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$\mathrm{p}$ is the length of perpendicular from the origin to the line whose intercepts on the axes are a and $\mathrm{b}$ respectively, then $\frac{1}{\mathrm{a}^2}+\frac{1}{\mathrm{~b}^2}$ equals
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Verified Answer
The correct answer is:
$\frac{1}{\mathrm{p}^2}$
Let the equation of the line be $\frac{x}{\mathrm{a}}+\frac{y}{\mathrm{~b}}=1$
According to the given condition,
$$
\begin{aligned}
& \mathrm{p}=\left|\frac{\mathrm{ab}}{\sqrt{\mathrm{a}^2+b^2}}\right| \\
& \Rightarrow \frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{a}^2 \mathrm{~b}^2}=\frac{1}{\mathrm{p}^2} \\
& \Rightarrow \frac{1}{\mathrm{a}^2}+\frac{1}{\mathrm{~b}^2}=\frac{1}{\mathrm{p}^2}
\end{aligned}
$$
According to the given condition,
$$
\begin{aligned}
& \mathrm{p}=\left|\frac{\mathrm{ab}}{\sqrt{\mathrm{a}^2+b^2}}\right| \\
& \Rightarrow \frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{a}^2 \mathrm{~b}^2}=\frac{1}{\mathrm{p}^2} \\
& \Rightarrow \frac{1}{\mathrm{a}^2}+\frac{1}{\mathrm{~b}^2}=\frac{1}{\mathrm{p}^2}
\end{aligned}
$$
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