If $2 p$ is the length of perpendicular from the origin to the line whose intercepts on the axes are $2 a$ and $b$ , then show that $\frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}$ .

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Solution

Equation of line with intercepts $a$ and $b$ is,
$\frac{x}{a} + \frac{y}{b} = 1 \Rightarrow b x + a y - a b = 0$ ....(1)
Now if $p$ is the length of the perpendicular from point $(0, 0)$ to line (1),then,
$p = \frac{| b (0) + a (0) - a b |}{\sqrt{b^{2} + a^{2}}}$
$\Rightarrow p = \frac{| - a b |}{\sqrt{a^{2} + b^{2}}}$
On squaring both sides, we obtain
$p^{2} = \frac{{| - a b |}^{2}}{({\sqrt{a^{2} + b^{2})}}^{2}}$
$\Rightarrow p^{2} (a^{2} + b^{2}) = a^{2} b^{2}$
$\Rightarrow \frac{a^{2} + b^{2}}{a^{2} b^{2}} = \frac{1}{p^{2}}$
$\Rightarrow \frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}}$

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