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.

Answer 1

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.