show that the orbit represented by the function r() is an ellipse

Question

let $r(θ)=a(1-β^2)/(1+β\cos\theta)$ representing the distance from the Sun to a planet. With $0<β<1$, show that the orbit represented by this function $r(θ)$ is an ellipse described by $(x+\sqrt{a^2-b^2})^2/a^2 + y^2/b^2 =1$. With b satisfying $β=\sqrt{1-a^2/b^2}$.

I tried to approach the answer by substitution... but I don't think it was the right method... Please help me!

Answer 1

There must be an error in the text: $\beta$ should be given by $β=\sqrt{1-b^2/a^2}$. First of all multiply your equation by $(1+β\cos\theta)$ to get $$ r+βr\cos\theta=a(1-β^2). $$ Now substitute $r=\sqrt{x^2+y^2}$, $r\cos\theta=x$ $$ \sqrt{x^2+y^2}=a(1-β^2)-βx, $$ and then square both sides: $$ x^2+y^2=a^2(1-β^2)^2-2a(1-β^2)βx+β^2x^2. $$ Now it's only a matter of reorganizing this equation: carry all $x$ terms on the left, divide both sides by $(1-β^2)$ and complete the square in $x$. You should end with only $a^2$ on the right hand side, so that a final division by $a^2$ will give the desired result.

Answer 2

Only it needs some care with signs during algebraic substitution /simplification:

Define semi-latus rectum and $c$, $ p= b^2/a, 1- \beta^2 = (b/a)^2, c^2 = a^2-b^2 $ where $\beta$ is eccentricity.These classical quantities help to simplify faster.

Polar coordinates are $ r, \theta$ with $ x =r \cos \theta , y =y_1= r \sin \theta $, for central ellipse Cartesian coordinates $ x_1,y_1.$

Standard ellipse equation in Polar/focal/Newtonian form:

$$ \frac{p}{r} = 1- \epsilon \cos \theta \tag{1}$$

or so very simply!

$$ p = r- \epsilon\, x \tag{2}$$

You want to shift from focal/Polar coordinates to Cartesian. $$ x = x_1 + c,\, y =y_1\tag{3} $$

Plug this into (2) $$ a (1 -\epsilon^2) = b^2/a=\sqrt{(x_1+c)^2 + y_1^2} + ( x_1 +c) \epsilon \tag{4}$$ or, $$ b^2/a=\sqrt{(x_1+c)^2 + y_1^2} + -x_1 \epsilon -a + b^2/a \tag{5}$$ Cancell $ b^2/a$ on both sides $$ \sqrt{(x_1+c)^2 + y_1^2} = a + x_1\, \epsilon \tag{6}$$ Now square both sides and cancel the common term $ 2 c x_1$ on either side

$$ x_1^2 (1-\epsilon^2) +y_1^2 = a^2- c^2 = b^2 \tag{7}$$

Divide by $b^2$

$$ { \frac{x_1^2}{b^2/(1-\epsilon^2)}} + { \left(\frac{y_1}{b}\right)}^2 = 1 \tag{8}$$ $$ { \frac{x_1^2}{a^2}} + { \frac{y_1^2}{b^2}} = 1 \tag{9}$$

$$ { \frac{(x-c)^2}{a^2}} + { \frac{y^2}{b^2}} = 1 \tag{10}.$$