If p is the length of the perpendicular from the origin to the line \(x\over a\) + \(y\over b\) = 1, then prove that \(1\over p^2\) = \(1\over a^2\) + \(1\over b^2\)

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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