Problem 16 - Thomas Calculus

Question

Exercises \(13-16\) give foci and corresponding directrices of ellipses centered at the origin of the \(x y\) -plane. In each case, use the dimensions in Figure 10.19 to find the eccentricity of the ellipse. Then find the ellipse's standard-form equation. $$ \begin{array}{l}{\text { Focus: }(-\sqrt{2}, 0)} \\ {\text { Directrix: } \quad x=-2 \sqrt{2}}\end{array} $$

Step-by-Step Solution

Answer

The eccentricity is \(\frac{1}{2}\), and the standard-form equation is \(\frac{x^2}{8} + \frac{y^2}{6} = 1\).

Step 1: Identify Key Elements

The focus is given as \((-\sqrt{2}, 0)\) and the directrix equation is \(x = -2\sqrt{2}\). The ellipse is centered at the origin, meaning \((h, k) = (0, 0)\). The relationship between the focus \(c\) and the directrix line \(d\) can be used to find eccentricity \(e\).

Step 2: Define Eccentricity Formula

The eccentricity \(e\) of an ellipse is given by the formula \(e = \frac{c}{d} = \frac{c}{a}\), where \(d\) (the distance to the directrix from the center) is \(|x - h|\). Here, \(d = 2\sqrt{2}\); substitute into the formula once \(c\) is determined.

Step 3: Determine Distance \(c\)

The focus distance \(c\) from the center is the absolute value of the \(x\)-coordinate of the focus point, which is \(|-\sqrt{2}| = \sqrt{2}\).

Step 4: Calculate Eccentricity

Substitute \(c\) and \(d\) into the eccentricity equation: \(e = \frac{\sqrt{2}}{2\sqrt{2}} = \frac{1}{2}\).

Step 5: Calculate Semi-Major Axis \(a\)

Using the relationship \(e = \frac{c}{a}\), rearrange to find \(a\): \(a = \frac{c}{e} = \frac{\sqrt{2}}{1/2} = 2\sqrt{2}\).

Step 6: Determine Semi-Minor Axis \(b\)

Use the relationship \(b = \sqrt{a^2 - c^2}\) to find \(b\). First, calculate \(a^2 = (2\sqrt{2})^2 = 8\), then calculate \(c^2 = (\sqrt{2})^2 = 2\). Substitute into the formula: \(b = \sqrt{8 - 2} = \sqrt{6}\).

Step 7: Write the Standard Form Equation of the Ellipse

The standard form equation for an ellipse centered at the origin is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). Substitute the values found: \(\frac{x^2}{8} + \frac{y^2}{6} = 1\).

Key Concepts

Eccentricity

Eccentricity is a fundamental property of an ellipse that indicates how much it deviates from being a circle. It is denoted by the letter "e". The eccentricity of an ellipse can be understood as the ratio of the distance from any point on the ellipse to its focus, divided by the perpendicular distance to the closest directrix.
The formula to find the eccentricity is given by:

• \[ e = \frac{c}{a} \]

where "c" is the distance from the center to a focus, and "a" is the distance from the center to a vertex on the major axis.
In our exercise, the focus is at \(-\sqrt{2}, 0\), which tells us that the distance "c" is \(\sqrt{2}\). The distance "d" to the directrix is given as \(2\sqrt{2}\). By substituting these values, we calculated the eccentricity to be \( \frac{1}{2} \), indicating a moderately stretched ellipse.

Foci

The foci of an ellipse are two special points located along the major axis, symmetrically placed about the center. Unlike a circle, an ellipse has two foci that help define its shape.
The distance from the center of the ellipse to each focus is represented as "c". This distance is crucial in deriving other properties of the ellipse, such as its eccentricity.

• For our given ellipse, the focus is located at \((-\sqrt{2}, 0)\).

The position of the focus helps determine how elongated the ellipse is. If the foci are closer to each other, the ellipse is more circular. If they are further apart, the ellipse is more elongated.

Directrices

Directrices are a pair of parallel lines associated with an ellipse that helps in defining its eccentricity. They are located outside the ellipse on either side and perpendicular to the major axis.
The distance from the center of the ellipse to the directrix is denoted by "d". This value is significant when calculating the eccentricity of the ellipse using the formula:

• \[ e = \frac{c}{d} \]

For our exercise, the directrix provided is at \(x = -2\sqrt{2}\), indicating a horizontal position relative to the center. This length "2\sqrt{2}" plays a pivotal role in determining how much the the ellipse is stretched from a perfect circle.

Standard-form Equation of an Ellipse

The standard-form equation of an ellipse centered at the origin is a crucial formula that helps describe its geometric properties. This equation is usually portrayed as:

• \[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]

where "a" represents the semi-major axis length, and "b" represents the semi-minor axis length.
It's important to remember that for an ellipse:

• \[ a > b \]

In our case, given values for the axes are \(a = 2\sqrt{2}\) and \(b = \sqrt{6}\). Thus, the standard-form equation is: \[\frac{x^2}{8} + \frac{y^2}{6} = 1\].
This equation satisfies all properties and dimensions of the given ellipse and helps us understand its overall structure.