Polar equation of a hyperbola with one focus at the origin

[Adgorn]

Homework Statement

Hello everyone,
I have an assignment (Spivak's Calculus) to show that the polar equation of a hyperbola with the right focus in the origin is $r=\frac {±\Lambda} {1+εcos(\theta)}$, but the equation I reached was slightly yet somewhat disturbingly different, and I'm not sure where the problem is.

Homework Equations

Distance between 2 points equation: $r^2=x^2+y^2$
Conversion from Cartesian to polar coordinates: $x=rcos(\theta)$, $y=rsin(\theta)$
Condition for a point being on a hyperbola: $r-s=±2a$
definition of $\Lambda$: $\Lambda=(1-ε^2)a$

The Attempt at a Solution

The development I did is as follows:

Say the difference between the distances from the foci is $2a$, the 1st focus is in the origin and the 2nd focus is in $(-2εa,0)$ where $ε\gt1$. The distance from the right focus will be denoted with $r$ and the distance from the left focus will be denoted with $s$
The distance of any point $(x,y)$ in the hyperbola from the left focus is $s=r±2a$ (Since $r-s=±2a$ by assumption). The same distance is also $s=\sqrt{(x+2εa)^2+y^2}$ (by the distance equation). Furthermore, since the right focus is the origin, $r^2=x^2+y^2$.
Thus:
i) $(r±a)^2=(x+2εa)^2+y^2$
ii) $r^2=x^2+y^2$

Solving the system gives the equation $±r=(1-ε^2)a-εx$. Denoting $\Lambda=(1-ε^2)a$ and substituting $x=rcos(\theta)$ gives
iii) $±r=\Lambda-εrcos(\theta)$.

Adding $εrcos(\theta)$ to the equation, taking out $r$ in the left side as a common factor and dividing yields the equation:
iv) $r=\frac \Lambda {±1+εcos(\theta)}$

Which is the equation I got to, putting the equation in a graphing program such as Desmos gives a correct graph as well. However if my answer and the book's answer really are 2 ways of expressing the same equation then it means:
$\frac {±\Lambda} {1+εcos(\theta)}=\frac \Lambda {±1+εcos(\theta)}$, in other terms (the minus case), $\frac {-\Lambda} {1+εcos(\theta)}=\frac {\Lambda} {-1+εcos(\theta)}$. This in turn implies $1+εcos(\theta)=1-εcos(\theta)$ which is bizarre since... well... $cos(\theta)≠0$ for all $\theta$.

So what am I missing here? help and clarification would be very appreciated.

 

[Mark44]

Which is the equation I got to, putting the equation in a graphing program such as Desmos gives a correct graph as well. However if these really are 2 ways of expressing the same equations then it means:
$\frac {±\Lambda} {1+εcos(\theta)}=\frac \Lambda {±1+εcos(\theta)}$,

Shouldn't that be $\frac {±\Lambda} {1+εcos(\theta)}=\frac \Lambda {±(1+εcos(\theta))}$?
IOW, the denominator on the right will be either $1 + \epsilon\cos(\theta)$ or $-1 - \epsilon\cos(\theta)$.

in other terms, $\frac {-\Lambda} {1+εcos(\theta)}=\frac {\Lambda} {-1+εcos(\theta)}$. This in turn implies $1+εcos(\theta)=1-εcos(\theta)$ which is bizarre since... well... $cos(\theta)≠0$ for all $\theta$.

So what am I missing here? help and clarification would be very appreciated.

 

[Adgorn]

Shouldn't that be $\frac {±\Lambda} {1+εcos(\theta)}=\frac \Lambda {±(1+εcos(\theta))}$?
IOW, the denominator on the right will be either $1 + \epsilon\cos(\theta)$ or $-1 - \epsilon\cos(\theta)$.

Not if I know my algebraic operations:
I start with
iii) $±r=\Lambda-εrcos(\theta)$
adding $εrcos(\theta)$ to both sides yields
$±r+εrcos(\theta)=\Lambda$
taking out $r$ as a common factor on the left side yields
$r(±1+εcos(\theta))=\Lambda$
and lastly dividing by $(±1+εcos(\theta))$ yields
iv) $r=\frac \Lambda {±1+εcos(\theta)}$.
So $εcos(\theta)$ remains with a plus sign in both cases, and my issue still remains...

