If for θ ∈ [-(π)/3, 0 ] , the points (x, y)= (3 tan (θ+(π)/3 ), 2 tan (θ+(π)/6 ) ) lie on x y+α x+β y+γ=0 , then α^2+β^2+γ^2 is equal to

Question

If for θ ∈ [-(π)/3, 0 ] , the points (x, y)= (3 tan (θ+(π)/3 ), 2 tan (θ+(π)/6 ) ) lie on x y+α x+β y+γ=0 , then α^2+β^2+γ^2 is equal to :

MARKS Content Team · Accepted Answer

Correct answer: (D) 75. x=3 ((tan θ+√3)/(1-√3 tan θ) ) x-√3 tan θ=3 tan θ+3 √3 tan θ=(x-3 √3)/(3+√3 x) …(1) 2 ((tan θ+1/(√3))/(1-(tan θ)/(√3))=y ) 2(√3 tan θ+1)=y(√3-tan θ) … . using (1) and (2) 2 ((x-3 √3)/(√3+x)+1 )=y (√3-((x-3 √3))/(√3(√3+x)) ) 2 √3(x-3 √3+x+√3)=y(3(√3+x)-x+3 √3) 4 √3 x-12=y(2 x+6 √3) x y-2 √3 x+3 √3 y-6=0 ⇒ α=-2 √3, β=3 √3, γ=-6 α^2+β^2+γ^2=12+27+36=75

If for $\theta \in\left[-\frac{\pi}{3}, 0\right]$, the points $(x, y)=\left(3 \tan \left(\theta+\frac{\pi}{3}\right), 2 \tan \left(\theta+\frac{\pi}{6}\right)\right)$ lie on $x y+\alpha x+\beta y+\gamma=0$, then $\alpha^2+\beta^2+\gamma^2$ is equal to :

Solution

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