If θ1 and θ2 be the angles which the lines (x2+y2)(cos2 θ sin2α+ sin2θ)= (x tanθ - y sinθ)2 make with the x axis and θ= π/6 , then what is the value of (tan θ1+ tan θ2) ?
answer:(-8/3 cosec 2α)
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3 Answers
62(x2+y2)(cos2 θ sin2α+ sin2θ)= (x tanθ - y sinθ)2
thus,
(x2+y2)(cot2 θ sin2α+ 1)= (x secθ - y)2
(x2+y2)(cot2 θ sin2α) +(x2+y2) = x2sec2θ + y2 - 2xysecθ
thus,
(x2+y2)(cot2 θ sin2α) = x2tan2θ - 2xysecθ
thus,
(x2+y2)(3sin2α) = x2/3 - 4/√3 xy
3sin2αx2+3sin2α y2 = x2/3 - 4/√3 xy
now this is a quadratic in m
y/x=mm
then,
3sin2α m2 + 4/√3m + 3sin2α -1/3 = 0
we have to find m1 + m2 = -4/(3√3) cosec2α
I dont know where i am going wrong :(