If for θ∈[−3π,0], the points (x,y)=(3tan(θ+3π),2tan(θ+6π)) lie on xy+αx+βy+γ=0, then α2+β2+γ2 is equal to:
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Step-by-Step Solution
Step 1: Define x and y in terms of new variables
Let's simplify the expressions for x and y by introducing new variables. We define A=θ+3π and B=θ+6π. This makes the expressions for x and y more manageable.
Step 2: Find the relationship between A and B
Now, let's find the relationship between A and B. Subtracting B from A, we get (θ+3π)−(θ+6π)=3π−6π=62π−π=6π. This constant difference will be useful.
Step 3: Apply the tangent subtraction formula
We know the value of A - B, so we can use the tangent subtraction formula. The formula states that tan(A−B) is equal to (tanA−tanB) divided by (1+tanAtanB).
Step 4: Substitute and rearrange the equation
Substitute tan(A−B)=tan(6π)=31, tanA=3x, and tanB=2y into the tangent formula. After cross-multiplication and rearrangement, we get xy−2x+3y−18=0. This is the equation of the locus.
Step 5: Compare with the given equation to find alpha, beta, gamma
The given equation is xy+αx+βy+γ=0. By comparing this with our derived equation xy−2x+3y−18=0, we can identify the values of α, β, and γ.
Step 6: Calculate the final expression
Finally, we need to calculate α2+β2+γ2. Substitute the values we found: (−2)2+(3)2+(−18)2=4+9+324=337.