Question

A rod of length $l$ moves such that its ends $A$ and $B$ always lie on the lines $3x-y+5=0$ and $y+5=0$ respectively. The locus of the point $P,$ which divides $AB$ internally in the ratio $2:1,$ is $l^2=\frac{1}{k}(ax-by-5{)}^2+9(y+5{)}^2.$ Then

• A. $k=4,a+b=6$
• B. $k=3,a+b=5$
• C. $k=4,a+b=0$
• D. $k=3,a+b=4$

Answer 1

The correct option is A $k=4,a+b=6$


By internal section formula,
$x=\frac{2\beta +\alpha }{3},y=\frac{-10+3\alpha +5}{3}$
$\Rightarrow \alpha =\frac{3y+5}{3}$ and $\beta =\frac{9x-3y-5}{6}$

Also, $l^2=A{B}^2=(\alpha -\beta {)}^2+(3\alpha +10{)}^2$
$\Rightarrow l^2=\frac{1}{4}(3x-3y-5{)}^2+9(y+5{)}^2$