Problem 69 - 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. $$\frac{(x-3)^{2}}{10}+\frac{(y-9)^{2}}{40}=1$$

Step-by-Step Solution

Answer

Answer: The eccentricity of the ellipse is $\frac{\sqrt{3}}{2}$.

Step 1: Identify the semi-major and semi-minor axes

Find the values of \(a^2\) and \(b^2\) in the given ellipse equation. $$\frac{(x-3)^{2}}{10}+\frac{(y-9)^{2}}{40}=1$$ Here, \(a^2\) and \(b^2\) are the denominators of the \(x\) and \(y\) terms, respectively. Therefore, we have: $$a^2 = 40 \, \text{ and } \, b^2 = 10$$

Step 2: Calculate the eccentricity using the formula

Substitute the values of \(a^2\) and \(b^2\) in the eccentricity formula. $$e = \frac{\sqrt{a^2 - b^2}}{a}$$ So we have, $$e = \frac{\sqrt{40 - 10}}{\sqrt{40}}$$ Now, simplify the expression: $$e = \frac{\sqrt{30}}{\sqrt{40}} = \frac{\sqrt{30}}{2\sqrt{10}} = \frac{\sqrt{3}}{2}$$

Step 3: State the eccentricity of the ellipse

The eccentricity of the ellipse with the given equation is: $$e = \frac{\sqrt{3}}{2}$$

Key Concepts

Semi-Major Axis

In an ellipse, the semi-major axis is half of the longest diameter. Imagine the ellipse as a stretched circle. The semi-major axis stretches from the center of the ellipse to the farthest point on its perimeter. This axis is crucial as it determines many of the ellipse's properties, including its size and shape.

• You can find the semi-major axis by taking the square root of the larger of the two denominators in the standard form of the ellipse's equation.
• In the standard form, the equation is on the right with '1', while each term on the left has a denominator. One term is for 'x' and another for 'y'.

Remember, the semi-major axis doesn't need to align with the horizontal axis. It just has to be the longest axis.

Semi-Minor Axis

The semi-minor axis is the shorter counterpart to the semi-major axis. It stretches from the center to the edge of the ellipse, perpendicular to the semi-major axis. This axis determines the width of the ellipse.

• With the ellipse equation, you pinpoint the semi-minor axis by identifying the smaller denominator under the square term.
• The semi-minor axis helps define the overall compactness or narrowness of the ellipse.

Both axes, the semi-major and semi-minor, form the backbone of ellipse geometry. Knowing their lengths helps in computing other characteristics like area or eccentricity.

Ellipse Equation

The standard equation of an ellipse appears as follows: \[ \frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1 \] This equation describes every point \((x, y)\) on the ellipse related to its center. Here \(h\) and \(k\) are the coordinates of the center.

• 'a' and 'b' depict the axes. 'a' corresponds to the semi-major axis.
• 'b' corresponds to the semi-minor axis unless reversed depending on which denominator is larger.

When you identify the values of 'a' and 'b', you reveal the scaling on the x and y-axes that gives the ellipse its shape.

Eccentricity Formula

Eccentricity is a measure of how much an ellipse deviates from being circular. The formula to calculate eccentricity, given an ellipse with semi-major axis \(a\) and semi-minor axis \(b\), is \[ e = \frac{\sqrt{a^2 - b^2}}{a} \]

• If an ellipse has an eccentricity close to 0, it resembles a circle.
• If the eccentricity approaches 1, the ellipse becomes more elongated.

For ellipses, the eccentricity is always between 0 and 1. This formula allows you to understand the degree of stretch of the ellipse, helping in comprehending its geometric properties.