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Question Number 68110 by TawaTawa last updated on 05/Sep/19
Commented by kaivan.ahmadi last updated on 07/Sep/19
((y/2))^2 =1−((x/3))^2 ⇒y=±2(√(1−((x/3))^2 )) equation of BC: m=tg30=((√3)/2) y−0=((√3)/2)(x+3)⇒y=((√3)/2)x+((3(√3))/2) equation of OD: m=tg60=(√3) y=(√3)x we find point C ((x/3))^2 +(((√3)/4)x+((3(√3))/4))^2 =1⇒(1/9)x^2 +(3/(16))x^2 +(9/8)x+((27)/(16))=1⇒ 16x^2 +27x^2 +162x+243=144⇒43x^2 +162x+99=0 x_(1,2) =((−162±96)/(86))= { ((((−162+96)/(86))=((−66)/(86))=((−33)/(43)))),((((−162−96)/(86))=−3)) :} so C(((−33)/(43)),((−33(√3))/(86))+((3(√3))/2))=(((−33)/(43)),((96(√3))/(86)))=(((−33)/(43)),((48(√3))/(43))) so this diagram is not true. and now we find point D: ((x/3))^2 +(((√3)/2)x)^2 =1⇒(1/9)x^2 +(3/4)x^2 =1⇒13x^2 =36⇒ x^2 =((36)/(13))⇒x=±(√((36)/(13)))⇒D((√((36)/(13))),(√((108)/(13)))) so we have S_(OBCD) =∫_(−3) ^((−33)/(43)) (((√3)/2)x+((3(√3))/2))dx+∫_((−33)/(43)) ^0 2(√(1−((x/3))^2 ))dx +∫_0 ^(√((36)/(13))) (2(√(1−((x/3))^2 ))−(√3)x)dx
$$\left(\frac{{y}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{1}−\left(\frac{{x}}{\mathrm{3}}\right)^{\mathrm{2}} \Rightarrow{y}=\pm\mathrm{2}\sqrt{\mathrm{1}−\left(\frac{{x}}{\mathrm{3}}\right)^{\mathrm{2}} } \\ $$$${equation}\:{of}\:{BC}: \\ $$$${m}={tg}\mathrm{30}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${y}−\mathrm{0}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left({x}+\mathrm{3}\right)\Rightarrow{y}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{x}+\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${equation}\:{of}\:{OD}: \\ $$$${m}={tg}\mathrm{60}=\sqrt{\mathrm{3}} \\ $$$${y}=\sqrt{\mathrm{3}}{x} \\ $$$${we}\:{find}\:{point}\:{C} \\ $$$$\left(\frac{{x}}{\mathrm{3}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}{x}+\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{4}}\right)^{\mathrm{2}} =\mathrm{1}\Rightarrow\frac{\mathrm{1}}{\mathrm{9}}{x}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{16}}{x}^{\mathrm{2}} +\frac{\mathrm{9}}{\mathrm{8}}{x}+\frac{\mathrm{27}}{\mathrm{16}}=\mathrm{1}\Rightarrow \\ $$$$\mathrm{16}{x}^{\mathrm{2}} +\mathrm{27}{x}^{\mathrm{2}} +\mathrm{162}{x}+\mathrm{243}=\mathrm{144}\Rightarrow\mathrm{43}{x}^{\mathrm{2}} +\mathrm{162}{x}+\mathrm{99}=\mathrm{0} \\ $$$${x}_{\mathrm{1},\mathrm{2}} =\frac{−\mathrm{162}\pm\mathrm{96}}{\mathrm{86}}=\begin{cases}{\frac{−\mathrm{162}+\mathrm{96}}{\mathrm{86}}=\frac{−\mathrm{66}}{\mathrm{86}}=\frac{−\mathrm{33}}{\mathrm{43}}}\\{\frac{−\mathrm{162}−\mathrm{96}}{\mathrm{86}}=−\mathrm{3}}\end{cases} \\ $$$${so}\:{C}\left(\frac{−\mathrm{33}}{\mathrm{43}},\frac{−\mathrm{33}\sqrt{\mathrm{3}}}{\mathrm{86}}+\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\left(\frac{−\mathrm{33}}{\mathrm{43}},\frac{\mathrm{96}\sqrt{\mathrm{3}}}{\mathrm{86}}\right)=\left(\frac{−\mathrm{33}}{\mathrm{43}},\frac{\mathrm{48}\sqrt{\mathrm{3}}}{\mathrm{43}}\right) \\ $$$${so}\:{this}\:{diagram}\:{is}\:{not}\:{true}. \\ $$$${and}\:{now}\:{we}\:{find}\:{point}\:{D}: \\ $$$$\left(\frac{{x}}{\mathrm{3}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{x}\right)^{\mathrm{2}} =\mathrm{1}\Rightarrow\frac{\mathrm{1}}{\mathrm{9}}{x}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}{x}^{\mathrm{2}} =\mathrm{1}\Rightarrow\mathrm{13}{x}^{\mathrm{2}} =\mathrm{36}\Rightarrow \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{36}}{\mathrm{13}}\Rightarrow{x}=\pm\sqrt{\frac{\mathrm{36}}{\mathrm{13}}}\Rightarrow{D}\left(\sqrt{\frac{\mathrm{36}}{\mathrm{13}}},\sqrt{\frac{\mathrm{108}}{\mathrm{13}}}\right) \\ $$$${so}\:{we}\:{have} \\ $$$${S}_{{OBCD}} =\int_{−\mathrm{3}} ^{\frac{−\mathrm{33}}{\mathrm{43}}} \left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{x}+\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}\right){dx}+\int_{\frac{−\mathrm{33}}{\mathrm{43}}} ^{\mathrm{0}} \mathrm{2}\sqrt{\mathrm{1}−\left(\frac{{x}}{\mathrm{3}}\right)^{\mathrm{2}} }{dx} \\ $$$$+\int_{\mathrm{0}} ^{\sqrt{\frac{\mathrm{36}}{\mathrm{13}}}} \:\left(\mathrm{2}\sqrt{\mathrm{1}−\left(\frac{{x}}{\mathrm{3}}\right)^{\mathrm{2}} }−\sqrt{\mathrm{3}}{x}\right){dx} \\ $$$$ \\ $$
Commented by TawaTawa last updated on 08/Sep/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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