question_answer

The area bounded by the lines x = 1 and \[\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{z}}=4\]is

A) \[4\sqrt{3}\]                  

B) \[2\sqrt{3}\]

C) \[8\sqrt{3}\]                  

D)  4

Correct Answer: A

Solution :

[a] : \[\sqrt{\frac{x}{y}}+\sqrt{\frac{y}{x}}=4,\]let\[t=\sqrt{\frac{y}{x}}\] \[\Rightarrow t+\frac{1}{t}=4\Rightarrow {{t}^{2}}-4t+1=0\] \[\Rightarrow \]\[t=2\pm \sqrt{3}=\sqrt{\frac{y}{x}}\therefore y=x{{(2\pm \sqrt{3})}^{2}}\] or\[y=(7+4\sqrt{3})x,y=(7-4\sqrt{3})x\] Area\[=\frac{1}{2}\times 1\times AB=\frac{1}{2}\left[ 7+4\sqrt{3}-(7-4\sqrt{3}) \right]\] \[=4\sqrt{3}\].