$I f p i s t h e l e n g t h o f t h e p e r p e n d i c u l a r f r o m t h e o r i g i n t o t h e l i n e \frac{x}{a} + \frac{y}{b} = 1 t h e n$

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

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

$\frac{1}{a^{2}} - \frac{1}{b^{2}} = \frac{1}{p^{2}}$

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

answer is A.

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Detailed Solution

$T h e p e r p e n d i c u l a r d i s \tan c e f r o m o r i g i n t o t h e l i n e a x + b y + c = 0 i s |\frac{c}{\sqrt{a^{2} + b^{2}}}| T h e d i s \tan c e f r o m o r i g i n t o t h e l i n e \frac{x}{a} + \frac{y}{b} = 1 i s p g i v e n p = \frac{1}{\sqrt{\frac{1}{a^{2}} + \frac{1}{b^{2}}}} t a k e r e c i p r o c a l s a n d t h e n s q u a r i n g b o t h s i d e s \frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{p^{2}}$