Take two points F and F' and a strictly positive value 2a
such that 2a is greater than |F,F'|.
The locus of all points D such that |D,F| + |D,F'| = 2a is an ellipse.
We choose the line FF' as x-axis and the perpendicular bissector of the segment
[F,F'] as y-axis.
We give F and F' resp. coordinates (c,0) and (-c,0).

We see that a>c
The points F and F' are called the foci of the ellipse
D(x,y) is on the ellipse <=>|D,F| + |D,F'| = 2a
<=>_______________ _______________
| 2 2 | 2 2
\| (x - c) + y + \| (x + c) + y = 2a
Squaring <=> ________________________________2 2 2 2 | 2 2 2 2 2
(x - c) + y + (x + c) + y + 2 \| ((x + c) + y ) ((x - c) + y ) = 4a
<=> ... <=> ________________________________| 2 2 2 2
\| ((x + c) + y ) ((x - c) + y ) = 2 a2 - (x2 + y2 + c2 )
<=> _______________________________________________| 2 2 2 2 2 2
\| (x + c + y + 2 c x) (x + c + y - 2 c x) = 2 a2 - (x2 + y2 + c2 ) (*)
Squaring =>(x2 + y2 + c2 )2 - 4 x2 c2 = 4 a4 - 4 a2 (x2 + y2 + c2 ) + (x2 + y2 + c2 )2 (**)
<=>- 4 x2 c2 = 4 a4 - 4 a2 (x2 + y2 + c2 )
<=> ... <=>(a 2 - c2 ) x2 + a 2 y2 = a2 (a2 - c2 )
Since a > c , we can say a2 - c2 = b 2
<=>b2 x2 + a2 y2 = a2 b2
<=>2 2
x y
-- + -- = 1 (***)
2 2
a b
This
is the equation of the ellipse.
But we don't know if (*) and (**) are equivalent because we don't know if the
right side of (*) is a positive value.
From above we know that if |D,F| + |D,F'| = 2a then (***) holds.
To prove the reverse, it is sufficient to show that if (***) holds, then the
right side of (*) is a positive value.
Well, from (***) we have
x2 y2
-- < or = 1 and -- < or = 1
a2 b2
then x2< or = a2 and y2 < or = b2
then x2 + y2 < or = a2 + b2
then x2 + y2 + c2 < or = a2 + b2 + c2
then x2 + y2 + c2 < or = 2a2 since b2 + c2 = a2
then the right side of (*) is a positive value.
Now
|D,F| + |D,F'| = 2a and (***) are equivalent.
The intersection points of the ellipse with the x-axis are A'(-a,0) and A(a,0).
These are the vertices on the x-axis.
The intersection points of the ellipse with the y-axis are B'(-b,0) and B(b,0).
These are the vertices on the y-axis.
The segment [A',A] is called major axis of the ellipse.
The segment [B',B] is called minor axis of the ellipse.
The segments [D,F'] and [D,F] are the focal radii through point D.
If
F = F' then c = 0 and a = b and then |D,F| + |D,F| = 2a
Hence |D,F| = a .
The ellipse is a circle with equation
x2 + y2 = a2
The radius is a.
Take in a plane two lines l and m with resp. equations
x = a cos(t) (1)
y = b sin(t) (2)
The
real numer t is the parameter.
We know, from the theory of 'Elimination of parameters', that the intersection
points of the two associated lines constitute a curve. To obtain the equation
of that curve, we eliminate the parameter t from the two equations. This means
that we search for the condition such that (1) and (2) has a solution for t.
The simultaneous equations (1) and (2) are equivalent to
x / a = cos(t) y / b = sin(t)
This system has a solution for t if and only if
sin2 (t) + cos2 (t) = 1<=>
x2 y2
-- + -- = 1
a2 b2
Hence,
the two associated lines constitute a curve and that curve is the ellipse.
We say that (1) and (2) are parametric equations of the ellipse.
The point
D(a cos(t) , b sin(t))
is on the ellipse for each t-value and with each point of the ellipse corresponds a t-value. From this it follows, as a special case, that
x = a cos(t) y = a sin(t)
are
parametric equations of the circle with radius a.
Then, the point D(a cos(t) , a sin(t)) is a variable point of that circle.
Again,
take a variable point D(a cos(t) , a sin(t)) of the circle with radius a.
Now, we compress the circle in the y-direction with a factor b/a. The
coordinates of the variable point of the new curve are D(a cos(t) , b sin(t)).
From this, we see that the new curve is the ellipse
x2 y2
-- + -- = 1
a2 b2
Take the ellipse
x2 y2
-- + -- = 1
a2 b2
To obtain the slope of the tangent line we differentiate implicitly.
2x 2y y'
-- + ----- = 0a2 b2
Solving for y', we obtain
b2 x
y'= - ----
a2 y
Say
D(xo,yo) is a fixed point of the ellipse.
The slope of the tangent line in point D is
b2 xo y'= - ------ a2 yo
The equation of the tangent line is
b2 xo y - yo = - ----- (x - xo) a2 yo<=>
a2 yo y - a2 yo2 = b2 xo2 - b2 xo x
<=>
a2 yo y + b2 xo x = a2 yo2 + b2 xo
<=>
since D(xo,yo) is on the ellipsea2 yo y + b2 xo x = a2 b2
<=>
xo x yo y
---- + ---- = 1a2 b2
The last equation is the tangent line in point D(xo,yo) of an ellipse.
Take
the bisectors t and n of the lines DF and DF'.
Say F" is the reflection point of F with respect to t.
Take any point T on t different from D.

Since
|D,F| = |D,F"| , |F',F"| = 2a .
Now in the triangle F'TF" , we see that
|F',T| + |T,F"| > 2a=> |T,F'| + |T,F| > 2a
And
from the definition of ellipse, it follows that T is outside of the ellipse.
Hence all the points of t, different from D, are outside of the ellipse and
therefore the bissector t of the lines DF and DF' is a tangent line of the ellipse.
The line n is a normal of the ellipse.
Since
|F',F"| = 2a = constant, we see that the mirror image of F with respect to
a variable tangent line is on the circle with center F' and with radius 2a.
Call P the projection of F on the tangent line.
Point O is the midpoint of the segment [F,F'] and point P is midpoint of the
segment [F,F"]. Hence |O,P| = a .
The orthogonal projection of F on a variable tangent line is the circle with
center O and radius a.