Figure 1: Ellipse (red curve) at origin with major axis horizontal. Origin at point
O
{\displaystyle O}
:
(
0
,
0
)
{\displaystyle :(0,0)}
. Foci are points
F
1
(
−
c
,
0
)
,
F
2
(
c
,
0
)
.
O
F
1
=
O
F
2
=
c
.
{\displaystyle F_{1}(-c,0),\ F_{2}(c,0).OF_{1}=OF_{2}=c.}
Line segment
V
1
O
V
2
{\displaystyle V_{1}OV_{2}}
is the
m
a
j
o
r
a
x
i
s
.
{\displaystyle major\ axis.}
O
V
1
=
O
V
2
=
B
1
F
1
=
B
1
F
2
=
a
.
{\displaystyle \ \ \ \ \ OV_{1}=OV_{2}=B_{1}F_{1}=B_{1}F_{2}=a.}
Line segment
B
1
O
B
2
{\displaystyle B_{1}OB_{2}}
is the
m
i
n
o
r
a
x
i
s
.
O
B
1
=
O
B
2
=
b
.
{\displaystyle minor\ axis.\ OB_{1}=OB_{2}=b.}
Each line
L
1
F
1
R
1
,
L
2
F
2
R
2
{\displaystyle L_{1}F_{1}R_{1},L_{2}F_{2}R_{2}}
is a
l
a
t
u
s
r
e
c
t
u
m
.
{\displaystyle latus\ rectum.}
Each line
x
=
D
1
,
x
=
D
2
{\displaystyle x=D_{1},x=D_{2}}
is a
d
i
r
e
c
t
r
i
x
.
{\displaystyle directrix.}
P
F
1
+
P
F
2
=
2
a
.
{\displaystyle PF_{1}+PF_{2}=2a.}
In cartesian geometry in two dimensions the ellipse is the locus of a point
P
{\displaystyle P}
that moves relative to two fixed points called
f
o
c
i
{\displaystyle foci}
:
F
1
,
F
2
.
{\displaystyle :F_{1},F_{2}.}
The distance
F
1
F
2
{\displaystyle F_{1}F_{2}}
from one
f
o
c
u
s
(
F
1
)
{\displaystyle focus\ (F_{1})}
to the other
f
o
c
u
s
(
F
2
)
{\displaystyle focus\ (F_{2})}
is non-zero. The sum of the distances
(
P
F
1
,
P
F
2
)
{\displaystyle (PF_{1},PF_{2})}
from point to foci is constant.
P
F
1
+
P
F
2
=
K
.
{\displaystyle PF_{1}+PF_{2}=K.}
See figure 1.
The center of the ellipse is located at the origin
O
(
0
,
0
)
{\displaystyle O(0,0)}
and the foci
(
F
1
,
F
2
)
{\displaystyle (F_{1},F_{2})}
are on the
X
a
x
i
s
{\displaystyle X\ axis}
at distance
c
{\displaystyle c}
from
O
.
{\displaystyle O.}
F
1
{\displaystyle F_{1}}
has coordinates
(
−
c
,
0
)
.
F
2
{\displaystyle (-c,0).F_{2}}
has coordinates
(
c
,
0
)
{\displaystyle (c,0)}
. Line segments
O
F
1
=
O
F
2
=
c
.
{\displaystyle OF_{1}=OF_{2}=c.}
By definition
P
F
1
+
P
F
2
=
B
1
F
1
+
B
1
F
2
=
V
1
F
1
+
V
1
F
2
=
K
.
{\displaystyle PF_{1}+PF_{2}=B_{1}F_{1}+B_{1}F_{2}=V_{1}F_{1}+V_{1}F_{2}=K.}
V
1
F
1
=
V
2
F
2
.
∴
V
1
F
1
+
V
1
F
2
=
V
1
F
2
+
V
2
F
2
=
V
1
V
2
=
K
=
2
a
,
{\displaystyle V_{1}F_{1}=V_{2}F_{2}.\ \therefore V_{1}F_{1}+V_{1}F_{2}=V_{1}F_{2}+V_{2}F_{2}=V_{1}V_{2}=K=2a,}
the length of the
m
a
j
o
r
a
x
i
s
(
V
1
V
2
)
.
O
V
1
=
O
V
2
=
a
.
{\displaystyle major\ axis\ (V_{1}V_{2}).\ OV_{1}=OV_{2}=a.}
Each point
(
V
1
,
V
2
)
{\displaystyle (V_{1},V_{2})}
where the curve intersects the major axis is called a
v
e
r
t
e
x
.
V
1
,
V
2
{\displaystyle vertex.\ V_{1},V_{2}}
are the vertices of the ellipse.
Line segment
B
1
B
2
{\displaystyle B_{1}B_{2}}
perpendicular to the major axis at the midpoint of the major axis is the
m
i
n
o
r
a
x
i
s
{\displaystyle minor\ axis}
with length
2
b
.
O
B
1
=
O
B
2
=
b
.
{\displaystyle 2b.\ OB_{1}=OB_{2}=b.}
B
1
F
1
+
B
1
F
2
=
2
a
;
B
1
F
1
=
B
1
F
2
=
a
.
∴
a
2
=
b
2
+
c
2
.
{\displaystyle B_{1}F_{1}+B_{1}F_{2}=2a;\ B_{1}F_{1}=B_{1}F_{2}=a.\ \therefore a^{2}=b^{2}+c^{2}.}
Any line segment that intersects the curve in two places is a
c
h
o
r
d
.
{\displaystyle chord.}
A chord through the focus is a
f
o
c
a
l
c
h
o
r
d
.
{\displaystyle focal\ chord.}
Each focal chord
L
1
F
1
R
1
,
L
2
F
2
R
2
{\displaystyle L_{1}F_{1}R_{1},\ L_{2}F_{2}R_{2}}
perpendicular to the major axis is a
l
a
t
u
s
r
e
c
t
u
m
.
{\displaystyle latus\ rectum.}
P
F
1
+
P
F
2
=
2
a
{\displaystyle PF_{1}+PF_{2}=2a}
Let point
P
{\displaystyle P}
have coordinates
(
x
,
y
)
.
{\displaystyle (x,y).}
P
F
1
=
(
x
−
(
−
c
)
)
2
+
(
y
−
0
)
2
=
(
x
+
c
)
2
+
y
2
=
M
{\displaystyle PF_{1}={\sqrt {(x-(-c))^{2}+(y-0)^{2}}}={\sqrt {(x+c)^{2}+y^{2}}}=M}
P
F
2
=
(
x
−
c
)
2
+
y
2
=
N
{\displaystyle PF_{2}={\sqrt {(x-c)^{2}+y^{2}}}=N}
M
M
=
(
x
+
c
)
2
+
y
2
;
N
N
=
(
x
−
c
)
2
+
y
2
.
{\displaystyle MM=(x+c)^{2}+y^{2};\ NN=(x-c)^{2}+y^{2}.}
(
x
+
c
)
2
+
y
2
+
(
x
−
c
)
2
+
y
2
=
2
a
{\displaystyle {\sqrt {(x+c)^{2}+y^{2}}}+{\sqrt {(x-c)^{2}+y^{2}}}=2a}
M
+
N
=
2
a
{\displaystyle M+N=2a}
M
M
+
2
M
N
+
N
N
=
4
a
a
{\displaystyle MM+2MN+NN=4aa}
2
M
N
=
4
a
a
−
M
M
−
N
N
{\displaystyle 2MN=4aa-MM-NN}
4
M
M
N
N
=
(
4
a
a
−
M
M
−
N
N
)
2
{\displaystyle 4MMNN=(4aa-MM-NN)^{2}}
4
M
M
N
N
−
(
4
a
a
−
M
M
−
N
N
)
2
=
0
{\displaystyle 4MMNN-(4aa-MM-NN)^{2}=0}
Make appropriate substitutions, expand and the result is:
16
a
a
x
x
−
16
c
c
x
x
+
16
a
a
y
y
−
16
a
a
a
a
+
16
a
a
c
c
=
0
{\displaystyle 16aaxx-16ccxx+16aayy-16aaaa+16aacc=0}
(
a
a
−
c
c
)
x
x
+
a
a
y
y
−
a
a
(
a
a
−
c
c
)
=
0
{\displaystyle (aa-cc)xx+aayy-aa(aa-cc)=0}
b
2
x
2
+
a
2
y
2
−
a
2
b
2
=
0
{\displaystyle b^{2}x^{2}+a^{2}y^{2}-a^{2}b^{2}=0}
or
x
2
a
2
+
y
2
b
2
=
1
{\displaystyle {\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=1}
If the equation
b
2
x
2
+
a
2
y
2
−
a
2
b
2
=
0
{\displaystyle b^{2}x^{2}+a^{2}y^{2}-a^{2}b^{2}=0}
is expressed as
A
x
2
+
B
y
2
+
C
=
0
:
{\displaystyle Ax^{2}+By^{2}+C=0:}
a
2
=
−
C
A
;
b
2
=
−
C
B
.
