Solution :
The given line is bx + ay – ab = 0 ………….(i)
It is given that
p = Length of the perpendicular from the origin to line (i)
\(\implies\) p = \(|b(0) + a(0) – ab|\over {\sqrt{b^2+a^2}}\) = \(ab\over \sqrt{a^2+b^2}\)
\(\implies\) \(p^2\) = \(a^2b^2\over a^2+b^2\) \(\implies\) \(1\over p^2\) = \(a^2+b^2\over a^2b^2\) \(\implies\) \(1\over p^2\) = \(1\over a^2\) + \(1\over b^2\)
Hence Proved.
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