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If alpha comma beta are the eccentric angles of the extremities of a focal chord of the ellipse fraction numerator x to the power of 2 end exponent over denominator 16 end fraction plus fraction numerator y to the power of 2 end exponent over denominator 9 end fraction equals 1, then tan left parenthesis alpha divided by 2 right parenthesis t a n invisible function application left parenthesis beta divided by 2 right parenthesis equals

  1. fraction numerator square root of 7 plus 4 over denominator square root of 7 minus 4 end fraction    
  2. negative fraction numerator 9 over denominator 23 end fraction    
  3. fraction numerator square root of 5 minus 4 over denominator square root of 5 plus 4 end fraction    
  4. fraction numerator 8 square root of 7 minus 23 over denominator 9 end fraction    

The correct answer is: fraction numerator 8 square root of 7 minus 23 over denominator 9 end fraction


    The equation of the ellipse is of the form x to the power of 2 end exponent divided by a to the power of 2 end exponent plus y to the power of 2 end exponent divided by b to the power of 2 end exponent equals 1 text  where  end text a to the power of 2 end exponent equals 16 comma b to the power of 2 end exponent equals 9
    thereforethe eccentricity e = square root of 1 minus fraction numerator 9 over denominator 16 end fraction end root=blank fraction numerator square root of 7 over denominator 4 end fraction.
    Let P(4 cos alpha, 3 sin alpha) and Q (4 cos beta, 3 sin beta) be a focal chord of the ellipse passing through the focus at (square root of 7, 0).
    Then fraction numerator 3 sin invisible function application beta over denominator 4 cos invisible function application beta minus square root of 7 end fraction=fraction numerator 3 sin invisible function application alpha over denominator 4 cos invisible function application alpha minus square root of 7 end fraction
    rightwards double arrowfraction numerator sin invisible function application left parenthesis alpha minus beta right parenthesis over denominator sin invisible function application alpha minus sin invisible function application beta end fraction=fraction numerator square root of 7 over denominator 4 end fraction
    rightwards double arrowfraction numerator cos invisible function application left square bracket left parenthesis alpha minus beta right parenthesis divided by 2 right square bracket over denominator cos invisible function application left square bracket left parenthesis alpha plus beta right parenthesis divided by 2 right square bracket end fraction = fraction numerator square root of 7 over denominator 4 end fraction
    rightwards double arrowtan open parentheses fraction numerator alpha over denominator 2 end fraction close parenthesestan open parentheses fraction numerator beta over denominator 2 end fraction close parentheses = fraction numerator square root of 7 minus 4 over denominator square root of 7 plus 4 end fraction= fraction numerator 23 minus 8 square root of 7 over denominator negative 9 end fraction.

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