$\begin{array}{1 1}(a)\;\large\frac{1}{p^2}&(b)\;p^2\\(c)\;0&(d)\;1\end{array}$

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p=length of perpendicular from origin to $\large\frac{x}{a}+\frac{y}{6}$$=1$

$p=\large\frac{\mid 0+0-1\mid}{\sqrt{\Large\frac{1}{a^2}+\frac{1}{b^2}}}$

$\Rightarrow \large\frac{1}{\sqrt{\Large\frac{1}{a^2}+\frac{1}{b^2}}}$

$p=\large\frac{1}{\sqrt{\Large\frac{1}{a^2}+\frac{1}{b^2}}}$

$p^2(\large\frac{1}{a^2}+\frac{1}{b^2})$$=1$

$(\large\frac{1}{a^2}+\frac{1}{b^2})$$=\large\frac{1}{p^2}$

Hence (a) is the correct answer.

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