Coordinate Geometry Question 3
Question: If $ A(at^{2},2at),\ B(a/t^{2},-2a/t) $ and $ C(a,0) $ , then 2a is equal to
[RPET 2000]
Options:
A) A.M. of CA and CB
B) G.M. of CA and CB
C) H.M. of CA and CB
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ CA=\sqrt{{{(at^{2}-a)}^{2}}+{{(2at)}^{2}}}=a\sqrt{{{(t^{2}-1)}^{2}}+4t^{2}} $
$ =a\sqrt{(t^{2}+1+2t^{2})}=a(1+t^{2}) $
$ CB=\sqrt{{{( \frac{a}{t^{2}}-a )}^{2}}+{{( \frac{-2a}{t} )}^{2}}}=a( 1+\frac{1}{t^{2}} ) $ H.M. of CA and CB $ =\frac{2a^{2}(1+t^{2})( 1+\frac{1}{t^{2}} )}{a[ 1+t^{2}+1+\frac{1}{t^{2}} ]}=2a $ . $ [ \because H\text{.M}.\text{of }x \text{and }y=\frac{2xy}{x+y} ] $
BETA
