tanβ=4/3
sinβ = 4/5,cosβ = 3/5
a+b = (cosα+cosβ,(λ-1)sinβ+sinβ)
a-b = (cosα-cosβ,(λ-1)sinβ-sinβ)
(a+b).(a-b) =0
(cosα+cosβ,(λ-1)sinβ+sinβ).(cosα-cosβ,(λ-1)sinβ-sinβ)=0
(cosα)^2 - (cosβ)^2 + λ(λ-2)(sinβ)^2 =0
(cosα)^2 - 9/25 +λ(λ-2)(16/25) =0 (1)
a.b=4/5
(cosα,(λ-1)sinβ).(cosβ,sinβ) =4/5
cosαcosβ+(λ-1)(sinβ)^2 = 4/5
(3/5)cosα + (λ-1) (16/25) =4/5
15cosα + 16(λ-1) = 20 (2)
solving (1) and (2),we can get tanα
(cosα)^2 - 9/25 +λ(λ-2)(16/25) =0
((20 - 16(λ-1)) /15)^2 - 9/25 +λ(λ-2)(16/25) =0
(36-16λ)^2 - 81 + 144λ(λ-2) =0
400λ^2-1440λ+1215 =0
80λ^2-288+243=0
λ = (-80+72)/320 or (-80-72)/320
= -1/40 or 19/40