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 $A B$ internally in the ratio $2: 1$ is $9\left(x^2+\alpha y^2+\beta x y+\gamma x+28 y\right)-76=0$, then…

Question

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 $A B$ internally in the ratio $2: 1$ is $9\left(x^2+\alpha y^2+\beta x y+\gamma x+28 y\right)-76=0$, then $\alpha-\beta-\gamma$ is equal to :

Aicharya · Accepted Answer

$\begin{aligned} & A B=8 \ & A B^2=64\end{aligned}$ $\Rightarrow(a-b)^2+(b+4)^2=64$ ...(1) Now $P$ divides $A B$ in the ratio $2: 1$ internally $\Rightarrow h=\frac{2 a+b}{3}$ and $k=\frac{-4+b+2}{3}$ $\Rightarrow 2 a+b=3 h$ $\ldots$ (2) $k=\frac{b-2}{3}$ From equation (2) and (3) $\quad \Rightarrow b=3 \mathrm{k}+2$ $\begin{aligned} & \Rightarrow \quad 2 a=3 h-3 k-2 \ & \Rightarrow \quad a=\frac{3 h-3 k-2}{2} \end{aligned}$ Now by putting value of $a$ and $b$ in equation $\begin{aligned} & \Rightarrow\left(\frac{3 h-3 k-2}{2}-(3 k+2)ight)^2+(3 k+2+4)^2=64 \ & \Rightarrow\left(\frac{3 h-3 k-2-6 k-4}{2}ight)^2+(3 k+6)^2=64 \ & \Rightarrow(3 h-9 k-6)^2+4(3 k+6)^2=4 	imes 64 \ & \Rightarrow 9(h-3 k-2)^2+36(k+2)^2=256 \ & \Rightarrow 9\left(h^2+9 k^2+4-6 h k-4 h+12 kight) \ & \quad+36\left(k^2+4+4 kight)=256 \ & \Rightarrow 9\left(h^2+13 k^2+20-6 h k-4 h+28 kight)=256 \end{aligned}$ Replacing $h$ by $x$ and $k$ by $y$ $\begin{aligned} & \Rightarrow 9\left(x^2+13 y^2-6 x y-4 x+28 yight)+180-256=0 \ & \Rightarrow 9\left(x^2+13 y^2-6 x y-4 x+28 yight)-76=0 \end{aligned}$ By comparing $\alpha=13, \beta=-6, \gamma=-4$ $\alpha-\beta-\gamma=13+6+4=23$ .

A rod of length eight units moves such that its ends $A$ and $B$ always lie on the lines $x-y+2=0$ and…

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