Question

An ellipse has \( OB \) as the semi-minor axis, and \( S \), \( S' \) as the foci. If \( \angle SBS' \) is a right angle, then the eccentricity \( e \) of the ellipse is:

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

Hint:

When dealing with angles involving the foci of an ellipse, remember to use coordinate geometry and right-angle triangle properties. Use symmetry and the standard definitions of ellipse parameters: \( c^2 = a^2 - b^2 \) and \( e = \frac{c}{a} \).

Answer 1

The Correct Option is C

We are given: - \( OB \) is the semi-minor axis (\( b \)), - \( S \), \( S' \) are the foci of the ellipse, - \( \angle SBS' = 90^\circ \) We are to find the eccentricity \( e \) of the ellipse. Step 1: Basic properties of ellipse - For an ellipse with semi-major axis \( a \) and semi-minor axis \( b \), the focal length is: \[ c = ae = \sqrt{a^2 - b^2} \] - Coordinates: focus at \( (\pm c, 0) \), and point \( B \) on the ellipse is \( (0, b) \) Step 2: Geometry of triangle \( \triangle SBS' \) We are told \( \angle SBS' = 90^\circ \). That means triangle \( SBS' \) is a right-angled triangle at vertex \( B \). Use the converse of the circle property: If angle \( B \) in triangle \( SBS' \) is \( 90^\circ \), then \( SB \perp S'B \). Also, this triangle lies on a circle with diameter \( SS' \). So, apply the Pythagorean Theorem in triangle \( SBS' \): \[ SB^2 + S'B^2 = SS'^2 \] But due to symmetry: \( SB = S'B \), and \( SS' = 2c \) So: \[ 2 \cdot SB^2 = (2c)^2 = 4c^2 \Rightarrow SB^2 = 2c^2 \] But point \( B = (0, b) \), and focus \( S = (c, 0) \) Then: \[ SB^2 = (c - 0)^2 + (0 - b)^2 = c^2 + b^2 \Rightarrow c^2 + b^2 = 2c^2 \Rightarrow b^2 = c^2 \Rightarrow \frac{b^2}{a^2} = \frac{c^2}{a^2} = e^2 \Rightarrow e = \frac{b}{a} \] But since \( b^2 = a^2 - c^2 \) and \( c^2 = b^2 \), then: \[ b^2 = a^2 - b^2 \Rightarrow 2b^2 = a^2 \Rightarrow \frac{b^2}{a^2} = \frac{1}{2} \Rightarrow e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{1}{2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \] Hence, the eccentricity of the ellipse is \( \boxed{\frac{1}{\sqrt{2}}} \)