{8} = 1\), which reduces to \(\frac{x^2}{8} + \frac{y^2}{4} = 1\).

Step 4: Identify \(a^2\) and \(b^2\)

In the equation \(\frac{x^2}{8} + \frac{y^2}{4} = 1\), we identify \(a^2 = 8\) and \(b^2 = 4\). This gives \(a = \sqrt{8}\) and \(b = \sqrt{4} = 2\).

Step 5: Use eccentricity formula for ellipse

The eccentricity formula for an ellipse is \(e = \sqrt{1 - \frac{b^2}{a^2}}\). Substitute the values of \(a^2\) and \(b^2\) into the formula: \( e = \sqrt{1 - \frac{4}{8}} = \sqrt{1 - 0.5}\).

Step 6: Calculate eccentricity

Simplify the expression: \(e = \sqrt{0.5} = \frac{\sqrt{2}}{2}\). Thus, the eccentricity of the ellipse is \(\frac{\sqrt{2}}{2}\).

Key Concepts

Conic Sections

Conic sections are the shapes created when a plane intersects a double cone. The main types include circles, ellipses, parabolas, and hyperbolas. Each shape has distinct characteristics based on the angle and position of the intersecting plane.

• **Circles** are formed when the plane slices parallel to the base of the cone, creating a perfectly round shape.
• **Ellipses** appear when the plane cuts through the cone at an angle, forming an elongated circle.
• **Parabolas** are formed by planes that are parallel to the side of the cone.
• **Hyperbolas** are created when the plane cuts through both halves of the double cone.

The specific type of conic section is determined by the equation that describes it. In the exercise, we identified an ellipse because the coefficients of the squared terms were both positive but different, and this aligns with the properties of an ellipse.

Standard Form of Ellipse

The standard form of an ellipse's equation is vital as it allows us to easily identify the ellipse's characteristics, such as its orientation, center, and lengths of axes. The standard equation is given by:\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]where \(a\) and \(b\) represent the semi-major and semi-minor axes, respectively. For each ellipse:

• If \(a > b\), the ellipse extends more along the x-axis.
• If \(b > a\), it stretches more along the y-axis.

To transform the given equation \(x^2 + 2y^2 = 8\) to this form, we divided each part by 8. This gives us \(\frac{x^2}{8} + \frac{y^2}{4} = 1\), helping us find \(a^2 = 8\) and \(b^2 = 4\). The transformation to standard form is crucial for further calculations, such as finding eccentricity.

Ellipse Properties

Ellipses have unique properties that distinguish them from other conic sections. One of these properties is eccentricity, denoted by \(e\), which measures how much the ellipse deviates from being a circle.

The formula to calculate the eccentricity of an ellipse is:\[e = \sqrt{1 - \frac{b^2}{a^2}}\]This formula considers \(a\) and \(b\), where \(a\) is always greater than \(b\), aligning with the condition for ellipses. In our exercise, by substituting \(a^2 = 8\) and \(b^2 = 4\) into this formula, we found the eccentricity as \(\frac{\sqrt{2}}{2}\).

Ellipses also have two focal points. The distance between these foci is influenced by the eccentricity. Importantly, smaller eccentricity values indicate a shape that is closer to a circle, while values closer to 1 present a more elongated ellipse. This unique combination of properties helps define the overall shape and behavior of the ellipse in geometric and algebraic terms.