A rod of length eight units moves such that its ends A and B always lie on the lines x – y + 2 = 0 and y + 2 = 0, respectively. If the locus of the point P, that divides the rod AB internally in the ratio 2 : 1 is $9 (x^{2} + {αy}^{2} + βxy + γx + 28 y) - 76 = 0$ , then $α - β - γ$ is equal to :

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answer is B.

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Detailed Solution

$\begin{array}{l} h = \frac{3 β + α}{3} \\ k = \frac{- 4 + α + 2}{3} \\ α = 3 k + 2 \\ 2 β = 3 h - a = 3 h - 3 k - 2 \end{array}$
so AB = 8
$\begin{array}{l} (α - β)^{2} + (α + 4)^{2} = 64 \\ {(3 k + 2 - (\frac{3 h - 3 k - 2}{2}))}^{2} + (3 k + 2 + 4)^{2} = 64 \\ \frac{(9 k - 3 h + 6)^{2}}{4} + (3 k + 6)^{2} = 64 \\ 9 [(3 k - h + 2)^{2} + 4 (k + 2)^{2}] = 64 \times 4 \\ 9 (x^{2} + 13 y^{2} - 6 xy - 4 x + 28 y) = 76 \\ α - β - γ = 13 + 6 + 4 = 23 \end{array}$