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Question

If  alpha comma beta are the eccentric angles of the 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


    open parentheses a cos invisible function application alpha comma b sin invisible function application alpha close parentheses comma open parentheses a cos invisible function application beta comma b sin invisible function application beta close parentheses comma open parentheses a e comma 0 close parentheses are collinear.
    fraction numerator b left parenthesis sin invisible function application beta – sin invisible function application alpha right parenthesis over denominator a left parenthesis cos invisible function application beta – cos invisible function application alpha right parenthesis end fraction equals fraction numerator b sin invisible function application alpha – 0 over denominator a cos invisible function application alpha – a e end fraction
    therefore left parenthesis c o s invisible function application alpha minus e right parenthesis left parenthesis s i n invisible function application beta minus s i n invisible function application alpha right parenthesis equals s i n invisible function application alpha left parenthesis c o s invisible function application beta minus c o s invisible function application alpha right parenthesis
    therefore e =blank fraction numerator cos invisible function application alpha left parenthesis sin invisible function application beta – sin invisible function application alpha right parenthesis – sin invisible function application alpha left parenthesis cos invisible function application beta – cos invisible function application alpha right parenthesis over denominator sin invisible function application beta – sin invisible function application alpha end fraction = fraction numerator sin invisible function application left parenthesis alpha – beta right parenthesis over denominator sin invisible function application alpha – sin invisible function application beta end fraction blank= 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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