If p is the length of the perpendicular from the origin on the line xa+yb=1 and a2 , p2 , b2 are in A.P., then a4–2p2a2+2p4=
-1
0
1
none of these
We have p=|ab|a2+b2 ⇒ p2=a2b2a2+b2Also, since a2 , p2 , b2 are in A.P., a2+b2=2p2 and a2b2=p2a2+b2=2p4⇒ 2p4=a22p2−a2 ⇒ a4−2p2a2+2p4=0