{\displaystyle a^{2}={\frac {-C}{A}};\ b^{2}={\frac {-C}{B}}.}
Length of latus rectum
edit
b
2
x
2
+
a
2
y
2
−
a
2
b
2
=
0
{\displaystyle b^{2}x^{2}+a^{2}y^{2}-a^{2}b^{2}=0}
b
2
c
2
+
a
2
y
2
−
a
2
b
2
=
0
{\displaystyle b^{2}c^{2}+a^{2}y^{2}-a^{2}b^{2}=0}
b
2
(
a
2
−
b
2
)
+
a
2
y
2
−
a
2
b
2
=
0
{\displaystyle b^{2}(a^{2}-b^{2})+a^{2}y^{2}-a^{2}b^{2}=0}
b
2
a
2
−
b
4
+
a
2
y
2
−
a
2
b
2
=
0
{\displaystyle b^{2}a^{2}-b^{4}+a^{2}y^{2}-a^{2}b^{2}=0}
a
2
y
2
=
b
4
{\displaystyle a^{2}y^{2}=b^{4}}
y
2
=
b
4
a
2
{\displaystyle y^{2}={\frac {b^{4}}{a^{2}}}}
y
=
b
2
a
{\displaystyle y={\frac {b^{2}}{a}}}
Length of latus rectum
=
L
1
R
1
=
L
2
R
2
=
2
b
2
a
.
{\displaystyle =L_{1}R_{1}=L_{2}R_{2}={\frac {2b^{2}}{a}}.}
Second definition of ellipse
edit
Graph of ellipse
x
2
a
2
+
y
2
b
2
=
1
{\displaystyle {\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=1}
where
a
,
b
=
20
,
12
{\displaystyle a,b=20,12}
. At point
B
,
u
v
=
e
.
{\displaystyle B,\ {\frac {u}{v}}=e.}
At point
A
,
a
t
=
e
.
{\displaystyle A,\ {\frac {a}{t}}=e.}
Ellipse is path of point that moves so that ratio of distance to
fixed point and distance to fixed line is constant.
Let
p
a
=
e
{\displaystyle {\frac {p}{a}}=e}
where:
p
{\displaystyle p}
is non-zero,
a
>
p
,
{\displaystyle a>p,}
a
=
p
+
u
.
{\displaystyle a=p+u.}
Therefore,
1
>
e
>
0.
{\displaystyle 1>e>0.}
Let directrix have equation
x
=
t
{\displaystyle x=t}
where
a
t
=
e
.
{\displaystyle {\frac {a}{t}}=e.}
At point
B
:
{\displaystyle B:}
p
p
+
u
=
p
+
u
p
+
u
+
v
=
e
{\displaystyle {\frac {p}{p+u}}={\frac {p+u}{p+u+v}}=e}
(
p
+
u
)
2
=
p
(
p
+
u
+
v
)
{\displaystyle (p+u)^{2}=p(p+u+v)}
p
p
+
p
u
+
p
u
+
u
u
=
p
p
+
p
u
+
p
v
{\displaystyle pp+pu+pu+uu=pp+pu+pv}
p
u
+
u
u
=
p
v
{\displaystyle pu+uu=pv}
u
(
p
+
u
)
=
p
v
{\displaystyle u(p+u)=pv}
u
v
=
p
p
+
u
=
e
{\displaystyle {\frac {u}{v}}={\frac {p}{p+u}}=e}
distance to focus
distance to directrix
=
e
…
(
3
)
{\displaystyle {\frac {\text{distance to focus}}{\text{distance to directrix}}}=e\ \dots \ (3)}
Statement
(
3
)
{\displaystyle (3)}
is true at point
A
{\displaystyle A}
also.
Section under "Proof" below proves that statement (3) is true for any point
P
{\displaystyle P}
on ellipse.
Figure 1: Ellipse (red curve) at origin with major axis horizontal. Origin at point
O
{\displaystyle O}
:
(
0
,
0
)
{\displaystyle :(0,0)}
. Foci are points
F
1
(
−
c
,
0
)
,
F
2
(
c
,
0
)
.
O
F
1
=
O
F
2
=
c
.
{\displaystyle F_{1}(-c,0),\ F_{2}(c,0).OF_{1}=OF_{2}=c.}
Directrices are lines
x
=
D
1
,
x
=
D
2
,
D
1
=
−
D
2
.
{\displaystyle x=D_{1},\ x=D_{2},\ D_{1}=-D_{2}.}
O
V
1
=
O
V
2
=
a
.
{\displaystyle OV_{1}=OV_{2}=a.}
e
c
c
e
n
t
r
i
c
i
t
y
e
=
c
a
=
F
2
V
2
V
2
D
2
=
V
1
F
2
V
1
D
2
=
O
V
2
O
D
2
{\displaystyle eccentricity\ e={\frac {c}{a}}={\frac {F_{2}V_{2}}{V_{2}D_{2}}}={\frac {V_{1}F_{2}}{V_{1}D_{2}}}={\frac {OV_{2}}{OD_{2}}}}
e
=
P
F
1
P
E
1
{\displaystyle e={\frac {PF_{1}}{PE_{1}}}}
(yellow lines);
e
=
P
F
2
P
E
2
{\displaystyle e={\frac {PF_{2}}{PE_{2}}}}
(purple lines) To define ellipse specify
f
o
c
u
s
,
d
i
r
e
c
t
r
i
x
,
e
.
{\displaystyle focus,\ directrix,\ e.}
See Figure 1. The vertical line through
D
2
{\displaystyle D_{2}}
with equation
x
=
D
2
{\displaystyle x=D_{2}}
is a
d
i
r
e
c
t
r
i
x
{\displaystyle directrix}
of the ellipse. Likewise for the vertical line
x
=
D
1
;
D
1
=
−
D
2
.
{\displaystyle x=D_{1};\ D_{1}=-D_{2}.}
A second definition of the ellipse: the locus of a point that moves relative to a point, the focus, and a fixed line called the directrix, so that the distance from point to focus and the distance from point to line form a constant ratio
e
=
e
c
c
e
n
t
r
i
c
i
t
y
,
0
<
e
<
1.
{\displaystyle e=eccentricity,\ 0<e<1.}
Consider the point
V
2
{\displaystyle V_{2}}
in the figure.
F
2
V
2
V
2
D
2
=
e
.
{\displaystyle {\frac {F_{2}V_{2}}{V_{2}D_{2}}}=e.}
Let the length
V
2
D
2
=
d
.
a
−
c
d
=
e
.
d
=
a
−
c
e
{\displaystyle V_{2}D_{2}=d.\ {\frac {a-c}{d}}=e.\ d={\frac {a-c}{e}}}
Consider the point
V
1
{\displaystyle V_{1}}
:
V
1
F
2
V
1
D
2
=
a
+
c
2
a
+
d
=
e
=
(
a
+
c
)
/
(
2
a
+
a
−
c
e
)
=
(
a
+
c
)
/
(
2
a
e
+
a
−
c
e
)
=
e
(
a
+
c
)
2
a
e
+
a
−
c
.
