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(C) The length of the perpendicular from the origin $(0,0)$ to the line $\frac{x}{a} + \frac{y}{b} - 1 = 0$ is given by $2p = \left| \frac{-1}{\sqrt{\

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$H.P.$

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(C) The length of the perpendicular from the origin $(0,0)$ to the line $\frac{x}{a} + \frac{y}{b} - 1 = 0$ is given by $2p = \left| \frac{-1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} \right|$.Squaring both sides,we get $4p^2 = \frac{1}{\frac{1}{a^2} + \frac{1}{b^2}}$,which implies $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{4p^2}$.Multiplying by $2$,we get $\frac{2}{a^2} + \frac{2}{b^2} = \frac{2}{4p^2} = \frac{1}{2p^2} = \frac{4}{8p^2}$.This can be rewritten as $\frac{1}{a^2} + \frac{1}{b^2} = \frac{2}{8p^2}$.This shows that $\frac{1}{a^2}, \frac{1}{8p^2}, \frac{1}{b^2}$ are in $A.P.$Therefore,$a^2, 8p^2, b^2$ are in $H.P.$

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

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