transform-the-ellipse-x-2-a-2-y-2-b-2-1-to-the-polar-equation-r-a-1-e-2-1-ecos-a-semimajor-axis-e-eccentricity-

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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