Question

In an ellipse with semi-minor axis OB, S and S' are its foci, \(a>b\) and \(\angle SBS'\) is a right angle, then the eccentricity of the ellipse is..

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

Hint:

Use the right angle at B with BS = BS' = a and SS' = 2ae.

Answer 1

The Correct Option is C

Step 1: Slope Method
Right angle at B means that $BS$ and $BS'$ have slopes with product $-1$: $\left(-\dfrac{b}{ae}\right)\left(\dfrac{b}{ae}\right)=-1$.

Step 2: Simplify
$b^2=a^2e^2$. Using $b^2=a^2(1-e^2)$, we get $1-e^2=e^2$.

Step 3: Result
$e^2=\dfrac{1}{2}$, so $e=\dfrac{1}{\sqrt{2}}$, option (C).

Final Answer:
Option (C). \[ \boxed{\text{(C) }\dfrac{1}{\sqrt{2}}} \]