<s>[INST] Hyperbolas: Deriving $\frac{x^2}{a^2} + \frac{y^2}{a^2 - c^2} = 1$ from $\sqrt{(x + c)^2 + y^2} - \sqrt{(x - c)^2 + y^2} = \pm2a$ My textbook's section on Hyperbolas states the following:

If the foci are $F_1(-c, 0)$ and $F_2(c, 0)$ and the constant difference is $2a$, then a point $(x, y)$ lies on the hyperbola if and only if $\sqrt{(x + c)^2 + y^2} - \sqrt{(x - c)^2 + y^2} = \pm2a$.
To simplify this equation, we move the second radical to the right-hand side, square, isolate the remaining radical, and square again, obtaining $\dfrac{x^2}{a^2} + \dfrac{y^2}{a^2 - c^2} = 1$.

I've attempted to "move the second radical to the right-hand side, square, isolate the remaining radical, and square again", but I cannot derive $\dfrac{x^2}{a^2} + \dfrac{y^2}{a^2 - c^2} = 1$. I could be misunderstanding the instructions, but my attempts to derive the textbook's solution by precisely following the instructions have not been successful.
I would greatly appreciate it if people could please take the time to demonstrate the derivation of $\dfrac{x^2}{a^2} + \dfrac{y^2}{a^2 - c^2} = 1$ from $\sqrt{(x + c)^2 + y^2} - \sqrt{(x - c)^2 + y^2} = \pm2a$, as mentioned in the textbook. Are the textbook's instructions incorrect/insufficient or am I simply misunderstanding them?
 [/INST] Let's do the case
$$
\sqrt{(x + c)^2 + y^2} - \sqrt{(x - c)^2 + y^2} = 2a
$$
which holds provided $x>0$.
Move the second radical to the right-hand side and square:
$$
(x + c)^2 + y^2=4a^2+(x - c)^2 + y^2 + 4a\sqrt{(x - c)^2 + y^2}
$$
Lots of terms cancel:
$$
2cx=4a^2-2cx+4a\sqrt{(x - c)^2 + y^2}
$$
which becomes
$$
cx-a^2=a\sqrt{(x - c)^2 + y^2}
$$
Note that $cx\ge a^2$, so we can square again and arrive easily to the result.
The case $x<0$ is identical.
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