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# If p is the length of the perpendicular from the origin to the line $\large\frac{x}{a}+\frac{y}{6}$$=1 then \large\frac{1}{a^2}+\frac{1}{b^2}=? \begin{array}{1 1}(a)\;\large\frac{1}{p^2}&(b)\;p^2\\(c)\;0&(d)\;1\end{array} Can you answer this question? ## 1 Answer 0 votes p=length of perpendicular from origin to \large\frac{x}{a}+\frac{y}{6}$$=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.