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If the tangent at Point P to the ellipse 16x2 + 11y2 = 256 is also the tangent to the circle x2 + y2 - 2x = 15, then the eccentric angle of point P is

  1. plus-or-minus fraction numerator pi over denominator 2 end fraction  
  2. plus-or-minus fraction numerator pi over denominator 4 end fraction  
  3. plus-or-minus fraction numerator pi over denominator 3 end fraction   
  4. plus-or-minus fraction numerator pi over denominator 6 end fraction  

The correct answer is: plus-or-minus fraction numerator pi over denominator 3 end fraction


    The equation of tangent at point to the ellipse
    16x2 + 11y2 = 256 is
    16x (4cosq) + 11y = 256
    4xcosq + y sinq = 16
    This touches the circle
    (x + 1)2 + y2 = 16
    So, = 4
    Þ (cosq - 4)2 = 11 + 5cos2q
    4cos2q + 8cosq - 5 = 0
    \ cosq =
    \ q = ±

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