Today while proving the equation of hyperbolas,$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\tag{1}$$ I came across this expression $$b^2=c^2-a^2\tag 2$$ Though this expression seems much like Pythagorean Theorem, I couldn't find any relation between the two (sorry for that). I'm using my math textbook to prove the hyperbola equation but even there nothing is mentioned about that "Pythagorean-like" equation. It's simply written there.
My question is :
Could you please give me an intuitive proof of that equation? I don't need the full hyperbola proof. I'll work it out myself. Just want to know how is the equation in $(2)$ derived.
P.S-I know a healthy bit of geometry so you can use it in your answers.
From focus $F_2$ draw a perpendicular to $x$-axis, to intersect the hyperbola at $P=(c,y)$. By definition we know that $PF_1=y+2a$, hence by Pythagoras' theorem: $$ (y+2a)^2=y^2+(2c)^2, \quad\text{that is:}\quad y={c^2-a^2\over a}. $$ Inserting then the coordinates of $P$ into the hyperbola equation gives: $$ {c^2\over a^2}-{(c^2-a^2)^2\over a^2b^2}=1, \quad\text{that is:}\quad c^2-a^2=b^2. $$