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I have a problem with the answer provided in the solution manual of Calculus, Michael Spivak, 3rd ed,
The Problem:
Consider a hyperbola, where the difference of the distance between the two foci is the constant 2a, and choose one focus at O and the other at (-2 $\varepsilon$,0). (In this case, we must have $\varepsilon > 1$). Show that we obtain the exact same equation in polar coordinates:
$$r=\frac{\Lambda}{1+\varepsilon cos\theta}$$ And here is the solution from the solution manual: The distance r from (x,y) to O is given by $$r^2=x^2+y^2{} \ (1).$$ While the distance s to f is given by $$s^2=(x+2\varepsilon a)^2+y^2.$$ Now writing the condition $$r-s=2a$$ as $$r-2a=s$$ and squaring, we get the equation:$$4a^2-4ar+r^2=x^2+4\varepsilon ax+4\varepsilon^2 a^2+y^2{} \ (2)$$ so subtracting (1) equation from (2) gives: $$a-r=\varepsilon x+\varepsilon ^2 a$$ and thus $r=\Lambda -\varepsilon x$, for $\Lambda=(1-\varepsilon ^2)a$, then $$r=\frac{\Lambda}{1+\varepsilon cos\theta}.$$ It remains to consider the points satisfying $$s-r=2a,$$ or $$r+2a=s$$ Squaring we now obtain $$r^2+4ar+4a^2=x^2+4\varepsilon ax+4\varepsilon ^2 a^2+y^2{} \ (\acute{2}).$$
Subtracting (1) from ($\acute{2}$) gives
$$a+r=\varepsilon x+\varepsilon ^2a,$$ or
$$r=(\varepsilon ^2-1) a + \varepsilon x = -(\Lambda-\varepsilon x),$$ Which is simply the negative of the r found previously;thus, the other branch of the hyperbola is obtained by choosing $-\Lambda$ for $\Lambda$.
The hyperbola

Now I think there is something wrong. Because if $-\Lambda =\Lambda$, then $$r=\Lambda + \varepsilon x \Longrightarrow r=\Lambda + \varepsilon rcos \theta \Longrightarrow r=\frac{\Lambda}{1-\varepsilon cos\theta}.$$
And simply this is different from the formula for the other branch of the hyperbola.

Any response would be appreciated.

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1 Answer 1

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The proof breaks down near the end. We have $r = (\varepsilon ^2-1) a +\varepsilon x = -\Lambda + \varepsilon r cos \theta$. It follows that $r = -\Lambda / (1 - \varepsilon cos \theta)$. Now replace $r$ by $-r_1$ and $\theta$ by $180 + \theta_1$ to get $r_1 = \Lambda / (1 + \varepsilon cos \theta_1)$, which is the same as the earlier formula.

Note that the left half of the hyperbola corresponds to positive values of $r$, while the right half corresponds to negative ones.

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