This page assumes you know nothing about ellipses. Before you read the parent note, we build every symbol it uses, from the ground up. Each item gives you: the plain-words meaning → the picture → why the topic needs it.
Every point in this chapter is written (x,y): walk x steps right (or left if negative) from the centre, then y steps up (or down if negative). Two crossing number-lines — the x-axis (horizontal) and y-axis (vertical) — meet at the origin(0,0).
Figure s01 shows the point P=(3,2): a dashed black arrow runs 3 right along the x-axis, then 2 up, landing on the red dot P. It is the visual dictionary for "(x,y) means go right then up".
The whole ellipse is defined by distances. So the very first tool we need is: given two points, how far apart are they?
Why this formula and not something simpler? Because a straight gap that goes both sideways and up cannot be measured by adding the two gaps — that would be the L-shaped walking distance, not the flying distance. The flying distance is the hypotenuse, and the only rule that turns two legs into a hypotenuse is the Pythagorean theorem(leg)2+(leg)2=(hyp)2. Pythagoras gives us the square of the distance, so to recover the plain length we must take the square root of both sides — that is exactly why a sits in front of the whole formula.
Figure s02 makes this concrete for P=(1,1) and Q=(5,3): the horizontal leg (x2−x1) and vertical leg (y2−y1) are drawn in black, and the red hypotenuse is the distance PQ we are after.
Here P is now a general point on the ellipse (any spot the pencil visits), and PF1, PF2 mean the distances from that P to each focus — using exactly the two-letter distance notation from §1.
A rule we must respect: the string must be longer than the nail gap, otherwise the pencil cannot leave the line between the nails. In symbols 2a>2c, i.e. a>c. Hold onto this — it is why b2 below is positive.
Figure s03 shows the two red foci F1=(−c,0), F2=(c,0), an arbitrary point P on the black oval, and the two black segments PF1 and PF2 whose lengths always add to 2a no matter where P sits.
Once a and c exist, a third length appears for free from a right triangle.
Why the hypotenuse of that triangle equals a (the crucial step, spelled out). Take P to be the top point(0,b). By left–right symmetry it is equally far from both foci, so PF1=PF2. But the defining rule says PF1+PF2=2a; two equal things adding to 2a means each one is a, so PF1=a. Now look at the right triangle with corners centre (0,0), focus (c,0), and top (0,b): its legs are c (along the axis) and b (straight up), and its hypotenuse is the segment from focus to top — which we just proved has length a. Pythagoras on that triangle gives the relation below.
Figure s04 draws that triangle right on top of a faint ellipse: leg c to a focus, leg b up to the co-vertex, and the red hypotenuse labelled a — the picture-proof of a2=b2+c2.