Question

A rod of length l slides between the two perpendicular lines.Find the locus of the point on the rod which divedes it in the ratio 1 : 2.

Answer 1

$\text{Let the two perpendicular lines be the coodrinates axes.Let AB be a rod length l.Let the coordinates of A and B be(a,0) and (0,b) respectively.As the rod slides the value of a and b change. So, a and b are two variables.}\text{Let P(h,k) be the point on the locus.Then}h=\frac{2\times a+1\times 0}{2+1}\Rightarrow h=\frac{2a}{3}\Rightarrow a=\frac{3h}{2}andk=\frac{2\times 0+b\times 1}{2+1}\Rightarrow k=\frac{b}{3}\Rightarrow b=3kfrom\Delta AOB,wehaveA{B}^2=O{A}^2+O{B}^2\Rightarrow l^2=\left[\right(a-0{)}^2+(0-0{)}^2]+\left[\right(0-0{)}^2+(b-0{)}^2]\Rightarrow l^2=a^2+b^2\Rightarrow a^2+b^2=l^2\Rightarrow {\left(\frac{3h}{2}\right)}^2+(3k{)}^2=l^2\Rightarrow \frac{9h^2}{4}+9k^2=l^2\Rightarrow \frac{h^2}{4}+k^2=\frac{l^2}{9}Hence,thelocusof(h,k)is\frac{x^2}{4}+y^2=\frac{l^2}{9}.$