Maths-
General
Easy

Question

A tangent to the ellipse fraction numerator x to the power of 2 end exponent over denominator 9 end fraction+ fraction numerator y to the power of 2 end exponent over denominator 4 end fraction = 1 is cut by the tangent at the extremities of the major axis at T and T' and the circle on TT' as diameter passes through the point Q, then Q may be -

  1. (–square root of 5, 0)    
  2. (2, 3)    
  3. (0, 0)    
  4. (3, 2)    

The correct answer is: (–square root of 5, 0)


    fraction numerator x to the power of 2 end exponent over denominator 9 end fraction plus fraction numerator y to the power of 2 end exponent over denominator 4 end fraction equals 1

    equation of tangent
    fraction numerator x left parenthesis 3 cos invisible function application theta right parenthesis over denominator 9 end fraction times plus times fraction numerator y left parenthesis 2 sin invisible function application theta right parenthesis over denominator 4 end fraction times equals times 1
    fraction numerator x cos invisible function application theta over denominator 3 end fraction + fraction numerator y sin invisible function application theta over denominator 2 end fraction = 1
    T open parentheses 3 comma fraction numerator 2 left parenthesis 1 – cos invisible function application theta right parenthesis over denominator sin invisible function application theta end fraction close parentheses, T ´ open parentheses – 3 comma fraction numerator 2 left parenthesis 1 plus cos invisible function application theta right parenthesis over denominator sin invisible function application theta end fraction close parentheses
    equation of circle TT' as diameter
    (x+3) (x – 3) + open parentheses y – fraction numerator 2 left parenthesis 1 – cos invisible function application theta right parenthesis over denominator sin invisible function application theta end fraction close parentheses open parentheses y – fraction numerator 2 left parenthesis 1 plus cos invisible function application theta right parenthesis over denominator sin invisible function application theta end fraction close parentheses = 0
    x2 – 9 + y2 + fraction numerator 4 over denominator sin invisible function application theta end fractionsin2theta – 4cosec y = 0
    x2 + y2 – 4y cosectheta – 5 = 0
    which satisfied by 1st option only.

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