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Question Number 83590 by Tony Lin last updated on 04/Mar/20
transform the ellipse (x^2 /a^2 )+(y^2 /b^2 )=1 to the polar equation r= ((a(1−e^2 ))/(1+ecosθ)) a: semimajor axis e: eccentricity
$${transform}\:{the}\:{ellipse}\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:{to} \\ $$$${the}\:{polar}\:{equation}\:{r}=\:\frac{{a}\left(\mathrm{1}−{e}^{\mathrm{2}} \right)}{\mathrm{1}+{ecos}\theta} \\ $$$${a}:\:{semimajor}\:{axis} \\ $$$${e}:\:{eccentricity} \\ $$
Commented by mr W last updated on 04/Mar/20
(((x+((a+b)/2))^2 )/a^2 )+(y^2 /b^2 )=1 ⇔ r= ((a(1−e^2 ))/(1+ecosθ))
$$\frac{\left({x}+\frac{{a}+{b}}{\mathrm{2}}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:\Leftrightarrow\:{r}=\:\frac{{a}\left(\mathrm{1}−{e}^{\mathrm{2}} \right)}{\mathrm{1}+{ecos}\theta} \\ $$
Commented by Tony Lin last updated on 04/Mar/20
(x^2 /a^2 )+(y^2 /b^2 )=1 b^2 x^2 +a^2 y^2 =a^2 b^2 (a^2 −c^2 )x^2 +a^2 y^2 =a^2 (a^2 −c^2 ) a^2 r^2 −c^2 x^2 =a^4 −c^2 a^2 e^2 =((c/a))^2 =((r^2 −a^2 )/(x^2 −a^2 )) e^2 =((1−(a^2 /r^2 ))/(cos^2 θ−(a^2 /r^2 ))) r^2 e^2 cos^2 θ−e^2 a^2 =r^2 −a^2 r^2 (e^2 cos^2 θ−1)=a^2 (e^2 −1) r=±a(√((1−e^2 )/((1−ecosθ)(1+ecosθ)))) Am I wrong?
$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${b}^{\mathrm{2}} {x}^{\mathrm{2}} +{a}^{\mathrm{2}} {y}^{\mathrm{2}} ={a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$$\left({a}^{\mathrm{2}} −{c}^{\mathrm{2}} \right){x}^{\mathrm{2}} +{a}^{\mathrm{2}} {y}^{\mathrm{2}} ={a}^{\mathrm{2}} \left({a}^{\mathrm{2}} −{c}^{\mathrm{2}} \right) \\ $$$${a}^{\mathrm{2}} {r}^{\mathrm{2}} −{c}^{\mathrm{2}} {x}^{\mathrm{2}} ={a}^{\mathrm{4}} −{c}^{\mathrm{2}} {a}^{\mathrm{2}} \\ $$$${e}^{\mathrm{2}} =\left(\frac{{c}}{{a}}\right)^{\mathrm{2}} =\frac{{r}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} } \\ $$$${e}^{\mathrm{2}} =\frac{\mathrm{1}−\frac{{a}^{\mathrm{2}} }{{r}^{\mathrm{2}} }}{{cos}^{\mathrm{2}} \theta−\frac{{a}^{\mathrm{2}} }{{r}^{\mathrm{2}} }} \\ $$$${r}^{\mathrm{2}} {e}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta−{e}^{\mathrm{2}} {a}^{\mathrm{2}} ={r}^{\mathrm{2}} −{a}^{\mathrm{2}} \\ $$$${r}^{\mathrm{2}} \left({e}^{\mathrm{2}} {cos}^{\mathrm{2}} \theta−\mathrm{1}\right)={a}^{\mathrm{2}} \left({e}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$${r}=\pm{a}\sqrt{\frac{\mathrm{1}−{e}^{\mathrm{2}} }{\left(\mathrm{1}−{ecos}\theta\right)\left(\mathrm{1}+{ecos}\theta\right)}} \\ $$$${Am}\:{I}\:{wrong}? \\ $$
Commented by mr W last updated on 04/Mar/20
correct!
$${correct}! \\ $$

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