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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

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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.

This problem has been solved

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