Question

An ellipse has OB as semi-minor axis. F and F' are its foci and the angle FBF' is a right angle. Then eccentricity of the ellipse is

• A. $\frac{1}{\sqrt{3}}$
• B. $\frac{1}{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. None of these

Answer 1

Answer: (C)

$\frac{1}{\sqrt{2}}$

Answer 2

Let the ellipse be $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$. Then co-ordinates of F, B & F' are (ae, 0), (0, b) (-ae, 0) respectively.

The slope of BF = $\frac{b - 0}{0 - ae} - = \frac{b}{ae}$

& slope of $BF' = \frac{0 - b}{- ae - 0} = \frac{b}{ae}$

$\because\angle FBF' = 90^{0}$

⇒ Slope of BF x slope of BF' = -1

⇒ $- \frac{b^{2}}{a^{2}e^{2}} = - 1$

⇒ $b^{2} = a^{2}e^{2}$

⇒ $a^{2}\left( 1 - e^{2} \right) = a^{2}e^{2}$

$\therefore e = \frac{1}{\sqrt{2}}$