Problem 57 - Precalculus with Limits

Question

In Exercises 57-60, find the eccentricity of the ellipse. \(\dfrac{x^2}{4}+\dfrac{y^2}{9}=1\)

Step-by-Step Solution

Answer

The eccentricity of the given ellipse is \(\dfrac{\sqrt{5}}{3}\).

Step 1: Determine values of a and b

Comparing \(\dfrac{x^2}{4}+\dfrac{y^2}{9}=1\) with \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\), it is clear that \(a^2 = 9\) and \(b^2 = 4\). Then, \(a = 3\) and \(b = 2\). Where, \(a\) is the semi-major axis and \(b\) is the semi-minor axis.

Step 2: Calculate the eccentricity

Now, eccentricity (e²) of the ellipse can be calculated using the formula \(e = \sqrt{1 - (\dfrac{b^2}{a^2})}\). So, \(e = \sqrt{1 - (\dfrac{2^2}{3^2})} = \sqrt{1 - \dfrac{4}{9}} = \sqrt{\dfrac{5}{9}} = \dfrac{\sqrt{5}}{3}\)

Key Concepts

Semi-Major Axis

When we talk about the ellipse defined by the equation \( \frac{x^2}{4} + \frac{y^2}{9} = 1 \), we identify the semi-major axis by comparing it to the standard form equation of an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Here, \( a^2 = 9 \), which means that \( a = 3 \). The semi-major axis is the longest radius of the ellipse, and in this case, it is along the y-axis. This tells us that the ellipse is vertically oriented because \( y^2 \) is divided by the larger number (9). Understanding the semi-major axis is important because:

• It defines the stretch of the ellipse in its major direction.
• It is always the square root of the larger denominator in the equation.
• In vertical ellipses like this one, it relates directly to the maximum y-value for any point on the ellipse.

The concept of the semi-major axis is crucial for calculating the eccentricity of an ellipse, as it represents one of the key distances involved.

Semi-Minor Axis

The semi-minor axis of an ellipse is the shorter radius that runs perpendicular to the semi-major axis. For the ellipse represented by \( \frac{x^2}{4} + \frac{y^2}{9} = 1 \), the semi-minor axis can be determined by examining the equation's standard form. Here, \( b^2 = 4 \), giving us \( b = 2 \), which means it stretches along the x-axis. This axis is always aligned with the smaller denominator of the equation. Here are some key points about the semi-minor axis:

• It defines the extent of the ellipse's spread in its minor direction.
• It directly impacts the shape's roundness, with a smaller \( b \) resulting in a more elongated ellipse.
• In horizontally aligned ellipses, the semi-minor axis would correspond to the y-axis, but here it's on the x-axis due to vertical orientation.

Understanding both the semi-major and semi-minor axes helps grasp the full dimensions of the ellipse, which is vital for calculating properties like the area and eccentricity.

Eccentricity Formula

The eccentricity of an ellipse is a number between 0 and 1 that describes how elongated the shape is. It's important because it tells us how much the ellipse deviates from being a perfect circle. The formula for eccentricity \( e \) is:\[ e = \sqrt{1 - \left( \frac{b^2}{a^2} \right)} \]In our example, \( a = 3 \) and \( b = 2 \), allowing us to calculate the eccentricity as follows:- Substitute the values: \( e = \sqrt{1 - \left( \frac{2^2}{3^2} \right)} \)- Simplify inside the square root: \( e = \sqrt{1 - \frac{4}{9}} \)- Calculate further: \( e = \sqrt{\frac{5}{9}} \)- Finalize: \( e = \frac{\sqrt{5}}{3} \)The steps illustrate that the eccentricity \( \frac{\sqrt{5}}{3} \approx 0.745 \), indicating the ellipse is moderately elongated. Eccentricity provides essential insights into:

• The overall shape, where a lower value is circular and higher values indicate more stretch.
• Helps in comparing various elliptical orbits and geometrical shapes quickly.

Comprehending the eccentricity formula and its calculation is key to mastering the understanding of elliptical shapes.