{\displaystyle {\frac {V_{1}F_{2}}{V_{1}D_{2}}}={\frac {a+c}{2a+d}}=e=(a+c)/(2a+{\frac {a-c}{e}})=(a+c)/({\frac {2ae+a-c}{e}})={\frac {e(a+c)}{2ae+a-c}}.}
2
a
e
+
a
−
c
=
a
+
c
;
2
a
e
=
2
c
.
{\displaystyle 2ae+a-c=a+c;\ 2ae=2c.}
e
=
c
a
.
{\displaystyle e={\frac {c}{a}}.}
Length
O
D
2
=
a
+
d
=
a
+
a
−
c
e
=
a
+
a
−
c
c
/
a
=
a
+
a
(
a
−
c
)
c
=
a
c
+
a
a
−
a
c
c
=
a
2
c
{\displaystyle OD_{2}=a+d=a+{\frac {a-c}{e}}=a+{\frac {a-c}{c/a}}=a+{\frac {a(a-c)}{c}}={\frac {ac+aa-ac}{c}}={\frac {a^{2}}{c}}}
.
Distance from center to directrix
=
O
D
2
=
a
2
c
=
a
2
e
a
=
a
e
=
c
e
2
.
e
=
O
V
2
O
D
2
.
{\displaystyle =OD_{2}={\frac {a^{2}}{c}}={\frac {a^{2}}{ea}}={\frac {a}{e}}={\frac {c}{e^{2}}}.\ e={\frac {OV_{2}}{OD_{2}}}.}
Distance from center to directrix =
O
F
2
+
F
2
D
2
=
c
+
d
0
=
c
e
2
{\displaystyle OF_{2}+F_{2}D_{2}=c+d_{0}={\frac {c}{e^{2}}}}
.
e
2
c
+
e
2
d
0
=
c
;
c
−
e
2
c
=
e
2
d
0
;
c
=
e
2
d
0
1
−
e
2
{\displaystyle e^{2}c+e^{2}d_{0}=c;\ c-e^{2}c=e^{2}d_{0};\ c={\frac {e^{2}d_{0}}{1-e^{2}}}}
.
1
−
e
2
=
1
−
c
2
a
2
=
a
2
−
c
2
a
2
=
b
2
a
2
{\displaystyle 1-e^{2}=1-{\frac {c^{2}}{a^{2}}}={\frac {a^{2}-c^{2}}{a^{2}}}={\frac {b^{2}}{a^{2}}}}
.
e
2
1
−
e
2
=
c
2
a
2
/
b
2
a
2
=
c
2
b
2
{\displaystyle {\frac {e^{2}}{1-e^{2}}}={\frac {c^{2}}{a^{2}}}/{\frac {b^{2}}{a^{2}}}={\frac {c^{2}}{b^{2}}}}
.
Distance from focus to directrix
=
F
2
D
2
=
b
2
c
{\displaystyle =F_{2}D_{2}={\frac {b^{2}}{c}}}
.
Ellipse using focus and directrix
edit
Figure 2: Ellipse (red curve) at origin with major axis vertical. Origin at point
O
{\displaystyle O}
:
(
0
,
0
)
{\displaystyle :(0,0)}
. Focus at point
F
{\displaystyle F}
:
(
0
,
−
c
)
.
{\displaystyle :(0,-c).}
Directrix is line
D
E
{\displaystyle DE}
:
y
=
−
c
e
2
.
{\displaystyle :\ y={\frac {-c}{e^{2}}}.}
Point
P
{\displaystyle P}
has coordinates:
(
x
,
y
)
.
{\displaystyle (x,y).}
e
c
c
e
n
t
r
i
c
i
t
y
e
=
P
F
P
E
{\displaystyle eccentricity\ e={\frac {PF}{PE}}}
(yellow lines)
See Figure 2. Let the point
P
{\displaystyle P}
be
(
x
,
y
)
{\displaystyle (x,y)}
, the focus
F
{\displaystyle F}
be
(
0
,
−
c
)
{\displaystyle (0,-c)}
and the directrix
D
E
{\displaystyle DE}
have equation
y
=
−
c
e
2
{\displaystyle y={\frac {-c}{e^{2}}}}
where
1
>
e
>
0.
{\displaystyle 1>e>0.}
The directrix is horizontal and the major axis vertical through the origin
(
0
,
0
)
{\displaystyle (0,0)}
.
P
F
=
x
2
+
(
y
+
c
)
2
{\displaystyle PF={\sqrt {x^{2}+(y+c)^{2}}}}
P
E
=
y
+
c
e
2
{\displaystyle PE=y+{\frac {c}{e^{2}}}}
e
=
P
F
P
E
;
P
F
=
e
P
E
.
{\displaystyle e={\frac {PF}{PE}};\ PF=ePE.}
x
2
+
(
y
+
c
)
2
=
e
(
y
+
c
e
2
)
{\displaystyle {\sqrt {x^{2}+(y+c)^{2}}}=e(y+{\frac {c}{e^{2}}})}
e
x
2
+
(
y
+
c
)
2
=
e
e
(
y
+
c
e
2
)
=
e
e
y
+
c
{\displaystyle e{\sqrt {x^{2}+(y+c)^{2}}}=ee(y+{\frac {c}{e^{2}}})=eey+c}
e
e
(
x
2
+
(
y
+
c
)
2
)
=
(
e
e
y
+
c
)
2
{\displaystyle ee(x^{2}+(y+c)^{2})=(eey+c)^{2}}
e
e
(
x
2
+
(
y
+
c
)
2
)
−
(
e
e
y
+
c
)
2
=
0
{\displaystyle ee(x^{2}+(y+c)^{2})-(eey+c)^{2}=0}
Expand and the result is:
+
e
e
x
x
+
e
e
y
y
−
e
e
e
e
y
y
+
c
c
e
e
−
c
c
=
0
{\displaystyle +eexx+eeyy-eeeeyy+ccee-cc=0}
x
x
+
y
y
−
e
e
y
y
+
c
c
−
c
c
e
e
=
0
{\displaystyle xx+yy-eeyy+cc-{\frac {cc}{ee}}=0}
x
x
+
(
1
−
e
e
)
y
y
+
c
c
−
a
a
=
0
{\displaystyle xx+(1-ee)yy+cc-aa=0}
x
x
+
(
1
−
e
e
)
y
y
−
(
a
a
−
c
c
)
=
0
{\displaystyle xx+(1-ee)yy-(aa-cc)=0}
x
x
+
(
1
−
e
e
)
y
y
−
b
b
=
0
{\displaystyle xx+(1-ee)yy-bb=0}
1
−
e
e
=
1
−
c
c
a
a
=
a
a
−
c
c
a
a
=
b
b
a
a
{\displaystyle 1-ee=1-{\frac {cc}{aa}}={\frac {aa-cc}{aa}}={\frac {bb}{aa}}}
x
x
+
(
b
b
a
a
)
y
y
−
b
b
=
0
{\displaystyle xx+({\frac {bb}{aa}})yy-bb=0}
a
2
x
2
+
b
2
y
2
−
a
2
b
2
=
0
{\displaystyle a^{2}x^{2}+b^{2}y^{2}-a^{2}b^{2}=0}
Compare this equation with the equation generated earlier:
b
2
x
2
+
a
2
y
2
−
a
2
b
2
=
0
{\displaystyle b^{2}x^{2}+a^{2}y^{2}-a^{2}b^{2}=0}
.
When the equation is
b
2
x
2
+
a
2
y
2
−
a
2
b
2
=
0
,
{\displaystyle b^{2}x^{2}+a^{2}y^{2}-a^{2}b^{2}=0,}
the major axis is horizontal.
When the equation is
a
2
x
2
+
b
2
y
2
−
a
2
b
2
=
0
,
{\displaystyle a^{2}x^{2}+b^{2}y^{2}-a^{2}b^{2}=0,}
the major axis is vertical.
General ellipse at origin
edit
Figure 1: Ellipse (red curve) at origin with major axis oblique. Origin at point
O
{\displaystyle O}
:
(
0
,
0
)
{\displaystyle :(0,0)}
. Foci are points
F
1
(
−
p
,
−
q
)
,
F
2
(
p
,
q
)
.
{\displaystyle F_{1}(-p,-q),\ F_{2}(p,q).}
O
F
1
=
O
F
2
=
c
=
p
2
+
q
2
.
