- #1
Adgorn
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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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