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 $A B$ 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 :

Accepted Answer

$$\begin{aligned}
& \mathrm{h}=\frac{3 \beta+\alpha}{3} \
& \mathrm{k}=\frac{-4+\alpha+2}{3} \
& \alpha=3 \mathrm{k}+2 \
& 2 \beta=3 \mathrm{~h}-\mathrm{a}=3 \mathrm{~h}-3 \mathrm{k}-2 \
& \text { so } \mathrm{AB}=8 \
& (\alpha-\beta)^2+(\alpha+4)^2=64 \
& \left(3 \mathrm{k}+2-\left(\frac{3 \mathrm{~h}-3 \mathrm{k}-2}{2}\right)\right)^2+(3 \mathrm{k}+2+4)^2=64 \
& \frac{(9 \mathrm{k}-3 \mathrm{~h}+6)^2}{4}+(3 \mathrm{k}+6)^2=64 \
& 9\left[(3 \mathrm{k}-\mathrm{h}+2)^2+4(\mathrm{k}+2)^2\right]=64 \times 4 \
& 9\left(\mathrm{x}^2+13 \mathrm{y}^2-6 \mathrm{xy}-4 \mathrm{x}+28 \mathrm{y}\right)=76 \
& \alpha-\beta-\gamma=13+6+4=23
\end{aligned}$$

Solutions · 1