{\displaystyle OF_{1}=OF_{2}=c={\sqrt {p^{2}+q^{2}}}.}
Line segment
V
1
O
V
2
{\displaystyle V_{1}OV_{2}}
is the
m
a
j
o
r
a
x
i
s
.
{\displaystyle major\ axis.}
O
V
1
=
O
V
2
=
a
.
{\displaystyle \ \ \ \ \ OV_{1}=OV_{2}=a.}
Line
D
1
O
D
2
{\displaystyle D_{1}OD_{2}}
is the major axis extended. Lines perpendicular to
D
1
O
D
2
{\displaystyle D_{1}OD_{2}}
at
D
1
,
F
1
,
O
,
F
2
,
D
2
{\displaystyle D_{1},F_{1},O,F_{2},D_{2}}
are directrix, latus rectum. minor axis, latus rectum, directrix.
O
D
1
=
O
D
2
=
a
2
c
.
{\displaystyle OD_{1}=OD_{2}={\frac {a^{2}}{c}}.}
F
1
D
1
=
F
2
D
2
=
b
2
c
.
{\displaystyle F_{1}D_{1}=F_{2}D_{2}={\frac {b^{2}}{c}}.}
P
F
1
+
P
F
2
=
2
a
.
{\displaystyle PF_{1}+PF_{2}=2a.}
The general ellipse allows for the major axis to have slope other than horizontal or vertical. See Figure 1.
The line
V
1
V
2
{\displaystyle V_{1}V_{2}}
is the major axis of the ellipse shown in red. Points
F
1
,
F
2
{\displaystyle F_{1},F_{2}}
are the foci with
coordinates
(
−
p
,
−
q
)
,
(
p
,
q
)
{\displaystyle (-p,-q),(p,q)}
respectively.
Point
P
(
x
,
y
)
{\displaystyle P(x,y)}
is any point on the curve. By definition
P
F
1
+
P
F
2
=
2
a
.
{\displaystyle PF_{1}+PF_{2}=2a.}
Length
P
F
1
=
(
x
−
(
−
p
)
)
2
+
(
y
−
(
−
q
)
)
2
=
(
x
+
p
)
2
+
(
y
+
q
)
2
=
M
.
{\displaystyle PF_{1}={\sqrt {(x-(-p))^{2}+(y-(-q))^{2}}}={\sqrt {(x+p)^{2}+(y+q)^{2}}}=M.}
Length
P
F
2
=
(
x
−
p
)
2
+
(
y
−
q
)
2
=
N
.
{\displaystyle PF_{2}={\sqrt {(x-p)^{2}+(y-q)^{2}}}=N.}
(
x
+
p
)
2
+
(
y
+
q
)
2
+
(
x
−
p
)
2
+
(
y
−
q
)
2
=
2
a
.
{\displaystyle {\sqrt {(x+p)^{2}+(y+q)^{2}}}+{\sqrt {(x-p)^{2}+(y-q)^{2}}}=2a.}
M
+
N
=
2
a
{\displaystyle M+N=2a}
M
M
+
2
M
N
+
N
N
=
4
a
a
{\displaystyle MM+2MN+NN=4aa}
2
M
N
=
4
a
a
−
M
M
−
N
N
{\displaystyle 2MN=4aa-MM-NN}
4
M
M
N
N
=
(
4
a
a
−
M
M
−
N
N
)
2
{\displaystyle 4MMNN=(4aa-MM-NN)^{2}}
4
M
M
N
N
−
(
4
a
a
−
M
M
−
N
N
)
2
=
0.
{\displaystyle 4MMNN-(4aa-MM-NN)^{2}=0.}
Make appropriate substitutions, expand and the result is:
(
a
a
−
p
p
)
x
2
−
2
p
q
x
y
+
(
a
a
−
q
q
)
y
2
+
a
a
p
p
+
a
a
q
q
−
a
a
a
a
=
0
{\displaystyle (aa-pp)x^{2}-2pqxy+(aa-qq)y^{2}+aapp+aaqq-aaaa=0}
.
This equation has the form
A
x
2
+
B
x
y
+
C
y
2
+
D
x
+
E
y
+
F
=
0
{\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0}
where:
A
=
a
2
−
p
2
{\displaystyle A=a^{2}-p^{2}}
B
=
−
2
p
q
{\displaystyle B=-2pq}
C
=
a
2
−
q
2
{\displaystyle C=a^{2}-q^{2}}
D
=
E
=
0
{\displaystyle D=E=0}
because center is at origin,
F
=
a
2
p
2
+
a
2
q
2
−
a
4
=
a
2
(
p
2
+
q
2
−
a
2
)
=
a
2
(
c
2
−
a
2
)
=
−
a
2
b
2
.
{\displaystyle F=a^{2}p^{2}+a^{2}q^{2}-a^{4}=a^{2}(p^{2}+q^{2}-a^{2})=a^{2}(c^{2}-a^{2})=-a^{2}b^{2}.}
An example
Let
(
p
,
q
)
{\displaystyle (p,q)}
be (3,4) and
a
=
8.
{\displaystyle a=8.}
The ellipse is:
880
x
2
−
384
x
y
+
768
y
2
+
(
0
)
x
+
(
0
)
y
−
39936
=
0
,
{\displaystyle 880x^{2}-384xy+768y^{2}+(0)x+(0)y-39936=0,}
or
55
x
2
−
24
x
y
+
48
y
2
+
(
0
)
x
+
(
0
)
y
−
2496
=
0.
{\displaystyle 55x^{2}-24xy+48y^{2}+(0)x+(0)y-2496=0.}
Reverse-engineering ellipse at origin
Given an ellipse in format
A
x
2
+
B
x
y
+
C
y
2
+
(
0
)
x
+
(
0
)
y
+
F
=
0
,
{\displaystyle Ax^{2}+Bxy+Cy^{2}+(0)x+(0)y+F=0,}
calculate
p
,
q
,
a
.
(
B
{\displaystyle p,q,a.\ (B}
is non-zero.
)
{\displaystyle )}
A
=
a
2
−
p
2
{\displaystyle A=a^{2}-p^{2}}
B
=
−
2
p
q
{\displaystyle B=-2pq}
C
=
a
2
−
q
2
{\displaystyle C=a^{2}-q^{2}}
F
=
a
2
p
2
+
a
2
q
2
−
a
4
.
{\displaystyle F=a^{2}p^{2}+a^{2}q^{2}-a^{4}.}
Coefficients provided could be, for example,
(
55
,
−
24
,
48
,
0
,
0
,
−
2496
)
{\displaystyle (55,-24,48,0,0,-2496)}
or
(
110
,
−
48
,
96
,
0
,
0
,
−
4992
)
{\displaystyle (110,-48,96,0,0,-4992)}
or
(
55
k
,
−
24
k
,
48
k
,
0
,
0
,
−
2496
k
)
{\displaystyle (55k,-24k,48k,0,0,-2496k)}
where
k
{\displaystyle k}
is an arbitrary constant and all groups of coefficients define the same ellipse.
To produce consistent, correct values for
p
,
q
,
a
{\displaystyle p,q,a}
the equations become:
K
A
=
a
2
−
p
2
{\displaystyle KA=a^{2}-p^{2}}
K
B
=
−
2
p
q
{\displaystyle KB=-2pq}
K
C
=
a
2
−
q
2
{\displaystyle KC=a^{2}-q^{2}}
K
F
=
a
2
p
2
+
a
2
q
2
−
a
4
.
{\displaystyle KF=a^{2}p^{2}+a^{2}q^{2}-a^{4}.}
or:
K
A
−
(
a
2
−
p
2
)
=
0
{\displaystyle KA-(a^{2}-p^{2})=0}
K
B
−
(
−
2
p
q
)
=
0
{\displaystyle KB-(-2pq)=0}
K
C
−
(
a
2
−
q
2
)
=
0
{\displaystyle KC-(a^{2}-q^{2})=0}
K
F
−
(
a
2
p
2
+
a
2
q
2
−
a
4
)
=
0.
