A line AB of length 2l moves with the end A always on the x-axis and the end B on the line y = 6x. The equation of the locus of the middle point of AB is

Let A=(a, 0), B=(b, 6b) which lies on the line y=6x

Let M=(h, k) be mid point of AB

$$\begin{gathered} \Rightarrow (\mathrm{h}, \mathrm{k})=\left(\frac{\mathrm{a}+\mathrm{b}}{2}, \frac{6\mathrm{b}}{2}\right) \\ \Rightarrow \mathrm{h}=\frac{\mathrm{a}+\mathrm{b}}{2}, \mathrm{k}=3\mathrm{b} \\ \Rightarrow \mathrm{a}=2\mathrm{h}-\mathrm{b}, \mathrm{b}=\frac{\mathrm{k}}{3} \\ \Rightarrow \mathrm{a}=2\mathrm{h}-\frac{\mathrm{k}}{3}, \mathrm{b}=\frac{\mathrm{k}}{3} \\ \mathrm{Given} \mathrm{AB}=2\mathrm{l} \\ \Rightarrow {\mathrm{AB}}^2=4{\mathrm{l}}^2 \\ \Rightarrow {(\mathrm{a}-\mathrm{b})}^2+{\left(6\mathrm{b}\right)}^2=4{\mathrm{l}}^2 \\ \Rightarrow {\left(2\mathrm{h}-\frac{\mathrm{k}}{3}-\frac{\mathrm{k}}{3}\right)}^2+4{\mathrm{k}}^2=4{\mathrm{l}}^2 \\ \Rightarrow {\left(2\mathrm{h}-\frac{2\mathrm{k}}{3}\right)}^2+4{\mathrm{k}}^2=4{\mathrm{l}}^2 \\ \Rightarrow {\left(\mathrm{h}-\frac{\mathrm{k}}{3}\right)}^2+{\mathrm{k}}^2={\mathrm{l}}^2 \\ \Rightarrow {\mathrm{h}}^2+\frac{{\mathrm{k}}^2}{9}-\frac{2\mathrm{hk}}{3}+{\mathrm{k}}^2={\mathrm{l}}^2 \\ \Rightarrow 9{\mathrm{h}}^2+{\mathrm{k}}^2-6\mathrm{hk}+9{\mathrm{k}}^2=9{\mathrm{l}}^2 \\ \therefore \mathrm{locus} \mathrm{of} (\mathrm{h}, \mathrm{k}) \mathrm{is} 9{\mathrm{x}}^2+10y^2-6\mathrm{xy}-9{\mathrm{l}}^2=0 \end{gathered}$$