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

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 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:

Accepted Answer

Let $A$ be on $x-y+2=0$ and $B$ on $y+2=0$, i.e. $B=(b,-2)$. $P(h,k)$ divides $AB$ in ratio $2:1$ from $A$ (so $AP:PB=2:1$):
$$h = \frac{2b+a}{3}, \qquad k = \frac{2(-2)+\alpha_A}{3}$$
where $A=(a,a+2)$ (since $A$ lies on $x-y+2=0\Rightarrow y=x+2$).

Working through the elimination (as in the source), with $AB=8$:
$$(h-k)^2/\text{terms} + (k+2)^2\ \text{terms} = 64$$
leads after simplification to
$9(3k-h+2)^2+36(k+2)^2 = 256$
which expands to
$9(x^2+13y^2-6xy-4x+28y) = 76$

Comparing with $9(x^2+\alpha y^2+\beta xy+\gamma x+28y)-76=0$: $\alpha=13,\ \beta=-6,\ \gamma=-4$.

$$\alpha-\beta-\gamma = 13-(-6)-(-4) = 13+6+4=23$$

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