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`.

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 A.P., then show that a^(4) + b^(4) =0 .

The length of perpendicular from the origin on the line x/a + y/b = 1 is -

If 2p is the length of perpendicular from the origin to the lines (x)/(a)+(y)/(b)=1, then a^(2),8p^(2),b^(2) are in

If p be the length of the perpendicular from the origin on the line x/a+y/b=1 , then

Length of perpendicular from origin to the line (x)/(a) + (y)/( b) = 1 is p . If a^(2) , p^(2) and b^(2) are in A.P. then prove that a^(4) + b^(4) =0 .

I.If p is the length of the perpendicular from the origin on the line (x)/(a)+(y)/(b)=1 and a^(2),p^(2),b^(2) are in A.P,then a^(4)-2p^(2)a^(2)+2p^(4)=

If 2p is the length of perpendicular from the origin to the lines x/a+ y/b = 1, then a^2,8p^2,b^2 are in

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 (A) a^(4)+b^(4)=0 (B) a^(4)-b^(4)=0 (C) a^(2)+b^(2)=0 (D) a^(2)-b^(2)=0

If p is the length of the perpendicular drawnfrom the origin upon the line (x)/(a)+(y)/(b)=1 and a^(2),p^(2),b^(2) are in arithmetic progression,) then |ab| is equal to