Understanding the Ellipse Eccentricity Problem

The question asks us to find the eccentricity of an ellipse. We are given a specific geometric condition: the angle formed by connecting the two foci (\(F_1, F_2\)) to one end of the minor axis (\(B\)) is \(90^\circ\). Let this angle be \(\angle F_1BF_2 = 90^\circ\).

Geometric Setup of the Ellipse

Let's represent the ellipse in a standard coordinate system:

• Assume the ellipse is centered at the origin \((0, 0)\).
• The foci are located at \(F_1 = (-c, 0)\) and \(F_2 = (c, 0)\), where \(c\) is the distance from the center to a focus.
• The major axis lies along the x-axis, and the semi-major axis length is \(a\).
• The minor axis lies along the y-axis, and the semi-minor axis length is \(b\).
• The extremities of the minor axis are at \(B = (0, b)\) and \(B' = (0, -b)\). We can use either \(B\) or \(B'\) for our calculation; let's use \(B = (0, b)\).

The relationship between \(a\), \(b\), and \(c\) in an ellipse is given by the formula: \(c^2 = a^2 - b^2\). The eccentricity, denoted by \(e\), is defined as \(e = \frac{c}{a}\).

Deriving the Relationship Using Geometry

Consider the triangle formed by the foci and the extremity of the minor axis, \(\triangle F_1BF_2\).

1. Vertices: \(F_1(-c, 0)\), \(F_2(c, 0)\), and \(B(0, b)\).
2. Side Lengths:
   • Length of \(F_1F_2\): The distance between \((-c, 0)\) and \((c, 0)\) is \(2c\).
   • Length of \(BF_1\): Using the distance formula, \(\sqrt{(-c - 0)^2 + (0 - b)^2} = \sqrt{c^2 + b^2}\).
   • Length of \(BF_2\): Using the distance formula, \(\sqrt{(c - 0)^2 + (0 - b)^2} = \sqrt{c^2 + b^2}\).
   Notice that \(BF_1 = BF_2\), confirming \(\triangle F_1BF_2\) is an isosceles triangle.
3. Applying the Pythagorean Theorem: We are given that \(\angle F_1BF_2 = 90^\circ\). In the right-angled triangle \(\triangle F_1BF_2\), the Pythagorean theorem states \((BF_1)^2 + (BF_2)^2 = (F_1F_2)^2\).
4. Substitution: \(( \sqrt{c^2 + b^2} )^2 + ( \sqrt{c^2 + b^2} )^2 = (2c)^2\) \((c^2 + b^2) + (c^2 + b^2) = 4c^2\) \(2(c^2 + b^2) = 4c^2\)
5. Simplifying: \(c^2 + b^2 = 2c^2\) Subtracting \(c^2\) from both sides gives: \(b^2 = c^2\)

Calculating the Eccentricity

Now we have a relationship (\(b^2 = c^2\)) derived from the given angle. We can use this with the fundamental ellipse equation \(c^2 = a^2 - b^2\) to find the eccentricity \(e = \frac{c}{a}\).

1. Substitute \(b^2 = c^2\) into \(c^2 = a^2 - b^2\): \(c^2 = a^2 - c^2\)
2. Solve for \(a^2\) in terms of \(c^2\): \(2c^2 = a^2\)
3. Find Eccentricity (\(e\)): We know \(e = \frac{c}{a}\). To find \(e^2\), we can write: \(e^2 = \frac{c^2}{a^2}\)
4. Substitute \(a^2 = 2c^2\): \(e^2 = \frac{c^2}{2c^2}\) \(e^2 = \frac{1}{2}\)
5. Solve for \(e\): Taking the square root of both sides (eccentricity is positive): \(e = \sqrt{\frac{1}{2}}\) \(e = \frac{1}{\sqrt{2}}\)

Conclusion

The eccentricity of the ellipse, given that the angle between the straight lines joining the foci to an extremity of the minor axis is \(90^\circ\), is \(\frac{1}{\sqrt{2}}\).