A) \[8\sqrt{3}\]
B) \[10\sqrt{3}\]
C) \[2\sqrt{3}\]
D) \[16\sqrt{3}\]
Correct Answer: B
Solution :
\[c=-29m-9{{m}^{3}}\] \[a=2\] Given \[{{(a{{t}^{2}}-a)}^{2}}+4{{a}^{2}}{{t}^{2}}=64\] \[\to (a({{t}^{2}}+1))=8\] \[\Rightarrow \,{{t}^{2}}+1=4={{t}^{2}}=3\] \[\Rightarrow t=\sqrt{3}\] \[\therefore \,\,C=-a\,[-2t\,-{{t}^{3}}]\,=2at\,(2+{{t}^{2}})\] \[=2\sqrt{3}\,(5)\] \[|C|\,=10\sqrt{3}\]
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