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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Aicharya · Accepted Answer

Refer image Equation of given line is Distance of line (1) from origin is $\begin{aligned} & \frac{|-a b|}{\sqrt{a^2+b^2}}=p \quad \Rightarrow a^2+b^2=\frac{a^2 b^2}{p^2} \ & 	herefore \frac{a^2+b^2}{a^2 b^2}=\frac{1}{p^2} \Rightarrow \frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{p^2} \end{aligned}$

If $\mathrm{p}$ is the length of the perpendicular from origin to the whose intercepts on the axes are $a$…

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