Question

If the angle between the lines joining the foci and the ends of the minor axis of the ellipse $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$ is $ 90^\circ $, then its **eccentricity** is:

• A. \( \frac{1}{2} \)
• B. \( \frac{1}{4} \)
• C. \( \frac{1}{3} \)
• D. \( \frac{1}{\sqrt{2}} \)

Hint:

Use vector dot product and properties of ellipse to connect eccentricity with geometry.

Answer 1

The Correct Option is D

For ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \): - Foci: \( (\pm ae, 0) \) - Ends of minor axis: \( (0, \pm b) \) Given: angle between line segments from foci to ends of minor axis is \( 90^\circ \) So, vectors from focus \( F = (ae, 0) \) to: - \( B = (0, b) \) → vector \( \vec{v}_1 = (-ae, b) \) - \( B' = (0, -b) \) → vector \( \vec{v}_2 = (-ae, -b) \) Angle between vectors \( \vec{v}_1 \) and \( \vec{v}_2 \) is \( 90^\circ \) Hence, their dot product: \[ \vec{v}_1 \cdot \vec{v}_2 = (-ae)^2 + b \cdot (-b) = a^2 e^2 - b^2 = 0 \Rightarrow a^2 e^2 = b^2 \] But \( b^2 = a^2(1 - e^2) \Rightarrow a^2 e^2 = a^2(1 - e^2) \) \[ \Rightarrow e^2 = 1 - e^2 \Rightarrow 2e^2 = 1 \Rightarrow e = \boxed{ \frac{1}{\sqrt{2}} } \]