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 $9(x^2+\alpha y^2+\beta xy+\gamma x+28y)-76 = 0$, then $\alpha-\beta-\gamma$ is equal to:
Step-by-Step Solution
Key Concept: Parametrise $A=(b,b+2)$ on $x-y+2=0$ and $B=(a,-2)$ on $y=-2$; apply $AB^2=64$ and the section formula for $P(h,k)$ dividing $AB$ in ratio $2:1$ to get $a,b$ in terms of $h,k$; substitute into the distance condition.
$A=(b,b+2)$, $B=(a,-2)$. $P(h,k)$ divides $AB$ in $2:1$: $h=\tfrac{2a+b}{3}$, $k=\tfrac{-4+b+2}{3}=\tfrac{b-2}{3}$.
So $b=3k+2$, $2a=3h-b=3h-3k-2$, $a=\tfrac{3h-3k-2}{2}$.
$AB^2=64$: $(a-b)^2+(b+2+2)^2=64$... After algebra:
$$9(h^2+13k^2-6hk-4h+28k)-76=0.$$
$\alpha=13$, $\beta=-6$, $\gamma=-4$.
$$\alpha-\beta-\gamma = 13+6+4 = 23.$$
Correct Answer: 3