Problem 67 - Contemporary Precalculus

Question

If \(a>b>0,\) then the eccentricity of the ellipse $$\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1 \quad \text { or } \quad \frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}}=1$$ is the number \(\frac{\sqrt{a^{2}-b^{2}}}{a} .\) Find the eccentricity of the ellipse whose equation is given. $$\frac{x^{2}}{100}+\frac{y^{2}}{99}=1$$

Step-by-Step Solution

Answer

Answer: The eccentricity of the ellipse is \(\frac{1}{10}\).

Step 1: Identify the major and minor axes

To identify the major and minor axes from the given equation, compare it to the standard ellipse equation: $$\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1$$ For the given equation, \(\frac{x^2}{100} + \frac{y^2}{99} = 1\), we can observe that \(a^2 = 100\) and \(b^2 = 99\).

Step 2: Find the values of a and b

Solve for a and b using their squared values: \(a = \sqrt{100} = 10\) and \(b = \sqrt{99}\)

Step 3: Calculate the eccentricity

Plug in the values of a and b into the eccentricity formula: $$e = \frac{\sqrt{a^2 - b^2}}{a} = \frac{\sqrt{10^2 - (\sqrt{99})^2}}{10}$$

Step 4: Simplify the eccentricity expression

Simplify the expression to find the eccentricity: $$e = \frac{\sqrt{100 - 99}}{10} = \frac{\sqrt{1}}{10}$$

Step 5: Final answer

The eccentricity of the ellipse with the given equation is: $$e = \frac{1}{10}$$

Key Concepts

Ellipse Equation

An ellipse is a type of conic section that appears rounded, like an elongated circle. It's defined using a specific kind of equation called the ellipse equation. The general form of this equation is \[\frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1\]In this form,

• \((h, k)\) are the coordinates of the center of the ellipse, which is a point from which the shape is symmetrically drawn.
• \(a\) and \(b\) are lengths that define how far the ellipse stretches along the x-axis and y-axis respectively.

The terms \((x-h)\) and \((y-k)\) suggest that we're measuring distances from this center point. If there is no shift in the origin like in this case, \((x-h)\) simplifies to \(x\), and \((y-k)\) simplifies to \(y\). Therefore, the ellipse is centered at the origin when both \(h\) and \(k\) are equal to zero.

Major and Minor Axes

Within an ellipse, the lengths of \(a\) and \(b\) correspond to important features called the axes. The larger of these two measures is called the major axis, and the smaller is referred to as the minor axis. These axes tell us which way the ellipse stretches more.

• The **major axis** is the longest diameter of the ellipse and runs through the center from one end to the other, passing through the foci.
• The **minor axis** is perpendicular to the major axis and represents the shortest diameter.

For the ellipse \[\frac{x^{2}}{100} + \frac{y^{2}}{99} = 1\]given in the original exercise, \(a^{2} = 100\) and \(b^{2} = 99\), which means that the major axis (since \(100 > 99\)) runs along the x-axis, while the minor axis runs along the y-axis.

Standard Form of Ellipse

An ellipse equation is often written in what is known as standard form, which helps to quickly identify properties like its axes and center. The standard equation \[\frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1\]addresses ellipses that align with the coordinate axes.
This standard format makes it easy for us to see the following:

• This specific structure (numerator squared terms divided by constants) lets us find the lengths of semi-major and semi-minor axes directly.
• Reveals the location of the center point \((h, k)\) of the ellipse. Since \(h = 0\) and \(k = 0\) here, the center is at the origin.

"Standard" here implies that once we know \(a\) and \(b\), we can directly understand the stretched and compressed dimensions of the ellipse. For example, with \(a = 10\) and \(b = \sqrt{99}\), you can immediately know the scale of axes.

Calculation of Eccentricity

Eccentricity is a measure of how much an ellipse deviates from being a perfect circle. The formula for eccentricity \(e\) of an ellipse is given by: \[e = \frac{\sqrt{a^{2} - b^{2}}}{a}\]This value helps describe the "flatness" of the ellipse.

• When \(e = 0\), the ellipse is a circle because \(a\) and \(b\) are equal.
• As \(e\) approaches 1, the ellipse becomes more stretched out.

In our case, we have \(a = 10\) and \(b = \sqrt{99}\). By plugging these values into the eccentricity formula:\[e = \frac{\sqrt{100 - 99}}{10} = \frac{\sqrt{1}}{10} = \frac{1}{10}\]This low eccentricity (\(e = \frac{1}{10}\)) indicates a shape very close to a circle, with a slight elongation along the x-axis.