Question

The area bounded by the line x=1 and the curve $\sqrt{\frac{y}{x}}+\sqrt{\frac{x}{y}}=4$ is

• A. $2\sqrt{3}$
• B. $\sqrt{3}$
• C. $3\sqrt{2}$
• D. $4\sqrt{3}$

Answer 1

Answer: (D)

$4\sqrt{3}$

Given $\frac{\sqrt{y}}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{y}}=4$

$\Rightarrow \frac{y}{x}+\frac{x}{y}+2=16$

$\Rightarrow \frac{y^2+x^2}{xy}=14$

$\Rightarrow y^2+x^2-14xy=0\}\begin{array}{c}\to y=m_{1}x\\ \to y=m_{2}x\end{array}$ (Equation of pain of straight lines)

In pair of straight lines:

$(y-m_{1}x)(y-m_{2}x)=0$

$\therefore y^2+x^2-14xy=y^2+\left(m_1{m}_2\right)x^2-(m_1+m_2)xy$

Comparing coefficient of $y^2,x^2$ and $xy$:

$m_1{m}_2=1m_1+m_2=14$ $\therefore (m_2-m_1{)}^2=(m_1+m_2{)}^2-4m_1{m}_2$

$=196-4=192$

$\therefore (m_2-m_1)=\sqrt{192}$

In dig area $\left(△OAB\right)=$ area ($△OAP)-$ area $\left(△OBP\right)$

$=\frac{1}{2}\times OP\times AP-\frac{1}{2}\times OP\times BP$

$\therefore \frac{1}{2}OP[AP-BP]$

$=\frac{1}{2}\times 1\times (m_2-m_1)$

$=\frac{1}{2}\times 1\times \sqrt{192}$

$=\frac{1}{2}\times 8\sqrt{3}=4\sqrt{3}$

Option $D$