{\displaystyle KF-(a^{2}p^{2}+a^{2}q^{2}-a^{4})=0.}
Solutions are:
K
=
4
F
B
2
−
4
A
C
{\displaystyle K={\frac {4F}{B^{2}-4AC}}}
. This formula for
K
{\displaystyle K}
is valid if both
D
,
E
{\displaystyle D,E}
are
0
{\displaystyle 0}
.
4
(
p
p
)
2
+
4
K
(
A
−
C
)
p
p
−
B
2
K
2
=
0
{\displaystyle 4(pp)^{2}+4K(A-C)pp-B^{2}K^{2}=0}
where
p
p
=
p
2
.
{\displaystyle pp=p^{2}.}
You should see one positive value
(
p
2
)
{\displaystyle (p^{2})}
and one negative value
(
−
(
q
2
)
)
{\displaystyle (-(q^{2}))}
for
p
p
.
{\displaystyle pp.}
Choose the positive value and
p
=
p
p
{\displaystyle p={\sqrt {pp}}}
.
q
=
−
K
B
2
p
{\displaystyle q={\frac {-KB}{2p}}}
a
=
A
K
+
p
2
{\displaystyle a={\sqrt {AK+p^{2}}}}
The solutions become simpler if K == 1.
if ( K != 1 ) { A ← KA; B ← KB; C ← KC; } and the solutions for
p
,
q
,
a
{\displaystyle p,q,a}
become:
4
(
p
p
)
2
+
4
(
A
−
C
)
p
p
−
B
2
=
0
{\displaystyle 4(pp)^{2}+4(A-C)pp-B^{2}=0}
where
p
p
=
p
2
.
{\displaystyle pp=p^{2}.}
q
=
−
B
2
p
{\displaystyle q={\frac {-B}{2p}}}
a
=
A
+
p
2
{\displaystyle a={\sqrt {A+p^{2}}}}
With values
p
,
q
,
a
{\displaystyle p,q,a}
available all the familiar values of the ellipse may be calculated:
c
=
p
2
+
q
2
{\displaystyle c={\sqrt {p^{2}+q^{2}}}}
Equation of major axis:
y
=
q
p
x
;
q
x
−
p
y
+
0
=
0
;
q
c
x
−
p
c
y
+
0
=
0
{\displaystyle y={\frac {q}{p}}x;\ qx-py+0=0;\ {\frac {q}{c}}x-{\frac {p}{c}}y+0=0}
in normal form.
Equation of minor axis:
p
c
x
+
q
c
y
+
0
=
0
{\displaystyle {\frac {p}{c}}x+{\frac {q}{c}}y+0=0}
in normal form.
Equations of directrices:
p
c
x
+
q
c
y
±
a
2
c
=
0
{\displaystyle {\frac {p}{c}}x+{\frac {q}{c}}y\pm {\frac {a^{2}}{c}}=0}
in normal form.
An example using focus and directrix
Figure 2: Ellipse (red curve) at origin with major axis oblique. Origin at point
O
{\displaystyle O}
:
(
0
,
0
)
{\displaystyle :(0,0)}
. Focus at point
F
{\displaystyle F}
:
(
−
20
,
15
)
{\displaystyle :(-20,15)}
Directrix is line
D
E
{\displaystyle DE}
:
4
5
x
−
3
5
y
+
36
=
0.
{\displaystyle :{\frac {4}{5}}x-{\frac {3}{5}}y+36=0.}
eccentricity
e
=
5
6
=
P
1
F
P
1
D
=
P
2
F
P
2
E
.
{\displaystyle e={\frac {5}{6}}={\frac {P_{1}F}{P_{1}D}}={\frac {P_{2}F}{P_{2}E}}.}
Ellipse has equation:
20
x
2
+
24
x
y
+
27
y
2
−
9900
=
0.
{\displaystyle 20x^{2}+24xy+27y^{2}-9900=0.}
Given focus
(
−
20
,
15
)
,
e
=
5
6
{\displaystyle (-20,15),\ e={\frac {5}{6}}}
and directrix with equation
4
5
x
−
3
5
y
+
36
=
0
{\displaystyle {\frac {4}{5}}x-{\frac {3}{5}}y+36=0}
,
calculate equation of the ellipse in form
A
x
2
+
B
x
y
+
C
y
2
+
D
x
+
E
y
+
F
=
0.
{\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0.}
See Figure 2. Let point
P
(
x
,
y
)
{\displaystyle P\ (x,y)}
be any point on the ellipse.
Distance from
P
{\displaystyle P}
to focus
=
(
x
+
20
)
2
+
(
y
−
15
)
2
{\displaystyle ={\sqrt {(x+20)^{2}+(y-15)^{2}}}}
Distance from
P
{\displaystyle P}
to directrix
=
0.8
x
−
0.6
y
+
36
{\displaystyle =0.8x-0.6y+36}
.
(
x
+
20
)
2
+
(
y
−
15
)
2
=
5
6
(
0.8
x
−
0.6
y
+
36
)
{\displaystyle {\sqrt {(x+20)^{2}+(y-15)^{2}}}={\frac {5}{6}}(0.8x-0.6y+36)}
.
6
(
x
+
20
)
2
+
(
y
−
15
)
2
=
5
(
0.8
x
−
0.6
y
+
36
)
=
4
x
−
3
y
+
180
{\displaystyle 6{\sqrt {(x+20)^{2}+(y-15)^{2}}}=5(0.8x-0.6y+36)=4x-3y+180}
.
36
(
(
x
+
20
)
2
+
(
y
−
15
)
2
)
=
(
4
x
−
3
y
+
180
)
2
{\displaystyle 36((x+20)^{2}+(y-15)^{2})=(4x-3y+180)^{2}}
.
36
(
(
x
+
20
)
2
+
(
y
−
15
)
2
)
−
(
4
x
−
3
y
+
180
)
2
=
0
{\displaystyle 36((x+20)^{2}+(y-15)^{2})-(4x-3y+180)^{2}=0}
.
Expand and the result is:
20
x
2
+
24
x
y
+
27
y
2
+
(
0
)
x
+
(
0
)
y
−
9900
=
0.
{\displaystyle 20x^{2}+24xy+27y^{2}+(0)x+(0)y-9900=0.}
Because
D
=
E
=
0
,
{\displaystyle D=E=0,}
the center of the ellipse is at the origin and the various lines have equations as follows.
Minor axis:
4
5
x
−
3
5
y
+
0
=
0
{\displaystyle {\frac {4}{5}}x-{\frac {3}{5}}y+0=0}
Other directrix:
4
5
x
−
3
5
y
−
36
=
0
{\displaystyle {\frac {4}{5}}x-{\frac {3}{5}}y-36=0}
Major axis:
3
5
x
+
4
5
y
+
0
=
0
{\displaystyle {\frac {3}{5}}x+{\frac {4}{5}}y+0=0}
If you reverse-engineer the ellipse using the method above,
K
=
25
{\displaystyle K=25}
,
the expression
K
A
=
a
2
−
p
2
{\displaystyle KA=a^{2}-p^{2}}
becomes
(
25
)
(
20
)
=
30
2
−
20
2
{\displaystyle (25)(20)=30^{2}-20^{2}}
, and
the expression
K
F
=
−
a
2
b
2
{\displaystyle KF=-a^{2}b^{2}}
becomes
(
25
)
(
−
9900
)
=
−
(
30
2
)
(
275
)
{\displaystyle (25)(-9900)=-(30^{2})(275)}
.
Figure 1: The general ellipse (red curve). Foci at
S
1
,
S
2
,
{\displaystyle S_{1},S_{2},}
center at
C
2
.
{\displaystyle C_{2}.}
Point
V
{\displaystyle V}
is a vertex. Length
V
C
2
=
a
.
{\displaystyle VC_{2}=a.}
P
S
1
+
P
S
2
=
2
a
.
{\displaystyle PS_{1}+PS_{2}=2a.}
The ellipse may assume any position and any orientation in Cartesian two-dimensional space. See the red curve in Figure 1.
