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Question

Let P be a variable point on the ellipse fraction numerator x to the power of 2 end exponent over denominator a to the power of 2 end exponent end fraction+ fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction= 1 with foci F1 and F2. If A is the area of the triangle PF1 F2, then the maximum value of A is-

  1. 2abe    
  2. abe    
  3. fraction numerator 1 over denominator 2 end fractionabe    
  4. None of these    

hintHint:

assume an arbitrary point on the curve and find the value of A

The correct answer is: abe


    abe
    Let p be  (a cos t , b sin t)
    focii are (ae,0) and (-ae,0)

    Area A = ½(bsin t x 2ae) = abe sin t
    This is maximum when sin t = 1
    Maximum value of A is abe

    Area of triangle with given vertices can be calculated using the matrix determimnant method.

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