Maths-
General
Easy

Question

Set of values of ‘h’ for which the number of distinct common normals of (x – 2)2 = 4(y – 3) and x2 + y2 – 2x – hy – c = 0 (c > 0) is 3, is -

  1. (2, straight infinity)    
  2. (4, straight infinity)    
  3. (2, 4)    
  4. (10, straight infinity)    

The correct answer is: (10, straight infinity)


    The equation of any normal
    (x – 2)2 = 4(y – 3) is x – 2 = m (y – 3) – 2m – m3
    If it passes through (1, h/2), then
    1 – 2 = m open parentheses fraction numerator h over denominator 2 end fraction minus 3 close parentheses – 2m – m3 rightwards double arrow 2m3 + m
    (10 – h) – 2 = 0
    This equation will give three distinct values of m, if ƒ' (m) = 0 has two distinct roots,
    whereƒ(m) = 2m3 + m (10 – h) – 2
    Nowƒ'(m) = 6m2 + (10 – h)
    ƒ' (m) = 0 rightwards double arrow m ± square root of fraction numerator h minus 10 over denominator 6 end fraction end root
    The values of m are real and distinct if h > 10 i.e. h element of (10, straight infinity).
    Hence (D) is correct answer.

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