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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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