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x-2y-3-2-3x-4y-1-2-100-what-is-the-area-of-the-ellipse-

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Question Number 32501 by riza kesk last updated on 26/Mar/18
(x−2y+3)^2 +(3x+4y−1)^2 =100 what is the area of the ellipse?
$$\left(\mathrm{x}−\mathrm{2y}+\mathrm{3}\right)^{\mathrm{2}} +\left(\mathrm{3x}+\mathrm{4y}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{100} \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{ellipse}? \\ $$
Answered by MJS last updated on 27/Mar/18
the center of ax^2 +bxy+cy^2 +dx+ey+f=0 is M= ((((be−2cd)/(4ac−b^2 ))),(((bd−2ae)/(4ac−b^2 ))) ) (x−2y+3)^2 +(3x+4y−1)^2 =100 x^2 +2xy+2y^2 −2y−9=0 M= (((−1)),(1) ) x → x−1 y → y+1 x^2 +2xy+2y^2 −10=0 tan 2θ=(b/(a−c))=−2 ⇒ tan θ=(1/2)−((√5)/2)∨tan θ=(1/2)+((√5)/2) axis a: y=((1/2)−((√5)/2))x x^2 +2((1/2)−((√5)/2))x^2 +2((1/2)−((√5)/2))^2 x^2 −10=0 x=±(√(10+4(√5))) y=∓(√(5+(√5))) a=(√(x^2 +y^2 ))=((√2)/2)(5+(√5)) axis b: y=((1/2)+((√5)/2))x x^2 +2((1/2)+((√5)/2))x^2 +2((1/2)+((√5)/2))^2 x^2 −10=0 x=±(√(10−4(√5))) y=±(√(5−(√5))) b=(√(x^2 +y^2 ))=((√2)/2)(5−(√5)) A=abπ=10π
$$\mathrm{the}\:\mathrm{center}\:\mathrm{of} \\ $$$${ax}^{\mathrm{2}} +{bxy}+{cy}^{\mathrm{2}} +{dx}+{ey}+{f}=\mathrm{0} \\ $$$$\mathrm{is}\:{M}=\begin{pmatrix}{\frac{{be}−\mathrm{2}{cd}}{\mathrm{4}{ac}−{b}^{\mathrm{2}} }}\\{\frac{{bd}−\mathrm{2}{ae}}{\mathrm{4}{ac}−{b}^{\mathrm{2}} }}\end{pmatrix} \\ $$$$\left({x}−\mathrm{2}{y}+\mathrm{3}\right)^{\mathrm{2}} +\left(\mathrm{3}{x}+\mathrm{4}{y}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{100} \\ $$$${x}^{\mathrm{2}} +\mathrm{2}{xy}+\mathrm{2}{y}^{\mathrm{2}} −\mathrm{2}{y}−\mathrm{9}=\mathrm{0} \\ $$$${M}=\begin{pmatrix}{−\mathrm{1}}\\{\mathrm{1}}\end{pmatrix} \\ $$$${x}\:\rightarrow\:{x}−\mathrm{1} \\ $$$${y}\:\rightarrow\:{y}+\mathrm{1} \\ $$$${x}^{\mathrm{2}} +\mathrm{2}{xy}+\mathrm{2}{y}^{\mathrm{2}} −\mathrm{10}=\mathrm{0} \\ $$$$\mathrm{tan}\:\mathrm{2}\theta=\frac{{b}}{{a}−{c}}=−\mathrm{2} \\ $$$$\Rightarrow\:\mathrm{tan}\:\theta=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\vee\mathrm{tan}\:\theta=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\mathrm{axis}\:\boldsymbol{{a}}:\:{y}=\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right){x} \\ $$$${x}^{\mathrm{2}} +\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right){x}^{\mathrm{2}} +\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} {x}^{\mathrm{2}} −\mathrm{10}=\mathrm{0} \\ $$$${x}=\pm\sqrt{\mathrm{10}+\mathrm{4}\sqrt{\mathrm{5}}} \\ $$$${y}=\mp\sqrt{\mathrm{5}+\sqrt{\mathrm{5}}} \\ $$$$\boldsymbol{{a}}=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{5}+\sqrt{\mathrm{5}}\right) \\ $$$$\mathrm{axis}\:\boldsymbol{{b}}:\:{y}=\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right){x} \\ $$$${x}^{\mathrm{2}} +\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right){x}^{\mathrm{2}} +\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} {x}^{\mathrm{2}} −\mathrm{10}=\mathrm{0} \\ $$$${x}=\pm\sqrt{\mathrm{10}−\mathrm{4}\sqrt{\mathrm{5}}} \\ $$$${y}=\pm\sqrt{\mathrm{5}−\sqrt{\mathrm{5}}} \\ $$$$\boldsymbol{{b}}=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{5}−\sqrt{\mathrm{5}}\right) \\ $$$${A}=\boldsymbol{{ab}}\pi=\mathrm{10}\pi \\ $$

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