If $$2p$$  is the length of perpendicular from the origin to the lines $$\frac{x}{a} + \frac{y}{b} = 1,$$   then $${a^2},\,8{p^2},\,{b^2}$$   are in :

Solution :
We have
$$\eqalign{ & 2p = \left| {\frac{{0 + 0 - 1}}{{\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}} }}} \right| \cr & \Rightarrow \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} = \frac{1}{{4{p^2}}} \cr & \Rightarrow \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} = \frac{2}{{8{p^2}}} \cr & \Rightarrow \frac{1}{{{a^2}}},\,\frac{1}{{8{p^2}}},\,\frac{1}{{{b^2}}}\,{\text{ are in A}}{\text{.P}}{\text{.}} \cr & \Rightarrow {a^2},\,8{p^2},\,{b^2}{\text{ }}\,{\text{are in H}}{\text{.P}}{\text{.}} \cr} $$