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1-line-y-x-4-meets-xy-1-at-A-B-S-OA-B-O-origin-of-cordinates-2-find-center-area-of-region-bonded-by-corve-x-a-y-b-1-and-x-y-axes-a-b-R-

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Question Number 86206 by behi83417@gmail.com last updated on 27/Mar/20
1.line:y=−x+4 ,meets : xy=1 at:A,B. ⇒ S_(OA^△ B) =? (O=origin of cordinates) 2.find :center area of region bonded by corve: (√(x/a))+(√(y/b))=1,and x,y axes. (a≠b)∈R^+
$$\mathrm{1}.\mathrm{line}:\boldsymbol{\mathrm{y}}=−\boldsymbol{\mathrm{x}}+\mathrm{4}\:\:,\mathrm{meets}\::\:\boldsymbol{\mathrm{xy}}=\mathrm{1}\:\mathrm{at}:\boldsymbol{\mathrm{A}},\boldsymbol{\mathrm{B}}. \\ $$$$\:\:\:\:\:\:\Rightarrow\:\:\mathrm{S}_{\mathrm{O}\overset{\bigtriangleup} {\mathrm{A}B}} =?\:\left(\mathrm{O}=\mathrm{origin}\:\mathrm{of}\:\mathrm{cordinates}\right) \\ $$$$\mathrm{2}.\mathrm{find}\::\mathrm{center}\:\mathrm{area}\:\mathrm{of}\:\mathrm{region}\:\mathrm{bonded}\:\mathrm{by} \\ $$$$\mathrm{corve}:\:\:\sqrt{\frac{\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{a}}}}+\sqrt{\frac{\boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{b}}}}=\mathrm{1},\mathrm{and}\:\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}}\:\mathrm{axes}. \\ $$$$\left(\boldsymbol{\mathrm{a}}\neq\boldsymbol{\mathrm{b}}\right)\in\boldsymbol{\mathrm{R}}^{+} \\ $$
Commented by jagoll last updated on 27/Mar/20
(1) xy = −x^2 +4x x^2 −4x+1 = 0 ⇒ { ((x_A = 2+(√3))),((x_B = 2−(√3))) :} { ((y_A = (1/(2+(√3))) = 2−(√3))),((y_B = (1/(2−(√3))) = 2+(√3))) :} S_(OAB) = (1/2)∣ determinant (((2+(√3) 2−(√3))),((0 0)))+ determinant (((0 0)),((2−(√3) 2+(√3))))+ determinant (((2−(√3) 2+(√3))),((2+(√3) 2−(√3) )))∣ = (1/2)∣0+0+(2−(√3))^2 −(2+(√3))^2 ∣ = (1/2) ∣ (4)(−2(√3))∣ = 4(√3) sq units
$$\left(\mathrm{1}\right)\:\mathrm{xy}\:=\:−\mathrm{x}^{\mathrm{2}} +\mathrm{4x} \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{4x}+\mathrm{1}\:=\:\mathrm{0}\:\Rightarrow\:\begin{cases}{\mathrm{x}_{\mathrm{A}} \:=\:\mathrm{2}+\sqrt{\mathrm{3}}}\\{\mathrm{x}_{\mathrm{B}} \:=\:\mathrm{2}−\sqrt{\mathrm{3}}}\end{cases} \\ $$$$\begin{cases}{\mathrm{y}_{\mathrm{A}} \:=\:\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{3}}}\:=\:\mathrm{2}−\sqrt{\mathrm{3}}}\\{\mathrm{y}_{\mathrm{B}} \:=\:\frac{\mathrm{1}}{\mathrm{2}−\sqrt{\mathrm{3}}}\:=\:\mathrm{2}+\sqrt{\mathrm{3}}}\end{cases} \\ $$$$\mathrm{S}_{\mathrm{OAB}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\mid\:\begin{vmatrix}{\mathrm{2}+\sqrt{\mathrm{3}}\:\:\:\:\mathrm{2}−\sqrt{\mathrm{3}}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\end{vmatrix}+ \\ $$$$\begin{vmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{\mathrm{2}−\sqrt{\mathrm{3}}\:\:\:\:\mathrm{2}+\sqrt{\mathrm{3}}}\end{vmatrix}+\:\begin{vmatrix}{\mathrm{2}−\sqrt{\mathrm{3}}\:\:\:\:\:\mathrm{2}+\sqrt{\mathrm{3}}}\\{\mathrm{2}+\sqrt{\mathrm{3}}\:\:\:\:\:\mathrm{2}−\sqrt{\mathrm{3}}\:}\end{vmatrix}\mid \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\mid\mathrm{0}+\mathrm{0}+\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} −\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:\mid \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\:\mid\:\left(\mathrm{4}\right)\left(−\mathrm{2}\sqrt{\mathrm{3}}\right)\mid\:=\:\mathrm{4}\sqrt{\mathrm{3}}\:\mathrm{sq}\:\mathrm{units} \\ $$
Commented by jagoll last updated on 27/Mar/20
Commented by behi83417@gmail.com last updated on 27/Mar/20
right answer.thank you so much sir.
$$\mathrm{right}\:\mathrm{answer}.\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{sir}. \\ $$

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