1
$\begingroup$

$r(θ) = a(1 − β^2)/(1 + β \cos θ)$

and I want to show this $r(θ)$ is an ellipse described by

$\dfrac{(x+\sqrt{a^2 − b^2})^2}{a^2}+\dfrac{y^2}{b^2}= 1$, when $0<β<1$.

How can we show this?

$\endgroup$
1
$\begingroup$

With the relations $$ r=\sqrt{x^2+y^2}, r\cos\theta=x $$ we can rewrite the equation as $$ r=\frac{ar(1-\beta^2)}{r+\beta x} $$ or, equivalently (disregarding $r=0$ that's not a solution), $$ r+\beta x=a(1-\beta^2) $$ that becomes $r=a(1-\beta^2)-\beta x$; now square and get $$ x^2+y^2=a^2(1-\beta^2)^2-2a(1-\beta^2)\beta x+\beta^2x^2 $$ Reorder: $$ x^2(1-\beta^2)+2a(1-\beta^2)\beta x+y^2=a^2(1-\beta^2)^2 $$ Divide everything by $1-\beta^2$: $$ x^2+2a\beta x+\frac{y^2}{1-\beta^2}=a^2(1-\beta^2) $$ Complete the square: $$ x^2+2a\beta x+a^2\beta^2+\frac{y^2}{1-\beta^2}=a^2 $$ Set $c=a\beta$ and $a^2(1-\beta^2)=b^2$: $$ (x+c)^2+\frac{a^2}{b^2}y^2=a^2 $$ Divide by $a^2$: $$ \frac{(x+c)^2}{a^2}+\frac{y^2}{b^2}=1 $$ Note that $c=\sqrt{a^2-b^2}$.

$\endgroup$
  • $\begingroup$ I think you are wrong my friend. Your last c doesn't match with first one! Read my answer. Your relation between polar and Cartesian parameters are wrong. $\endgroup$ – A.F.23 Oct 10 '15 at 14:03
  • $\begingroup$ @A.F.23 There was a wrong division. The relation between polar and cartesian coordinates is of course correct. $\endgroup$ – egreg Oct 10 '15 at 14:16
  • $\begingroup$ Yes, I thought two equations are in two coordinate systems and we should use $x=r\cos\theta-f$. Sorry. $\endgroup$ – A.F.23 Oct 10 '15 at 18:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.