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 $9\left(x^2+\alpha y^2+\beta xy+\gamma x+28y\right)-76=0$, then $\alpha-\beta-\gamma$ is equal to :

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] Let $A(a,a+2)$, $B(b,-2)$ with $(a-b)^2+(a+4)^2=64$. $P=\frac{A+2B}3$ gives $y=\frac{a-2}3$, $x=\frac{a+2b}3$, so $a=3y+2$ and $b=\frac{3x-3y-2}2$. Then $a-b=\frac{9y-3x+6}2$ and $a+4=3y+6$. Substituting: $9(3y-x+2)^2+36(y+2)^2=256$, i.e. $9\left(x^2+13y^2-6xy-4x+28y\right)-76=0$. So $\alpha=13,\beta=-6,\gamma=-4$ and $\alpha-\beta-\gamma=23$. Answer (3).

Explanation

Reviewed by Dr. Abubacker

Want to attempt the full paper?