If OB is the semi-minor axis of an ellipse,F1 and F2 are its foci and the angle between F1B and F2B is a right angle, then the square of the eccentricity of the ellipse is :

  • Option 1)

    \frac{1}{2}

  • Option 2)

    \frac{1}{\sqrt{2}}\;

  • Option 3)

    \frac{1}{2\sqrt{2}}\;

  • Option 4)

    \frac{1}{4}

 

Answers (1)

As we learnt in

Length of major axis -

2b

- wherein

b\rightarrow Semi minor axis

 

 and

Coordinates of foci -

\pm ae,o

- wherein

For the ellipse  

\frac{x^{2}}{a^{2}}+ \frac {y^{2}}{b^{2}}= 1

 Given \angle F_{1}BF_{2}=90^{o }

Also  m_{BF_{1}}\times m_{BF_{2}}=-1

\frac{-b}{ae}\times \frac{-b}{ae}=-1

b^{2}=a^{2}e^{2}

Also for ellipse

b2 = a2(1 - e2)

a2e2 = a2 - a2e2

e2 = 1/2

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

 


Option 1)

\frac{1}{2}

This is incorrect option

Option 2)

\frac{1}{\sqrt{2}}\;

This is correct option

Option 3)

\frac{1}{2\sqrt{2}}\;

This is incorrect option

Option 4)

\frac{1}{4}

This is incorrect option

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