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Question
If p is the length of perpendicular from the origin on the line `x/a + y/b` = 1 and a2, p2, b2 are in A.P, then show that a4 + b4 = 0.
Solution
Given equation is `x/a + y/b` = 1
p = `|(0/a + 0/b - 1)/sqrt(1/a^2 + 1/b^2)|`
Squaring both sides, we have
p2 = `1/(1/a^2 + 1/b^2)`
⇒ `1/a^2 + 1/b^2 = 1/p^2` ......(i)
Since a2, p2, b2 are in A.P.
∴ 2p2 = a2 + b2
⇒ p2 = `(a^2 + b^2)/2`
⇒ `1/p^2 = 2/(a^2 + b^2)`
Putting the value of `1/p^2` is equation (i) we get,
`1/a^2 + 1/b^2 = 2/(a^2 + b^2)`
⇒ `(a^2 + b^2)/(a^2b^2) = 2/(a^2 + b^2)`
⇒ (a2 + b2)2 = 2a2b2
⇒ a4 + b4 + 2a2b2 = 2a2b2
⇒ a4 + b4 = 0.
Hence proved.
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