The equation of this curve may be derived as follows:
Given two
f
o
c
i
:
S
1
(
s
,
t
)
,
S
2
(
u
,
v
)
,
{\displaystyle foci:\ S_{1}(s,t),\ S_{2}(u,v),}
m
a
j
o
r
a
x
i
s
{\displaystyle major\ axis}
of length
2
a
{\displaystyle 2a}
and point
P
(
x
,
y
)
{\displaystyle P(x,y)}
P
S
1
+
P
S
2
=
2
a
{\displaystyle PS_{1}+PS_{2}=2a}
(
x
−
s
)
2
+
(
y
−
t
)
2
+
(
x
−
u
)
2
+
(
y
−
v
)
2
=
2
a
{\displaystyle {\sqrt {(x-s)^{2}+(y-t)^{2}}}+{\sqrt {(x-u)^{2}+(y-v)^{2}}}=2a}
The expansion of this expression is somewhat complicated because it contains 5 variables
s
,
t
,
u
,
v
,
a
.
{\displaystyle s,t,u,v,a.}
The expansion may be simplified by reducing the number of variables from
5
{\displaystyle 5}
to
3
{\displaystyle 3}
, the familiar
p
,
q
,
a
.
{\displaystyle p,q,a.}
Let
C
2
{\displaystyle C_{2}}
, the center of the ellipse, have coordinates
(
G
,
H
)
{\displaystyle (G,H)}
where
G
=
s
+
u
2
;
H
=
t
+
v
2
{\displaystyle G={\frac {s+u}{2}};\ H={\frac {t+v}{2}}}
and
p
=
u
−
G
;
q
=
v
−
H
.
{\displaystyle p=u-G;\ q=v-H.}
Then
(
s
,
t
)
=
(
G
−
p
,
H
−
q
)
,
{\displaystyle (s,t)=(G-p,H-q),}
(
u
,
v
)
=
(
G
+
p
,
H
+
q
)
,
{\displaystyle (u,v)=(G+p,H+q),}
and
(
x
−
(
G
−
p
)
)
2
+
(
y
−
(
H
−
q
)
)
2
+
(
x
−
(
G
+
p
)
)
2
+
(
y
−
(
H
+
q
)
)
2
=
2
a
{\displaystyle {\sqrt {(x-(G-p))^{2}+(y-(H-q))^{2}}}+{\sqrt {(x-(G+p))^{2}+(y-(H+q))^{2}}}=2a}
The expansion is:
A
x
2
+
B
x
y
+
C
y
2
+
D
x
+
E
y
+
F
=
0
{\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0}
where:
A
=
a
2
−
p
2
{\displaystyle A=a^{2}-p^{2}}
B
=
−
2
p
q
{\displaystyle B=-2pq}
C
=
a
2
−
q
2
{\displaystyle C=a^{2}-q^{2}}
D
=
2
(
G
p
2
+
H
p
q
−
G
a
2
)
{\displaystyle D=2(Gp^{2}+Hpq-Ga^{2})}
E
=
2
(
H
q
2
+
G
p
q
−
H
a
2
)
{\displaystyle E=2(Hq^{2}+Gpq-Ha^{2})}
F
=
(
a
p
)
2
+
(
a
q
)
2
−
a
4
+
(
G
2
+
H
2
)
a
2
−
(
G
p
+
H
q
)
2
{\displaystyle F=(ap)^{2}+(aq)^{2}-a^{4}+(G^{2}+H^{2})a^{2}-(Gp+Hq)^{2}}
A different approach
Begin with ellipse at origin:
A
x
2
+
B
x
y
+
C
y
2
+
F
=
0
{\displaystyle Ax^{2}+Bxy+Cy^{2}+F=0}
where:
A
=
a
2
−
p
2
{\displaystyle A=a^{2}-p^{2}}
B
=
−
2
p
q
{\displaystyle B=-2pq}
C
=
a
2
−
q
2
{\displaystyle C=a^{2}-q^{2}}
F
=
(
a
p
)
2
+
(
a
q
)
2
−
a
4
{\displaystyle F=(ap)^{2}+(aq)^{2}-a^{4}}
By translation of coordinates, move the ellipse so that the new center of the ellipse is:
(
G
,
H
)
.
x
{\displaystyle (G,H).\ x}
←
(
x
−
G
)
,
y
{\displaystyle (x-G),\ y}
←
(
y
−
H
)
.
{\displaystyle (y-H).}
The equation above becomes:
A
x
2
+
B
x
y
+
C
y
2
+
D
x
+
E
y
+
F
1
=
0
{\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F^{1}=0}
where:
D
=
−
(
2
A
G
+
B
H
)
;
E
=
−
(
B
G
+
2
C
H
)
;
F
1
=
A
G
2
+
B
G
H
+
C
H
2
+
F
.
{\displaystyle D=-(2AG+BH);\ E=-(BG+2CH);\ F^{1}=AG^{2}+BGH+CH^{2}+F.}
D
=
−
(
2
A
G
+
B
H
)
=
−
(
2
(
a
2
−
p
2
)
G
+
(
−
2
p
q
)
H
)
=
−
(
2
G
a
2
−
2
G
p
2
−
2
p
q
H
)
=
2
(
G
p
2
+
H
p
q
−
G
a
2
)
,
{\displaystyle D=-(2AG+BH)=-(2(a^{2}-p^{2})G+(-2pq)H)=-(2Ga^{2}-2Gp^{2}-2pqH)=2(Gp^{2}+Hpq-Ga^{2}),}
the same as the value of
D
{\displaystyle D}
in the method above.
The expansion of
E
,
F
1
{\displaystyle E,F^{1}}
will show that this method produces the same results as the method above.
Given foci and major axis, perhaps the simplest way to produce
A
x
2
+
B
x
y
+
C
y
2
+
D
x
+
E
y
+
F
1
{\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F^{1}}
is to calculate
A
x
2
+
B
x
y
+
C
y
2
+
F
{\displaystyle Ax^{2}+Bxy+Cy^{2}+F}
and move the center from
C
1
:
(
0
,
0
)
{\displaystyle C_{1}:\ (0,0)}
to
C
2
:
(
G
,
H
)
{\displaystyle C_{2}:\ (G,H)}
.
Center of ellipse
Given
A
x
2
+
B
x
y
+
C
y
2
+
D
x
+
E
y
+
F
=
0
,
{\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0,}
we know from above that
D
=
−
(
2
A
G
+
B
H
)
;
E
=
−
(
B
G
+
2
C
H
)
.
{\displaystyle D=-(2AG+BH);\ E=-(BG+2CH).}
Therefore
G
=
B
E
−
2
C
D
4
A
C
−
B
2
,
H
=
−
(
E
+
B
G
)
2
C
{\displaystyle G={\frac {BE-2CD}{4AC-B^{2}}},\ H={\frac {-(E+BG)}{2C}}}
where the point
(
G
,
H
)
{\displaystyle (G,H)}
is the center of the ellipse.
An Example of the General Ellipse
edit
Figure 1: Translation of coordinate axes. Red curve and green curve have same size, shape and orientation. The only difference is that center of green curve
C
1
(
0
,
0
)
{\displaystyle C_{1}(0,0)}
has been moved to
C
2
(
−
34
,
14
)
{\displaystyle C_{2}(-34,14)}
where it is the center of the red curve. In both curves
p
=
16
;
q
=
12.
V
1
C
1
=
V
2
C
2
=
a
=
25.
{\displaystyle p=16;\ q=12.\ V_{1}C_{1}=V_{2}C_{2}=a=25.}
See Figure 1. Given foci
(
−
50
,
2
)
,
(
−
18
,
26
)
{\displaystyle (-50,2),(-18,26)}
and
a
=
25
,
{\displaystyle a=25,}
calculate the equation of the ellipse in form
A
x
2
+
B
x
y
+
C
y
2
+
D
x
+
E
y
+
F
=
0.
{\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0.}
Calculate the center:
G
=
−
50
+
(
−
18
)
2
=
−
34
;
H
=
2
+
26
2
=
14.
{\displaystyle G={\frac {-50+(-18)}{2}}=-34;\ H={\frac {2+26}{2}}=14.}
p
=
−
18
−
(
−
34
)
=
16
;
q
=
26
−
14
=
12.
