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
Correct Answer: C
Solution :
\[CA=\sqrt{{{(a{{t}^{2}}-a)}^{2}}+{{(2at)}^{2}}}=a\sqrt{{{({{t}^{2}}-1)}^{2}}+4{{t}^{2}}}\] \[=a\sqrt{({{t}^{2}}+1+2{{t}^{2}})}=a\,(1+{{t}^{2}})\] \[CB=\sqrt{{{\left( \frac{a}{{{t}^{2}}}-a \right)}^{2}}+{{\left( \frac{-2a}{t} \right)}^{2}}}=a\,\left( 1+\frac{1}{{{t}^{2}}} \right)\] H.M. of CA and CB \[=\frac{2{{a}^{2}}\,(1+{{t}^{2}})\,\left( 1+\frac{1}{{{t}^{2}}} \right)}{a\,\left[ 1+{{t}^{2}}+1+\frac{1}{{{t}^{2}}} \right]}=2a\]. \[\left[ \because \,\,\text{H}\text{.M}.\,\text{of }x \text{and }y=\frac{2xy}{x+y} \right]\]
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