Q.183. Pair of lines

Question 183.
If θ is the acute angle between the lines 3x^2 + 4xy + by^2 = 0 and tan θ = 1/2,
then find b.

Answer 183.
3x^2 + 4xy + by^2 = 0
=> (y/x)^2 + (4/b) xy + 3/b = 0
Slopes, m1 and m2 of the lines are roots
of this quadratic eqn. in y/x
=> m1 + m2 = - 4/b and m1m2 = 3/b
=> m1 - m2 = √[(m1+m2)^2 - 4m1m2] = √(16/b^2 - 12/b)

tanθ = 1/2
=> (m1 - m2)/ (1 + m1m2) = 1/2
=> √(16/b^2 - 12/b) / (1 + 3/b) = 1/2
=> 4 (16/b^2 - 12/b) = (1 + 3/b)^2
=> 64/b^2 - 48/b = 1 + 6/b + 9/b^2
=> 64 - 48b = b^2 + 6b + 9
=> b^2 + 54b - 55 = 0
=> (b + 55) (b - 1) = 0
=> b = 1 or - 55.

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