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If $\mathrm{p}$ is the length of the perpendicular from origin to the whose intercepts on the axes are $a$ and $b$, then $\frac{1}{a^2}+\frac{1}{b^2}=$
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Verified Answer
The correct answer is:
$\frac{1}{\mathrm{p}^2}$
Refer image
Equation of given line is
Distance of line (1) from origin is
$$
\begin{aligned}
& \frac{|-a b|}{\sqrt{a^2+b^2}}=p \quad \Rightarrow a^2+b^2=\frac{a^2 b^2}{p^2} \\
& \therefore \frac{a^2+b^2}{a^2 b^2}=\frac{1}{p^2} \Rightarrow \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}
\end{aligned}
$$
Equation of given line is
Distance of line (1) from origin is
$$
\begin{aligned}
& \frac{|-a b|}{\sqrt{a^2+b^2}}=p \quad \Rightarrow a^2+b^2=\frac{a^2 b^2}{p^2} \\
& \therefore \frac{a^2+b^2}{a^2 b^2}=\frac{1}{p^2} \Rightarrow \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2}
\end{aligned}
$$
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