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,…

$\begin{aligned} & A B=8 \\ & A B^2=64\end{aligned}$​AB=8AB2=64​

$\Rightarrow(a-b)^2+(b+4)^2=64$⇒(a−b)2+(b+4)2=64 ...(1) Now $P$P divides $A B$AB in the ratio $2: 1$2:1 internally $\Rightarrow h=\frac{2 a+b}{3}$⇒h=32a+b​ and $k=\frac{-4+b+2}{3}$k=3−4+b+2​ $\Rightarrow 2 a+b=3 h$⇒2a+b=3h $\ldots$… (2) $k=\frac{b-2}{3}$k=3b−2​ From equation (2) and (3) $\quad \Rightarrow b=3 \mathrm{k}+2$⇒b=3k+2 $\begin{aligned}
& \Rightarrow \quad 2 a=3 h-3 k-2 \
& \Rightarrow \quad a=\frac{3 h-3 k-2}{2}
\end{aligned}$Now by putting value of$Nowbyputtingvalueofa$and$andb$in equation$inequation\begin{aligned}
& \Rightarrow\left(\frac{3 h-3 k-2}{2}-(3 k+2)\right)^2+(3 k+2+4)^2=64 \
& \Rightarrow\left(\frac{3 h-3 k-2-6 k-4}{2}\right)^2+(3 k+6)^2=64 \
& \Rightarrow(3 h-9 k-6)^2+4(3 k+6)^2=4 \times 64 \
& \Rightarrow 9(h-3 k-2)^2+36(k+2)^2=256 \
& \Rightarrow 9\left(h^2+9 k^2+4-6 h k-4 h+12 k\right) \
& \quad+36\left(k^2+4+4 k\right)=256 \
& \Rightarrow 9\left(h^2+13 k^2+20-6 h k-4 h+28 k\right)=256
\end{aligned}$Replacing$Replacingh$by$byx$and$andk$by$byy$$\begin{aligned}
& \Rightarrow 9\left(x^2+13 y^2-6 x y-4 x+28 y\right)+180-256=0 \
& \Rightarrow 9\left(x^2+13 y^2-6 x y-4 x+28 y\right)-76=0
\end{aligned}$By comparing$Bycomparing\alpha=13, \beta=-6, \gamma=-4$$\alpha-\beta-\gamma=13+6+4=23$