 

[Mark44]

I misunderstood. I thought you were just moving the $\pm$ from the numerator of the fraction on the left to its denominator.

 

[verty]

...
i) $(r±a)^2=(x+2εa)^2+y^2$
ii) $r^2=x^2+y^2$

Solving the system gives the equation $±r=(1-ε^2)a-εx$. Denoting $\Lambda=(1-ε^2)a$ and substituting $x=rcos(\theta)$ gives
iii) $±r=\Lambda-εrcos(\theta)$.
...

You get the formula $±r=(1-ε^2)a-εx$. Does this mean that the axis of symmetry is at $x = {\Lambda \over ε}$? Then you substitute in $x = r \cos{\theta}$. I wonder if this should be something like $x = {\Lambda \over ε} \pm ({\Lambda \over ε} + r \cos{\theta})$.

Anyway, that's the best I can do.

 

[Adgorn]

You get the formula $±r=(1-ε^2)a-εx$. Does this mean that the axis of symmetry is at $x = {\Lambda \over ε}$? Then you substitute in $x = r \cos{\theta}$. I wonder if this should be something like $x = {\Lambda \over ε} \pm ({\Lambda \over ε} + r \cos{\theta})$.

Anyway, that's the best I can do.

I'm not very experienced with hyperbolas at this point (as in, this is about the 3rd question I have ever did on them) so forgive me if I'm somewhat dull witted here. Since the foci of the described hyperbola are $(0,0)$ and $(-2εa,0)$, then the (vertical) axis of symmetry should be to my understanding in $x=-εa$, which is distinctly different than $x=\frac \Lambda ε={\frac {(1-ε^2)} ε}a$ for all possible ε.

Because I don't understand that I suppose it's not too surprising that I don't understand your 2nd claim about the formula for $x$, so I'm forced to rudely ask for a slightly more basics-covering explanation...

 

[verty]

Adgorn, I have checked your formula another way and I get the same result. I think the author was trying to be obtuse. He gave an answer not easily disproven. But it is not correct. And here is my philosophy about this. It's a hard question and proving him to be wrong is very difficult. I don't need to do it. I have corroborated your result and I trust you when you say it doesn't make sense. So I'm happy to move on.

Here is my derivation. Let $c = \sqrt{a^2 + b^2}$, $k = {a \over b}$:
$${(x + c)^2 \over a^2} - {y^2 \over b^2} = 1 \\
(x + c)^2 - k^2 y^2 - a^2 = 0 \\
(x^2 - k^2 y^2) + 2cx + b^2 = 0$$

Let $x = r \cos\theta$, $y = r \sin\theta$: $$r^2 (\cos^2\theta - k^2 \sin^2\theta) + 2cr \cos\theta + b^2 = 0 \\
(\cos^2\theta - k^2 \sin^2\theta) + {2c \cos\theta \over r} + {b^2 \over r^2} = 0$$

Let $z = {1 \over r}$: $$b^2 z^2 + 2c z \cos\theta + (\cos^2\theta - k^2 \sin^2\theta) = 0 \\
z = {1 \over 2 b^2} \bigg[-2 c \cos\theta \pm \sqrt{4 c^2 \cos^2\theta - 4 b^2 (\cos^2\theta - k^2 \sin^2\theta)}\ \bigg] \\
z = {1 \over b^2} \bigg[-c \cos\theta \pm \sqrt{a^2 \cos^2\theta + a^2 sin^2\theta)}\ \bigg] \\
z = {- c \cos\theta \pm a \over b^2} \\
r = {b^2 \over \pm a - c \cos\theta}$$

To remove b and c, let $\varepsilon = {c \over a}$. Then: $$\varepsilon^2 - 1 = {b^2 \over a^2} \\
r = {a(\varepsilon^2 - 1) \over \pm 1 - \varepsilon\cos\theta} = {a(1 - \varepsilon^2) \over \pm 1 + \varepsilon\cos\theta}$$