S and T are the foci of an ellipse and B is an end of the minor axis. If STB is an equilateral triangle, the eccentricity of the ellipse is
Question
S and T are the foci of an ellipse and B is an end of the minor axis. If STB is an equilateral triangle, the eccentricity of the ellipse is
Solution
The eccentricity of an ellipse is given by the formula:
e = √(1 - (b²/a²))
Where:
- a is the semi-major axis (half the length of the major axis)
- b is the semi-minor axis (half the length of the minor axis)
In this case, we have an equilateral triangle STB, where S and T are the foci of the ellipse and B is an end of the minor axis.
In an equilateral triangle, all sides are of equal length. Therefore, the distance from S to T (the distance between the foci) is equal to the distance from S to B (the length of the semi-minor axis).
Let's denote the length of the sides of the equilateral triangle as d. So, we have:
2a = d (because the distance between the foci is 2a) b = d (because the length of the semi-minor axis is b)
Substituting these values into the formula for eccentricity, we get:
e = √(1 - (d²/d²)) = √(1 - 1) = 0
So, the eccentricity of the ellipse is 0.
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