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Sum

If p is the length of perpendicular from the origin on the line `x/a + y/b` = 1 and a^{2}, p^{2}, b^{2} are in A.P, then show that a^{4} + b^{4} = 0.

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#### 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

p^{2} = `1/(1/a^2 + 1/b^2)`

⇒ `1/a^2 + 1/b^2 = 1/p^2` ......(i)

Since a^{2}, p^{2}, b^{2} are in A.P.

∴ 2p^{2} = a^{2} + b^{2}

⇒ p^{2} = `(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)`

⇒ (a^{2} + b^{2})^{2} = 2a^{2}b^{2}

⇒ a^{4} + b^{4} + 2a^{2}b^{2} = 2a^{2}b^{2}

⇒ a^{4} + b^{4} = 0.

Hence proved.

Concept: Slope of a Line

Is there an error in this question or solution?

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