Problem 32 - Algebra 2

Question

In an ellipse, the ratio \(\frac{c}{a}\) is called the eccentricity and is denoted by the letter \(e\) . Eccentricity measures the elongation of an ellipse. The closer \(e\) is to \(0,\) the more an ellipse looks like a circle. Pluto has the most eccentric orbit in our solar system with \(e \approx 0.25 .\) Find an equation to model the orbit of Pluto, given that the length of the major axis is about 7.34 billion miles. Assume that the major axis is horizontal and that the center of the orbit is the origin.

Step-by-Step Solution

Answer

The equation is \( \frac{x^2}{13.4689} + \frac{y^2}{12.6271} = 1 \).

Step 1: Understand the Relationship

The eccentricity of an ellipse is given by \( e = \frac{c}{a} \). Here, \( a \) is the semi-major axis, \( c \) is the distance from the center to a focus, and \( e \) is the eccentricity. Given that \( e = 0.25 \) and the length of the major axis is 7.34 billion miles, we can compute \( a = \frac{7.34}{2} = 3.67 \) billion miles.

Step 2: Calculate the Focal Distance

Using the formula \( c = ea \), we can calculate \( c \) by substituting the known values: \( c = 0.25 \times 3.67 = 0.9175 \) billion miles.

Step 3: Apply the Ellipse Equation

The general equation of an ellipse centered at the origin with a horizontal major axis is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). We need to find \( b \) using the relationship \( a^2 = b^2 + c^2 \).

Step 4: Calculate the Semi-minor Axis

Substitute the known values into the equation \( a^2 = b^2 + c^2 \): \[ (3.67)^2 = b^2 + (0.9175)^2 \] Calculating this gives: \[ 13.4689 = b^2 + 0.8418 \] \[ b^2 = 13.4689 - 0.8418 = 12.6271 \] Thus, \( b = \sqrt{12.6271} \approx 3.5537 \) billion miles.

Step 5: Formulate the Ellipse Equation

Substitute \( a \) and \( b \) back into the ellipse equation: \[ \frac{x^2}{(3.67)^2} + \frac{y^2}{(3.5537)^2} = 1 \] This simplifies to: \[ \frac{x^2}{13.4689} + \frac{y^2}{12.6271} = 1 \]

Key Concepts

Eccentricity

Eccentricity is a key feature of ellipses, describing how "stretched out" they appear. Unlike a circle, which has an eccentricity of 0, an ellipse has a non-zero eccentricity. The value ranges between 0 and 1.
An eccentricity close to 0 means the ellipse is circular, whereas a value approaching 1 indicates a highly elongated shape.
The formula to find the eccentricity is:

• \( e = \frac{c}{a} \)

Here, \( c \) is the focal distance — the distance from the center to one of its foci, and \( a \) is the semi-major axis.Using this formula, you can understand the orbit's shape. For instance, Pluto's orbit, with an eccentricity of 0.25, looks more stretched than a circle due to its elongated path around the sun.

Major Axis

The major axis is the longest diameter of an ellipse, passing through both foci and the center. This axis is crucial as it defines the extent of the ellipse across its widest part.
If you know the total length of the major axis, labeled as \( 2a \), it's straightforward to find the semi-major axis, \( a \), through the relation:

• \( a = \frac{\text{Total length of major axis}}{2} \)

The major axis in Pluto's orbit is 7.34 billion miles, implying each half, or semi-major axis, is 3.67 billion miles long. This gives insight into the size and reach of the orbit as it circles the sun.

Semi-minor Axis

The semi-minor axis is the shortest diameter of an ellipse, perpendicular to the major axis, spanning through the center. This segment helps us understand the breadth of the ellipse at its narrowest part.
To find the semi-minor axis, denoted as \( b \), we often use it in conjunction with the semi-major axis \( a \) and the focal distance \( c \) in the relationship:

• \( a^2 = b^2 + c^2 \)

By rearranging this, you can solve for the semi-minor axis:
\( b = \sqrt{a^2 - c^2} \)In our example, using the length of Pluto's semi-major axis and eccentricity-derived focal distance, we find the semi-minor axis to be approximately 3.5537 billion miles. This measure contributes to the elliptical shape's complete description.

Semi-major Axis

The semi-major axis, denoted \( a \), is a significant part of understanding ellipses because it represents half of the major axis. It not only helps calculate the ellipse's eccentricity but also plays a role in formulating the ellipse's equation.
For any ellipse with a given major axis length, the semi-major axis is half of that length:

• \( a = \frac{\text{Length of major axis}}{2} \)

In Pluto's elliptical orbit's context, this value is 3.67 billion miles. Important in both geometric and orbital contexts, the semi-major axis informs calculations like the ellipse's area and other orbit characteristics, providing a foundation for calculating all related measurements.