I'm trying to follow the derivation of equipotential lines surrounding two parallel, oppositely charged wires, found here. But I feel like the source is skipping a few critical steps in the derivation right at the end, and it's got me lost. I'm pretty sure the missing steps are just alegbra involving hyperbolic trig functions, which is why I'm asking here instead of on physics stack exchange.
The derivation arrives at the expression $\frac{(x+a)^2 + y^2}{(x-a)^2 + y^2} = e^{2 \eta} $. They next say,
"Then, introducing the hyperbolic trigonometric functions,
$2 \sinh \eta = e^\eta – e^{- \eta}$
and
$2 \cosh \eta = e^\eta + e^{- \eta}$,
where
$e^\eta = \cosh \eta + \sinh \eta$
and using
$\cosh^2 \eta – \sinh^2 \eta = 1$,
we obtain
$(x – a \coth \eta)^2 + y^2 = \left( \frac{a}{\sinh \eta} \right)^2 $."
All of this makes perfect sense to me right up to the very last line. I'm struggling to understand how they combine all this information to arrive at the final expression. Any insight would be greatly appreciated! Thank you so much in advance.
Best Answer
The way I should use $$\frac{(x+a)^2 + y^2}{(x-a)^2 + y^2} = k\implies (x+a)^2 + y^2=k(x-a)^2 + ky^2$$ that is to say $$(x+a)^2-k(x-a)^2=-(1-k)y^2$$ Develop the lhs $$(x+a)^2-k(x-a)^2=(1-k) x^2+2 a (k+1) x+(1-k)a^2$$ Divide everything by $(1-k)$ to get $$x^2-\frac{2 a (k+1) }{k-1}x+a^2=-y^2$$ Complete the square $$\left(x-\frac{ a (k+1) }{k-1}\right)^2-\left(\frac{ a (k+1) }{k-1}\right)^2+a^2=-y^2$$ that is to say $$\left(x-\frac{ a (k+1) }{k-1}\right)^2+y^2=\frac{4 a^2 k}{(k-1)^2}$$ Now, make $k=e^{2\eta}$ which gives $$\frac{ k+1}{k-1}=\coth (\eta )\qquad \text{and} \qquad \frac{ k}{(k-1)^2}=\frac{\text{csch}^2(\eta )}{4}$$ and you get the formula.