Question

If the angle between the lines joining the foci of an ellipse to an extremity of the minor axis is \(90^\circ\), and the major axis is \(2\sqrt{2}\), then the equation of ellipse and its eccentricity is:

• A. \(2x^2 + y^2 = 2; \frac{1}{2}\)
• B. \(2x^2 + y^2 = 1; \frac{1}{\sqrt{2}}\)
• C. \(x^2 + 2y^2 = 2; \frac{1}{\sqrt{2}}\)
• D. \(x^2 + y^2 = 2; \frac{1}{\sqrt{2}}\)

Hint:

Exam Tip:
For ellipses:

• \(c^2 = a^2 - b^2\).
• Eccentricity \(e = c/a\).
• If the angle between the lines from foci to an end of minor axis is \(90^\circ\), then \(b = c\).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We need to find the equation of an ellipse given the condition on the angle between the lines from the foci to an extremity of the minor axis.

Step 2: Key Formula or Approach:
Let the ellipse be \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) with \(a > b\). The foci are at \((\pm c, 0)\) where \(c^2 = a^2 - b^2\).
The extremity of the minor axis is \((0, b)\).
The lines from the foci to \((0, b)\) have slopes:
\(m_1 = \frac{b - 0}{0 - c} = -\frac{b}{c}\) and \(m_2 = \frac{b - 0}{0 - (-c)} = \frac{b}{c}\).
The angle between them is \(90^\circ\), so \(m_1 m_2 = -1\).
Thus, \(\left(-\frac{b}{c}\right)\left(\frac{b}{c}\right) = -1 \Rightarrow \frac{b^2}{c^2} = 1 \Rightarrow b^2 = c^2\).
Since \(c^2 = a^2 - b^2\), we get \(b^2 = a^2 - b^2 \Rightarrow a^2 = 2b^2\).
The major axis is \(2a = 2\sqrt{2} \Rightarrow a = \sqrt{2}\).
Then \(a^2 = 2 \Rightarrow b^2 = 1\).
So, the ellipse is \(\frac{x^2}{2} + \frac{y^2}{1} = 1 \Rightarrow x^2 + 2y^2 = 2\).
Eccentricity: \(e = \frac{c}{a} = \frac{b}{a} = \frac{1}{\sqrt{2}}\).
This matches option (C).

Step 4: Final Answer:
Therefore, option (C) is correct.