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Question

P is a 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 plus fraction numerator y to the power of 2 end exponent over denominator b to the power of 2 end exponent end fraction equals 1. The in-radius of ΔPSS’ (S and S’ are focii), where its area is maximum.

  1. fraction numerator b e over denominator 1 plus e end fraction    
  2. fraction numerator b left parenthesis 1 plus e right parenthesis over denominator e end fraction    
  3. fraction numerator a e over denominator 1 plus e end fraction    
  4. None of these    

hintHint:

find the dimensions of the  triangle and find the values of semiperimetre and Area. use the formula r = A/ s to find the in radius.

The correct answer is: fraction numerator b e over denominator 1 plus e end fraction


    fraction numerator b e over denominator 1 plus e end fraction
    let P be (a cos t, b sin t)
    area of triangle PSS’ = ½ (2ae)(b sin t) = abe (sin t)
    area is maximum when t= 90
    hence, point P is (0,b)
    area of triangle = abe
    length of side PS = √a2e2+b2
    semi perimeter = (2ae + 2√a2e2+b2)/2
    = ae + √a2e2+b2
    In radius = abe / (ae + √a2e2+b2)
    = abe /( ae + a ) = be /(1+e)

    in radius is the radius of the circle inscribed inside the triangle.

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