Polar equation of a hyperbola with one focus at the origin

  • #1
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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##
Definiton 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.
 
Last edited:

Answers and Replies

  • #2
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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)##.
Adgorn said:
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.
 
  • #3
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10
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...
 
  • #4
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I misunderstood. I thought you were just moving the ##\pm## from the numerator of the fraction on the left to its denominator.
 
  • #5
verty
Homework Helper
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...
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.
 
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  • #6
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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...
 
  • #7
verty
Homework Helper
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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: $$\varepsilon^2 - 1 = {b^2 \over a^2} \\
r = {a(\varepsilon^2 - 1) \over \pm 1 - \varepsilon\cos\theta} = {a(1 - \varepsilon^2) \over \pm 1 + \varepsilon\cos\theta}$$
 

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