If p is the length of perpendicular from the origin on the line `(x)/(a)+(y)/(b)=1` and `a^(2)`,`p^(2)` and `b^(2)` are in AP, the show that `a^(4)+b^(4)=0`.

Answer

N/a

Explanation

Given equation of line is,
`(x)/(a)+(y)/(b)=1`
Perpendicular length form the origin on the line (i) is given by P
i.e., `p=(1)/(sqrt((1)/(a^2)+(1)/(b^2)))=(ab)/(sqrt(a^2+b^2))`
`therefore p^2=(a^2b^2)/(a^2+b^2)`
Given that `a^2`,`p^2` and `b^2` are in AP.
`therefore 2p^2=a^2+b^2`
`rArr (2a^2b^2)/(a^2+b^2)=a^2+b^2`
` rArr2a^2b^2=(a^2+b^2)^2`
`rArr2a^2+b^2=a^4+b^4+2a^2b^2`
`rArr a^4+b^4=0`