Where does $b^2 = a^2 - c^2$ come from in derivation of the standard equation of an ellipse? At the level of precalculus, there's a certain derivation of the equation of an ellipse that follows the same path. See this for example. 
Inevitably at one point in the derivation there's this substitution, with no explanation given:
$b^2 = a^2 - c^2$
where $a$ is the length of a major axis, $b$ is the length of a minor axis, and $c$ is the length from center to focus.
How is that equation possible? It doesn't seem to make sense from just looking at the ellipse.
 A: Take a vertex of the ellipse on the $y$ axis. For this point the distance from  the origin of the axis is $b$, the distance from a focus is $a$ (by simmetry and noting, using the vertex on the $x$ axis, that the sum of the distances of a point from the two foci is $2a$) and the distance of the focus from the origin is $c$, and $b$ , $c$ are sides of a right-angled triangle with hypotenuse $a$.
You can see this image from here.

A: Draw a right triangle from $(0,0)$ to $(c,0)$ to $(0,b)$.  The pythagorean theorem tells us that $h^2 = b^2 + c^2$ where $h$, the hypotenuse, is the distance from $(c,0)$ to $(0,b)$. [Which by the distance formula is $\sqrt {(c-0)^2 + (0-b)^2} = \sqrt {c^2 + b^2}$.  But that is redundant because the distance formula IS the pythagorean theorem.]
Likewise the right triangle from $(0,0)$ to $(-c,0)$ to $(0,b)$ has the same hypotenuse.
Because this is an ellipse $h + h = 2h =D$, some constant. 
Now consider the distance from $(c,0)$ to $(a,0)$.  This is $d_1 = a-c$.  And consider the distance from $(c,0)$ to $(-a,0)$.  This is $c + a$.  
Because the is an ellipse $(a-c) + (c + a) = 2a = D = 2h$.
So $a = h$.
And $a^2 = c^2 + b^2$.
Or $c^2 = a^2 - b^2$.
