Question

If the distance between the foci of an ellipse is 6 and the length of the minor axis is 8, then the eccentricity is

• A. \(\frac{1}{\sqrt{5}}\)
• B. \(\frac{1}{2}\)
• C. \(\frac{3}{5}\)
• D. \(\frac{4}{5}\)

Hint:

\(b^2 = a^2(1 - e^2)\) and \(2ae\) = distance between foci.

Answer 1

The Correct Option is C

To find the eccentricity of an ellipse, we can use the relationship between its axes and foci. The standard equation of an ellipse with a horizontal major axis is given by:

\(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)

where \( a \) is the semi-major axis and \( b \) is the semi-minor axis.

The distance between the foci is represented by \( 2c \), where \( c = \sqrt{a^2 - b^2} \) and the eccentricity \( e \) is given by:

\(e = \frac{c}{a}\)

1. We are given that the distance between the foci \( 2c = 6 \). Therefore, \(c = 3\).
2. We are also told that the length of the minor axis is 8. Thus, the semi-minor axis \( b = \frac{8}{2} = 4 \).
3. Using the relationship \(c = \sqrt{a^2 - b^2}\), we substitute the known values to find \( a \):
   \(3 = \sqrt{a^2 - 4^2}\)
   Squaring both sides: \(9 = a^2 - 16\)
   \(a^2 = 25\)
   Thus, \(a = 5\).
4. Now calculate the eccentricity using the formula \(e = \frac{c}{a}\):
   \(e = \frac{3}{5}\).

Therefore, the eccentricity of the ellipse is \(\frac{3}{5}\). The correct answer is:

\(\frac{3}{5}\)