A variable line is at constant distance p from the origin and meets the co-ordinate axes in $A$ and $B.$ Show that the locus of the centroid of the$∆OAB$ is $\frac{1}{x^2}+\frac{1}{y^2}=\frac{9}{p^2}.$

Let the variable line be $\frac{x}{a}+\frac{y}{b}=1$ and the coordinates of $O, A, B$ are $(0,0), (a,0), (0,b).$

Let the centroid be $G(\alpha , \beta ).$

Thus, $\alpha =\frac{a}{3},\beta =\frac{b}{3}$

$\Rightarrow a=3\alpha ,b=3\beta$

It is given that 

$\begin{array}{l}OM=p\\ \Rightarrow \left|\frac{0+0-1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}\right|=p\\ \Rightarrow \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}\end{array}$

$\begin{array}{l}\Rightarrow \frac{1}{(3\alpha {)}^2}+\frac{1}{(3\beta {)}^2}=\frac{1}{p^2}\\ \Rightarrow \frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=\frac{9}{p^2}\end{array}$

Hence, the locus of $G\left(\alpha , \beta \right)$ is 

$\frac{1}{x^2}+\frac{1}{y^2}=\frac{9}{p^2}$