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Question Number 86206 by behi83417@gmail.com last updated on 27/Mar/20

$$1.\mathrm{line}:\mathbf{y}=-\mathbf{x}+4,\mathrm{meets}:\mathbf{xy}=1\mathrm{at}:\mathbf{A},\mathbf{B}.$$$$\Rightarrow {\mathrm{S}}_{\mathrm{O}\stackrel{△}{\mathrm{A}B}}=?(\mathrm{O}=\mathrm{origin}\mathrm{of}\mathrm{cordinates})$$$$2.\mathrm{find}:\mathrm{center}\mathrm{area}\mathrm{of}\mathrm{region}\mathrm{bonded}\mathrm{by}$$$$\mathrm{corve}:\sqrt{\frac{\mathbf{x}}{\mathbf{a}}}+\sqrt{\frac{\mathbf{y}}{\mathbf{b}}}=1,\mathrm{and}\mathbf{x},\mathbf{y}\mathrm{axes}.$$$$(\mathbf{a}\ne \mathbf{b})\in {\mathbf{R}}^{+}$$

Commented by jagoll last updated on 27/Mar/20

$$\left(1\right)\mathrm{xy}=-{\mathrm{x}}^2+4\mathrm{x}$$$${\mathrm{x}}^2-4\mathrm{x}+1=0\Rightarrow \{\begin{array}{l}{\mathrm{x}}_{\mathrm{A}}=2+\sqrt{3}\\ {\mathrm{x}}_{\mathrm{B}}=2-\sqrt{3}\end{array}$$$$\{\begin{array}{l}{\mathrm{y}}_{\mathrm{A}}=\frac{1}{2+\sqrt{3}}=2-\sqrt{3}\\ {\mathrm{y}}_{\mathrm{B}}=\frac{1}{2-\sqrt{3}}=2+\sqrt{3}\end{array}$$$${\mathrm{S}}_{\mathrm{OAB}}=\frac{1}{2}\mid \left|\begin{array}{c}2+\sqrt{3}2-\sqrt{3}\\ 00\end{array}\right|+$$$$\left|\begin{array}{c}00\\ 2-\sqrt{3}2+\sqrt{3}\end{array}\right|+\left|\begin{array}{c}2-\sqrt{3}2+\sqrt{3}\\ 2+\sqrt{3}2-\sqrt{3}\end{array}\right|\mid$$$$=\frac{1}{2}\mid 0+0+{(2-\sqrt{3})}^2-{(2+\sqrt{3})}^2\mid$$$$=\frac{1}{2}\mid \left(4\right)(-2\sqrt{3})\mid =4\sqrt{3}\mathrm{sq}\mathrm{units}$$

Commented by jagoll last updated on 27/Mar/20

Commented by behi83417@gmail.com last updated on 27/Mar/20

$$\mathrm{right}\mathrm{answer}.\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{sir}.$$

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