Q.

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=

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a

-1

b

0

c

1

d

none of these

answer is B.

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Detailed Solution

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
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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=