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'
-- + ----- = 0
a2 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 ellipse
a2 yo y + b2 xo x = a2 b2
<=>
xo x yo y
---- + ---- = 1
a2 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. |