{\displaystyle p=-18-(-34)=16;\ q=26-14=12.}
Equation of ellipse at origin:
369
x
2
−
384
x
y
+
481
y
2
−
140625
=
0.
{\displaystyle 369x^{2}-384xy+481y^{2}-140625=0.}
Move ellipse from origin
C
1
{\displaystyle C_{1}}
to
C
2
(
−
34
,
14
)
:
{\displaystyle C_{2}\ (-34,14):}
369
x
2
−
384
x
y
+
481
y
2
+
30468
x
−
26524
y
+
562999
=
0.
{\displaystyle 369x^{2}-384xy+481y^{2}+30468x-26524y+562999=0.}
Reverse-engineering the general ellipse
Given ellipse
369
x
2
−
384
x
y
+
481
y
2
+
30468
x
−
26524
y
+
562999
=
0
,
{\displaystyle 369x^{2}-384xy+481y^{2}+30468x-26524y+562999=0,}
calculate the foci and the major axis.
Calculate the center of the ellipse (point
C
2
{\displaystyle C_{2}}
):
(
G
,
H
)
=
(
−
34
,
14
)
.
{\displaystyle (G,H)=(-34,14).}
K
A
−
(
a
2
−
p
2
)
=
0
{\displaystyle KA-(a^{2}-p^{2})=0}
K
B
−
(
−
2
p
q
)
=
0
{\displaystyle KB-(-2pq)=0}
K
C
−
(
a
2
−
q
2
)
=
0
{\displaystyle KC-(a^{2}-q^{2})=0}
K
F
−
(
(
a
p
)
2
+
(
a
q
)
2
−
a
4
+
(
G
2
+
H
2
)
a
2
−
(
G
p
+
H
q
)
2
)
=
0.
{\displaystyle KF-((ap)^{2}+(aq)^{2}-a^{4}+(G^{2}+H^{2})a^{2}-(Gp+Hq)^{2})=0.}
Solutions are:
K
=
4
F
−
4
(
A
G
2
+
B
G
H
+
C
H
2
)
B
2
−
4
A
C
{\displaystyle K={\frac {4F-4(AG^{2}+BGH+CH^{2})}{B^{2}-4AC}}}
where
A
=
369
;
B
=
−
384
;
C
=
481
;
F
=
562999
;
G
=
−
34
;
H
=
14.
{\displaystyle A=369;\ B=-384;\ C=481;\ F=562999;\ G=-34;\ H=14.}
In this example,
K
=
1.
{\displaystyle K=1.}
If
(
K
{\displaystyle (K}
!=
1
)
{
A
{\displaystyle 1)\ \{A}
←
A
K
;
B
{\displaystyle AK;\ B}
←
B
K
;
C
{\displaystyle BK;\ C}
←
C
K
}
{\displaystyle CK\}}
The values
p
,
q
,
a
{\displaystyle p,q,a}
may be calculated as in "Reverse-engineering ellipse at origin" above.
p
=
16
;
q
=
12
;
a
=
25.
{\displaystyle p=16;\ q=12;\ a=25.}
Length of major axis
=
2
a
=
2
(
25
)
=
50.
{\displaystyle =2a=2(25)=50.}
Focus
S
1
=
(
s
,
t
)
=
(
−
34
−
16
,
14
−
12
)
=
(
−
50
,
2
)
.
{\displaystyle S_{1}=(s,t)=(-34-16,14-12)=(-50,2).}
Focus
S
2
=
(
u
,
v
)
=
(
−
34
+
16
,
14
+
12
)
=
(
−
18
,
26
)
{\displaystyle S_{2}=(u,v)=(-34+16,14+12)=(-18,26)}
Significant lines of the Ellipse
Figure 2. Graph of ellipse illustrating axes, each latus rectum, each directrix. Directrix through
D
1
:
3
5
x
+
4
5
y
+
145
=
0
{\displaystyle D_{1}:\ {\frac {3}{5}}x+{\frac {4}{5}}y+145=0}
Directrix through
D
2
:
3
5
x
+
4
5
y
−
105
=
0
{\displaystyle D_{2}:\ {\frac {3}{5}}x+{\frac {4}{5}}y-105=0}
Latus rectum through
F
1
:
3
5
x
+
4
5
y
+
100
=
0
{\displaystyle F_{1}:\ {\frac {3}{5}}x+{\frac {4}{5}}y+100=0}
Latus rectum through
F
2
:
3
5
x
+
4
5
y
−
60
=
0.
{\displaystyle F_{2}:\ {\frac {3}{5}}x+{\frac {4}{5}}y-60=0.}
Major axis
:
y
=
4
3
x
+
100.
{\displaystyle :\ y={\frac {4}{3}}x+100.}
Minor axis
:
3
5
x
+
4
5
y
+
20
=
0.
{\displaystyle :\ {\frac {3}{5}}x+{\frac {4}{5}}y+20=0.}
The significant lines of the ellipse are:
m
a
j
o
r
a
x
i
s
,
m
i
n
o
r
a
x
i
s
,
{\displaystyle major\ axis,\ minor\ axis,}
each
l
a
t
u
s
r
e
c
t
u
m
,
{\displaystyle latus\ rectum,}
each
d
i
r
e
c
t
r
i
x
.
{\displaystyle directrix.}
Consider the ellipse in Figure 2. Given foci
F
1
=
(
−
108
,
−
44
)
,
F
2
=
(
−
12
,
84
)
{\displaystyle F_{1}=(-108,-44),\ F_{2}=(-12,84)}
and
a
=
100
,
{\displaystyle a=100,}
calculate the equations of all the significant lines.
Slope of major axis
=
84
−
(
−
44
)
−
12
−
(
−
108
)
=
128
96
=
4
3
.
{\displaystyle ={\frac {84-(-44)}{-12-(-108)}}={\frac {128}{96}}={\frac {4}{3}}.}
Major axis has equation
y
=
4
3
x
+
g
{\displaystyle y={\frac {4}{3}}x+g}
and it passes through
F
1
.
{\displaystyle F_{1}.}
Therefore,
g
=
−
44
−
4
3
(
−
108
)
=
−
44
+
144
=
100.
{\displaystyle g=-44-{\frac {4}{3}}(-108)=-44+144=100.}
Major axis
V
1
V
2
{\displaystyle V_{1}V_{2}}
has equation:
y
=
4
3
x
+
100.
{\displaystyle y={\frac {4}{3}}x+100.}
Center of ellipse
C
=
(
−
108
+
(
−
12
)
2
,
−
44
+
84
2
)
=
(
−
60
,
20
)
.
{\displaystyle C=({\frac {-108+(-12)}{2}},{\frac {-44+84}{2}})=(-60,20).}
Minor axis is perpendicular to major axis. Therefore, minor axis has equation:
y
=
−
3
4
x
+
g
{\displaystyle y=-{\frac {3}{4}}x+g}
and it passes through the center
C
.
{\displaystyle C.}
g
=
20
+
3
4
(
−
60
)
=
20
−
45
=
−
25.
{\displaystyle g=20+{\frac {3}{4}}(-60)=20-45=-25.}
Equation of minor axis (orange line through
C
{\displaystyle C}
):
y
=
−
3
4
x
−
25
{\displaystyle y=-{\frac {3}{4}}x-25}
or
3
x
+
4
y
+
100
=
0
{\displaystyle 3x+4y+100=0}
or
3
5
x
+
4
5
y
+
20
=
0.
{\displaystyle {\frac {3}{5}}x+{\frac {4}{5}}y+20=0.}
F
1
C
=
F
2
C
=
c
=
(
−
60
−
(
−
108
)
)
2
+
(
20
−
(
−
44
)
)
2
=
48
2
+
64
2
=
±
80.
{\displaystyle F_{1}C=F_{2}C=c={\sqrt {(-60-(-108))^{2}+(20-(-44))^{2}}}={\sqrt {48^{2}+64^{2}}}=\pm 80.}
Using the equation of the minor axis, the fact that each latus rectum is parallel to the minor axis, and that the distance
from minor axis to latus rectum
=
c
=
80
,
{\displaystyle =c=80,}
each latus rectum has equation:
3
5
x
+
4
5
y
+
20
±
80
=
0.
