Understanding the Ellipse Problem

The question asks us to find the eccentricity of an ellipse given a specific geometric condition. The condition states that the angle formed by the lines connecting the endpoints of the minor axis to one of the foci is 90 degrees (\(\frac{\pi}{2}\) radians).

Let's consider a standard ellipse with the equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), where \(a\) is the semi-major axis length and \(b\) is the semi-minor axis length. We assume \(a > b\). The foci of this ellipse are located at \((\pm ae, 0)\), where \(e\) is the eccentricity. The endpoints of the minor axis (also called co-vertices) are located at \((0, \pm b)\).

We are given that the angle between the lines joining the endpoints of the minor axis with one of its foci is \(\frac{\pi}{2}\). Let's choose the focus \(F_2\) at \((ae, 0)\) and the minor axis endpoints \(B_1(0, b)\) and \(B_2(0, -b)\). The condition is that the angle \(\angle B_1F_2B_2 = \frac{\pi}{2}\).

Method 1: Using Slopes of Perpendicular Lines

If two lines are perpendicular, the product of their slopes is -1 (provided neither line is vertical). The lines in question are the line segment \(B_1F_2\) and the line segment \(B_2F_2\).

Let's find the slope of the line \(B_1F_2\) passing through points \((0, b)\) and \((ae, 0)\). The slope \(m_1\) is given by:

\(m_1 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - b}{ae - 0} = \frac{-b}{ae}\)

Next, let's find the slope of the line \(B_2F_2\) passing through points \((0, -b)\) and \((ae, 0)\). The slope \(m_2\) is given by:

\(m_2 = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - (-b)}{ae - 0} = \frac{b}{ae}\)

Since the angle between these two lines is \(\frac{\pi}{2}\), they are perpendicular. Therefore, the product of their slopes must be -1:

\(m_1 \times m_2 = -1\)

\(\left(\frac{-b}{ae}\right) \times \left(\frac{b}{ae}\right) = -1\)

\(\frac{-b^2}{a^2e^2} = -1\)

\(\frac{b^2}{a^2e^2} = 1\)

\(b^2 = a^2e^2\)

We know the relationship between \(a\), \(b\), and \(e\) for an ellipse is \(b^2 = a^2(1-e^2)\) (assuming \(a > b\)). Substituting this into the equation above:

\(a^2(1-e^2) = a^2e^2\)

Assuming \(a \neq 0\), we can divide both sides by \(a^2\):

\(1-e^2 = e^2\)

\(1 = 2e^2\)

\(e^2 = \frac{1}{2}\)

Since eccentricity \(e\) must be positive for an ellipse, we take the positive square root:

\(e = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}\)

Method 2: Using the Pythagorean Theorem

Consider the triangle formed by the points \(B_1(0, b)\), \(F_2(ae, 0)\), and \(B_2(0, -b)\). The angle at \(F_2\) is given as \(\frac{\pi}{2}\), which means \(\triangle B_1F_2B_2\) is a right-angled triangle with the right angle at \(F_2\).

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. The hypotenuse is the side opposite the right angle, which is \(B_1B_2\). The other two sides are \(F_2B_1\) and \(F_2B_2\).

Length of \(B_1B_2\): The distance between \((0, b)\) and \((0, -b)\) is \(b - (-b) = 2b\). So, \((B_1B_2)^2 = (2b)^2 = 4b^2\).

Length of \(F_2B_1\): The distance between \((ae, 0)\) and \((0, b)\) is \(\sqrt{(ae-0)^2 + (0-b)^2} = \sqrt{a^2e^2 + b^2}\). So, \((F_2B_1)^2 = a^2e^2 + b^2\).

Length of \(F_2B_2\): The distance between \((ae, 0)\) and \((0, -b)\) is \(\sqrt{(ae-0)^2 + (0-(-b))^2} = \sqrt{a^2e^2 + b^2}\). So, \((F_2B_2)^2 = a^2e^2 + b^2\).

Applying the Pythagorean theorem \((B_1B_2)^2 = (F_2B_1)^2 + (F_2B_2)^2\):

\(4b^2 = (a^2e^2 + b^2) + (a^2e^2 + b^2)\)

\(4b^2 = 2a^2e^2 + 2b^2\)

Subtract \(2b^2\) from both sides:

\(2b^2 = 2a^2e^2\)

\(b^2 = a^2e^2\)

Again, substitute the relation \(b^2 = a^2(1-e^2)\) for an ellipse:

\(a^2(1-e^2) = a^2e^2\)

Divide by \(a^2\):

\(1-e^2 = e^2\)

\(1 = 2e^2\)

\(e^2 = \frac{1}{2}\)

Taking the positive square root for eccentricity:

\(e = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}\)

Conclusion on Eccentricity

Both methods demonstrate that the eccentricity of the ellipse under the given condition is \(\frac{1}{\sqrt{2}}\). This matches one of the provided options.

Revision Table: Key Ellipse Properties

Additional Information: Eccentricity and Ellipse Shape

The eccentricity \(e\) of an ellipse is a measure of how much the ellipse deviates from a perfect circle. It is defined as the ratio of the distance from the center to the focus (\(ae\)) to the distance from the center to the vertex (\(a\)), i.e., \(e = \frac{ae}{a}\). Its value always lies between 0 and 1 for an ellipse (\(0 < e < 1\)).

• If \(e = 0\), the foci coincide with the center, and the ellipse is a circle.
• As \(e\) increases from 0 towards 1, the ellipse becomes more elongated or 'flatter'.
• If \(e\) approaches 1, the ellipse becomes very elongated.
• The case \(e=1\) corresponds to a parabola (an unbounded curve).
• Values of \(e > 1\) correspond to a hyperbola (another type of unbounded curve).

In this problem, we found \(e = \frac{1}{\sqrt{2}}\), which is approximately 0.707. This value is between 0 and 1, as expected for an ellipse, indicating a moderately elongated shape.