Question

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

विकल्प

• `1/(2)`

• `1/2`

• `1/4`

• `1/sqrt(3)`

Answer 1

`1/(2)`

Explanation:

Since, ∠FBF' = 90°, then ∠OBF' = 45° and ∠BF'O = 45°

⇒ ae = b  .......[Since, ΔBOF' is an isosceles triangle]

And `e^2 = 1 - b^2/a^2`

⇒ `e^2 = 1 - (a^2e^2)/a^2`

⇒ `e^2 = 1 + e^2`

⇒ `2e^2` = 1

∴ `e = 1/sqrt(2)`  .......[Since, e cannot be negative]

Alternate solution: Since, F and F' are foci of an ellipse whose coordinates are (ae, 0) and (– ae, 0) respectively and coordinates of B are (0 , b)

∴ Slope of  BF = `b/(- ae)`

And slope of BF' = `b/(ae)`

∵ ∠FBF' = 90°

∴ `- b/(ae) * (b/ae)` = − 1

⇒ `b^2 = a^2e^2`

∴ `e^2 = 1 - b^2/a^2 = 1 - (a^2e^2)/a^2`

⇒ 2e² = 1

∴ e = `1/sqrt(2)`