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In an ellipse, the distance between its foci is 6, and the minor axis is 8. What is its eccentricity?

Sagot :

To find the eccentricity of an ellipse given the distance between its foci and the length of its minor axis, follow these steps:

1. Identify the given values:
- The distance between the foci is given as 6.
- The length of the minor axis is given as 8.

2. Determine the semi-minor axis (b):
- The semi-minor axis (b) is half of the minor axis.
[tex]\[ b = \frac{8}{2} = 4 \][/tex]

3. Determine the focal distance (c):
- The distance between the foci is twice the focal distance from the center (2c), so:
[tex]\[ c = \frac{6}{2} = 3 \][/tex]

4. Find the semi-major axis (a):
- Using the relationship between the semi-major axis, semi-minor axis, and the focal distance:
[tex]\[ a^2 = b^2 + c^2 \][/tex]
Plugging in the values for [tex]\(b\)[/tex] and [tex]\(c\)[/tex]:
[tex]\[ a^2 = 4^2 + 3^2 = 16 + 9 = 25 \][/tex]
Therefore:
[tex]\[ a = \sqrt{25} = 5 \][/tex]

5. Calculate the eccentricity (e):
- The eccentricity of an ellipse is given by the formula:
[tex]\[ e = \frac{c}{a} \][/tex]
Substituting the values of [tex]\(c\)[/tex] and [tex]\(a\)[/tex]:
[tex]\[ e = \frac{3}{5} = 0.6 \][/tex]

So, the eccentricity of the ellipse is [tex]\(0.6\)[/tex].
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