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What-is-the-equation-of-ellipse-with-center-1-2-excentricity-1-3-passing-through-2-2-2-2-3-

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Question Number 135695 by liberty last updated on 15/Mar/21
What is the equation of ellipse with center (1,−2) excentricity (1/( (√3))) , passing through (2, ((2(√2)−2)/3))
$${What}\:{is}\:{the}\:{equation}\:{of} \\ $$$${ellipse}\:{with}\:{center}\:\left(\mathrm{1},−\mathrm{2}\right)\: \\ $$$${excentricity}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:,\:{passing} \\ $$$${through}\:\left(\mathrm{2},\:\frac{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2}}{\mathrm{3}}\right)\: \\ $$
Answered by mr W last updated on 15/Mar/21
there is no unique solution! it depents on the direction of the major axis! since the major axis is not given, there are infinite possibilities. assume the major axis is parallel to x−axis, then the eqn. of ellipse is (((x−1)^2 )/a^2 )+(((y+2)^2 )/b^2 )=1 e=(√(1−((b/a))^2 ))=(1/( (√3))) ⇒(b^2 /a^2 )=(2/3) (((2−1)^2 )/a^2 )+(((((2(√2)−2)/3)+2)^2 )/b^2 )=1 (1/a^2 )+((8((√2)+1)^2 )/(9b^2 ))=1 (1/a^2 )+((8((√2)+1)^2 )/(9×(2/3)a^2 ))=1 a^2 =1+((4((√2)+1)^2 )/3)=5+((8(√2))/3) b^2 =(2/3)(5+((8(√2))/3)) ⇒(x−1)^2 +((3(y+2)^2 )/2)=5+((8(√2))/3) see the red one in diagram. but this is only one of infinite possibilities.
$${there}\:{is}\:{no}\:{unique}\:{solution}!\:{it}\: \\ $$$${depents}\:{on}\:{the}\:{direction}\:{of}\:{the} \\ $$$${major}\:{axis}!\:{since}\:{the}\:{major}\:{axis} \\ $$$${is}\:{not}\:{given},\:{there}\:{are}\:{infinite} \\ $$$${possibilities}. \\ $$$${assume}\:{the}\:{major}\:{axis}\:{is}\:{parallel} \\ $$$${to}\:{x}−{axis},\:{then}\:{the}\:{eqn}.\:{of}\:{ellipse} \\ $$$${is}\:\frac{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\left({y}+\mathrm{2}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${e}=\sqrt{\mathrm{1}−\left(\frac{{b}}{{a}}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\frac{\left(\mathrm{2}−\mathrm{1}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{\left(\frac{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2}}{\mathrm{3}}+\mathrm{2}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{8}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{9}{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{8}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{9}×\frac{\mathrm{2}}{\mathrm{3}}{a}^{\mathrm{2}} }=\mathrm{1} \\ $$$${a}^{\mathrm{2}} =\mathrm{1}+\frac{\mathrm{4}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{3}}=\mathrm{5}+\frac{\mathrm{8}\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$$${b}^{\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{5}+\frac{\mathrm{8}\sqrt{\mathrm{2}}}{\mathrm{3}}\right) \\ $$$$\Rightarrow\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\frac{\mathrm{3}\left({y}+\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{2}}=\mathrm{5}+\frac{\mathrm{8}\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$$${see}\:{the}\:{red}\:{one}\:{in}\:{diagram}.\:{but}\:{this} \\ $$$${is}\:{only}\:{one}\:{of}\:{infinite}\:{possibilities}. \\ $$
Commented by mr W last updated on 15/Mar/21
Commented by liberty last updated on 15/Mar/21
agree sir
$${agree}\:{sir} \\ $$

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