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If alpha and beta are the eccentric angles of extremities of a focal chord of an ellipse, then the eccentricity of the ellipse is -

  1. fraction numerator cos invisible function application alpha plus cos invisible function application beta over denominator cos invisible function application left parenthesis alpha plus beta right parenthesis end fraction    
  2. fraction numerator sin invisible function application alpha – sin invisible function application beta over denominator sin invisible function application left parenthesis alpha – beta right parenthesis end fraction    
  3. fraction numerator cos invisible function application alpha – cos invisible function application beta over denominator cos invisible function application left parenthesis alpha – beta right parenthesis end fraction    
  4. fraction numerator sin invisible function application alpha plus sin invisible function application beta over denominator sin invisible function application left parenthesis alpha plus beta right parenthesis end fraction    

The correct answer is: fraction numerator sin invisible function application alpha plus sin invisible function application beta over denominator sin invisible function application left parenthesis alpha plus beta right parenthesis end fraction


    Equation of chord be
    fraction numerator x over denominator a end fractioncos fraction numerator alpha plus beta over denominator 2 end fraction + fraction numerator y over denominator b end fractionsinfraction numerator alpha plus beta over denominator 2 end fraction = cosfraction numerator alpha – beta over denominator 2 end fraction
    since it passes through (ae, 0)
    so e cos fraction numerator alpha plus beta over denominator 2 end fraction = cosfraction numerator alpha – beta over denominator 2 end fraction
    e = fraction numerator cos invisible function application fraction numerator alpha – beta over denominator 2 end fraction over denominator cos invisible function application fraction numerator alpha plus beta over denominator 2 end fraction end fraction = fraction numerator 2 cos invisible function application fraction numerator alpha – beta over denominator 2 end fraction sin invisible function application fraction numerator alpha plus beta over denominator 2 end fraction over denominator sin invisible function application left parenthesis alpha plus beta right parenthesis end fraction
    e = fraction numerator sin invisible function application alpha plus sin invisible function application beta over denominator sin invisible function application left parenthesis alpha plus beta right parenthesis end fraction

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