If length of perpendicular drawn from origin on the line #x/a + y/b = 1# is #2p#, then #a^2,8p^2,b^2# are in?

Question

A) H.P
B) G.P
C) A.P
D) None of these

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Answer 1 · 2017-12-27T10:08:01

A) H.P. is the right option.

The #bot"-distance"# from the Origin #O(0,0)# to the line

# L : x/a+y/b-1=0, i.e., 2p, # is given by,

#2p=|0/a+0/b-1|/sqrt{(1/a)^2+(1/b)^2}=|ab|/sqrt(a^2+b^2)#.

#:. 4p^2=(a^2b^2)/(a^2+b^2), or, 8p^2=(2a^2b^2)/(a^2+b^2)#.

#:. 1/(8p^2)=(a^2+b^2)/(2a^2b^2)=1/2{a^2/(a^2b^2)+b^2/(a^2b^2)}, i.e.,#

#1/(8p^2)=1/2{1/a^2+1/b^2}#, meaning that, #1/a^2,1/(8p^2),1/b^2# are in AP.

Equivalently, #a^2,8p^2,b^2# are in HP.

Hence, A) H.P. is the right option.

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