Question

investigation plan: 1. measure the distance between the foci - the sun and the other point inside the ellipse. record in the first column below. 2. measure the length of the major axis - the longest diameter of the ellipse. record in the second column below. 3. calculate the eccentricity (\out of roundness\) using the equation below and write your answer in the third column. eccentricity = distance between foci / length of the major axis 4. continue steps one through three for the remaining ellipses. observations: ellipse distance between foci (mm) length of major axis (mm) show your work here eccentricity 1 2 3 4 5 6 application: 1. what would happen to the shape of the ellipse if the foci were farther apart? 2. what is the relationship between the eccentricity of an ellipse and how nearly round the ellipse appears to be? 3. of the six ellipses, which one do you think is closest to the earths eccentricity?

Explanation:

Step1: Recall ellipse - eccentricity concept

The eccentricity \(e\) of an ellipse is given by the formula \(e=\frac{d}{l}\), where \(d\) is the distance between the foci and \(l\) is the length of the major - axis.

Step2: Analyze the effect of foci distance on ellipse shape

If the foci are farther apart while the length of the major - axis remains constant, the value of \(e\) increases. As \(e\) approaches 1, the ellipse becomes more elongated.

Step3: Understand the relationship between eccentricity and roundness

When \(e = 0\), the ellipse is a circle. As \(e\) increases from 0 to 1, the ellipse becomes less round. So, a lower eccentricity means the ellipse appears more round.

Step4: Know Earth's eccentricity

The Earth's orbit has an eccentricity of approximately \(e\approx0.0167\). We need to find the ellipse among the six with an eccentricity closest to this value.

Answer:

1. The ellipse would become more elongated.
2. A lower eccentricity means the ellipse appears more round, and as the eccentricity increases from 0 to 1, the ellipse becomes less round.
3. We would need to calculate the eccentricities of the six ellipses using the formula \(e=\frac{\text{distance between foci}}{\text{length of the major axis}}\) and then compare them to 0.0167 to determine which one is closest.