Polar equation of a hyperbola with one focus at the origin

In summary: I'm not entirely sure how you got to the last equation, but if I understand correctly from the 1st line, the 2nd line of your derivation, you found the distance of any point on the hyperbola from the left focus. In my development, the distance from the right focus is denoted with ##r## and from the left focus with ##s##. My 1st equation is an application of the definition of a hyperbola, and my 2nd equation is an application of the distance equation. As I understand, your derivation gives the same equation as the first, whereas I do not understand what your last line is supposed to mean, nor how it's derived. I have some other issues with your
  • #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.
 
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  • #2
Adgorn said:
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
Mark44 said:
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
I misunderstood. I thought you were just moving the ##\pm## from the numerator of the fraction on the left to its denominator.
 
  • #5
Adgorn said:
...
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
verty said:
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
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: $$\varepsilon^2 - 1 = {b^2 \over a^2} \\
r = {a(\varepsilon^2 - 1) \over \pm 1 - \varepsilon\cos\theta} = {a(1 - \varepsilon^2) \over \pm 1 + \varepsilon\cos\theta}$$
 

1. What is the polar equation of a hyperbola with one focus at the origin?

The polar equation of a hyperbola with one focus at the origin is given by r = a/(1±e*cosθ), where r is the distance from the origin to a point on the hyperbola, a is the distance from the origin to the vertex of the hyperbola, e is the eccentricity, and θ is the angle between the polar axis and the line connecting the focus and the point on the hyperbola.

2. How do you graph a polar equation of a hyperbola with one focus at the origin?

To graph a polar equation of a hyperbola with one focus at the origin, first plot the focus at the origin. Then, use the eccentricity to determine the shape of the hyperbola. If e <1, the hyperbola will be a right hyperbola, if e >1, it will be a left hyperbola. Finally, plot points on the hyperbola by plugging in different values of θ and solving for r. Connect the points to sketch the hyperbola.

3. What is the significance of the focus in the polar equation of a hyperbola with one focus at the origin?

The focus in the polar equation of a hyperbola with one focus at the origin is the fixed point that determines the shape of the hyperbola. It is the point that the distance from any point on the hyperbola to the focus is constant.

4. How does the eccentricity affect the polar equation of a hyperbola with one focus at the origin?

The eccentricity, represented by the variable e, determines the distance between the focus and the vertex of the hyperbola. It also affects the shape of the hyperbola - if e <1, the hyperbola will be a right hyperbola, if e >1, it will be a left hyperbola.

5. Can the polar equation of a hyperbola with one focus at the origin be converted to a Cartesian equation?

Yes, the polar equation of a hyperbola with one focus at the origin can be converted to a Cartesian equation using the following formula: x^2/a^2 - y^2/b^2 = 1, where a is the distance from the origin to the vertex of the hyperbola and b = a*e.

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