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


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Question Number 83590 by Tony Lin last updated on 04/Mar/20

$t r a n s f o r m t h e e l l i p s e \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 t o$ $t h e p o l a r e q u a t i o n r = \frac{a (1 - e^{2})}{1 + e c o s θ}$ $a : s e m i m a j o r a x i s$ $e : e c c e n t r i c i t y$
Commented by mr W last updated on 04/Mar/20

$\frac{{(x + \frac{a + b}{2})}^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \Leftrightarrow r = \frac{a (1 - e^{2})}{1 + e c o s θ}$
Commented by Tony Lin last updated on 04/Mar/20

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $b^{2} x^{2} + a^{2} y^{2} = a^{2} b^{2}$ $(a^{2} - c^{2}) x^{2} + a^{2} y^{2} = a^{2} (a^{2} - c^{2})$ $a^{2} r^{2} - c^{2} x^{2} = a^{4} - c^{2} a^{2}$ $e^{2} = {(\frac{c}{a})}^{2} = \frac{r^{2} - a^{2}}{x^{2} - a^{2}}$ $e^{2} = \frac{1 - \frac{a^{2}}{r^{2}}}{{c o s}^{2} θ - \frac{a^{2}}{r^{2}}}$ $r^{2} e^{2} {c o s}^{2} θ - e^{2} a^{2} = r^{2} - a^{2}$ $r^{2} (e^{2} {c o s}^{2} θ - 1) = a^{2} (e^{2} - 1)$ $r = \pm a \sqrt{\frac{1 - e^{2}}{(1 - e c o s θ) (1 + e c o s θ)}}$ $A m I w r o n g ?$
Commented by mr W last updated on 04/Mar/20

$c o r r e c t!$
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