I understand that a hyperbola can be defined as the locus of all points on a plane such that the absolute value of the difference between the distance to the foci is $2a$, the distance between the two vertices.
In the simple case of a horizontal hyperbola centred on the origin, we have the following:
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$\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$
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$c = \sqrt{a^2 + b^2} = a\varepsilon = a\sqrt{1 + \frac{b^2}{a^2}}$
The foci lie at $(\pm c, 0)$.
Now, if I'm not wrong about that, then this should be pretty basic algebra, but I can't see how to get from the above to an equation given a point $(x,y)$ describing the difference in distances to the foci as being $2a$. While I actually do care about the final result, how to get there is more important.
Why do I want to know this? Well, I'd like to attempt trilateration based off differences in distance rather than fixed radii.
Best Answer
We will use a little trick to avoid work. We want to have $$\sqrt{(x+c)^2+y^2} -\sqrt{(x-c)^2+y^2}=\pm 2a.\qquad\text{(Equation 1)}$$
Rationalize the numerator, by multiplying "top" and "bottom" by $\sqrt{(x+c)^2+y^2} +\sqrt{(x-c)^2+y^2}.$ After the (not very dense) smoke clears, we get $$\frac{4xc}{\sqrt{(x+c)^2+y^2} +\sqrt{(x-c)^2+y^2}}=\pm 2a.$$ Flip it over, do some easy algebra. We get $$\sqrt{(x+c)^2+y^2} +\sqrt{(x-c)^2+y^2}=\pm \frac{2cx}{a}.\qquad\text{(Equation 2)}$$ From Equations 1 and 2, by adding, we get $$2\sqrt{(x+c)^2+y^2}=\pm 2\left(a+ \frac{xc}{a}\right).$$ Cancel the $2$'s, square. We get $$x^2+2cx+c^2+y^2=a^2+ 2cx+ \frac{c^2x^2}{a^2}.$$ Now it's basically over, the $2cx$ terms cancel. Multiply through by $a^2$, put $c^2=a^2+b^2$, and rearrange.