{\displaystyle {\frac {3}{5}}x+{\frac {4}{5}}y+20\pm 80=0.}
Equation of latus rectum (blue line) through
F
1
:
3
5
x
+
4
5
y
+
100
=
0
{\displaystyle F_{1}:\ {\frac {3}{5}}x+{\frac {4}{5}}y+100=0}
Equation of latus rectum (blue line) through
F
2
:
3
5
x
+
4
5
y
−
60
=
0.
{\displaystyle F_{2}:\ {\frac {3}{5}}x+{\frac {4}{5}}y-60=0.}
Using the equation of the minor axis, the fact that each directrix is parallel to the minor axis, and that the distance
from minor axis to directrix
=
D
1
C
=
D
2
C
=
a
2
c
=
100
2
80
=
125
,
{\displaystyle =D_{1}C=D_{2}C={\frac {a^{2}}{c}}={\frac {100^{2}}{80}}=125,}
each directrix has equation:
3
5
x
+
4
5
y
+
20
±
125
=
0.
{\displaystyle {\frac {3}{5}}x+{\frac {4}{5}}y+20\pm 125=0.}
Equation of directrix (red line) through
D
1
:
3
5
x
+
4
5
y
+
145
=
0
{\displaystyle D_{1}:\ {\frac {3}{5}}x+{\frac {4}{5}}y+145=0}
Equation of directrix (red line) through
D
2
:
3
5
x
+
4
5
y
−
105
=
0.
{\displaystyle D_{2}:\ {\frac {3}{5}}x+{\frac {4}{5}}y-105=0.}
Figure 1: Three ellipses illustrating "standard form." For green curve
K
=
25.
{\displaystyle K=25.}
For red curve
K
=
1
256
.
{\displaystyle K={\frac {1}{256}}.}
For blue curve
K
=
36.
{\displaystyle K=36.}
Everything about the ellipse can be derived from
G
,
H
,
p
,
q
,
a
{\displaystyle G,H,\ p,q,a}
the last three of which
(
p
,
q
,
a
)
{\displaystyle (p,q,a)}
are contained within:
A
=
a
2
−
p
2
{\displaystyle A=a^{2}-p^{2}}
B
=
−
2
p
q
{\displaystyle B=-2pq}
C
=
a
2
−
q
2
{\displaystyle C=a^{2}-q^{2}}
F
=
−
a
2
b
2
{\displaystyle F=-a^{2}b^{2}}
for ellipse at origin.
Consider the ellipse:
369
x
2
−
384
x
y
+
481
y
2
−
3
,
515
,
625
=
0
,
{\displaystyle 369x^{2}-384xy+481y^{2}-3,515,625=0,}
the green curve in Figure 1. It is tempting to say that
p
=
16
;
q
=
12
;
a
=
25.
{\displaystyle p=16;\ q=12;\ a=25.}
These values satisfy
A
=
a
2
−
p
2
=
25
2
−
16
2
=
369
;
B
=
−
2
p
q
=
−
2
(
16
)
(
12
)
=
−
384
;
C
=
a
2
−
q
2
=
25
2
−
12
2
=
481.
{\displaystyle A=a^{2}-p^{2}=25^{2}-16^{2}=369;\ B=-2pq=-2(16)(12)=-384;\ C=a^{2}-q^{2}=25^{2}-12^{2}=481.}
However,
c
2
=
400
;
b
2
=
a
2
−
c
2
=
625
−
400
=
225
;
F
=
−
a
2
b
2
=
−
(
625
)
(
225
)
=
−
140
,
625.
{\displaystyle c^{2}=400;\ b^{2}=a^{2}-c^{2}=625-400=225;\ F=-a^{2}b^{2}=-(625)(225)=-140,625.}
These values for
p
,
q
,
a
{\displaystyle p,q,a}
are not correct.
Put the equation of the ellipse into "standard form." In this context "standard form" means that
K
=
1.
{\displaystyle K=1.}
For ellipse at origin
K
=
4
F
B
2
−
4
A
C
=
4
(
−
3
,
515
,
625
)
(
−
384
)
2
−
4
(
369
)
(
481
)
=
−
14
,
062
,
500
−
562
,
500
=
25.
{\displaystyle K={\frac {4F}{B^{2}-4AC}}={\frac {4(-3,515,625)}{(-384)^{2}-4(369)(481)}}={\frac {-14,062,500}{-562,500}}=25.}
In fact
3
,
515
,
625
140
,
625
=
25
=
K
.
{\displaystyle {\frac {3,515,625}{140,625}}=25=K.}
A
{\displaystyle A}
←
A
K
;
B
{\displaystyle AK;\ B}
←
B
K
;
C
{\displaystyle BK;\ C}
←
C
K
;
F
{\displaystyle CK;\ F}
←
F
K
.
{\displaystyle FK.}
A
=
9
,
225
;
B
=
−
9
,
600
;
C
=
12
,
025
;
F
=
−
87
,
890
,
625.
{\displaystyle A=9,225;\ B=-9,600;\ C=12,025;\ F=-87,890,625.}
The equation of the ellipse becomes:
9
,
225
x
2
−
9
,
600
x
y
+
12
,
025
y
2
−
87
,
890
,
625
=
0
{\displaystyle 9,225x^{2}-9,600xy+12,025y^{2}-87,890,625=0}
and
K
=
4
F
B
2
−
4
A
C
=
−
351
,
562
,
500
(
−
9600
)
2
−
4
(
9225
)
(
12025
)
=
351562500
351562500
=
1.
{\displaystyle K={\frac {4F}{B^{2}-4AC}}={\frac {-351,562,500}{(-9600)^{2}-4(9225)(12025)}}={\frac {351562500}{351562500}}=1.}
The equation of the ellipse is in "standard form" and:
p
=
80
;
q
=
60
;
a
=
125.
{\displaystyle p=80;\ q=60;\ a=125.}
A
=
a
2
−
p
2
=
15
,
625
−
6
,
400
=
9
,
225.
{\displaystyle A=a^{2}-p^{2}=15,625-6,400=9,225.}
B
=
−
2
p
q
=
−
2
(
80
)
(
60
)
=
−
9
,
600.
{\displaystyle B=-2pq=-2(80)(60)=-9,600.}
C
=
a
2
−
q
2
=
15
,
625
−
3
,
600
=
12
,
025.
{\displaystyle C=a^{2}-q^{2}=15,625-3,600=12,025.}
c
2
=
p
2
+
q
2
=
80
2
+
60
2
=
100
2
;
b
2
=
a
2
−
c
2
=
15
,
625
−
10
,
000
=
5
,
625
;
{\displaystyle c^{2}=p^{2}+q^{2}=80^{2}+60^{2}=100^{2};\ b^{2}=a^{2}-c^{2}=15,625-10,000=5,625;}
F
=
−
a
2
b
2
=
−
15
,
625
(
5
,
625
)
=
−
87
,
890
,
625
{\displaystyle F=-a^{2}b^{2}=-15,625(5,625)=-87,890,625}
The values
p
=
80
;
q
=
60
;
a
=
125
{\displaystyle p=80;\ q=60;\ a=125}
are correct.
Example 2. Consider the ellipse
5
,
904
x
2
−
6
,
144
x
y
+
7
,
696
y
2
−
140
,
625
=
0
,
{\displaystyle 5,904x^{2}-6,144xy+7,696y^{2}-140,625=0,}
the red curve in Figure 1.
In this example,
K
=
1
256
=
0.003
,
906
,
25
{\displaystyle K={\frac {1}{256}}=0.003,906,25}
and the equation in "standard form" is:
23.0625
x
2
−
24
x
y
+
30.0625
y
2
−
549.316
,
406
,
25
=
0
{\displaystyle 23.0625x^{2}-24xy+30.0625y^{2}-549.316,406,25=0}
.
Example 3. Consider the ellipse
9
x
2
+
25
y
2
−
900
x
−
2000
y
+
54400
=
0
,
{\displaystyle 9x^{2}+25y^{2}-900x-2000y+54400=0,}
the blue curve in Figure 1.
The center
(
G
,
H
)
=
(
50
,
40
)
.
{\displaystyle (G,H)=(50,40).}
In this example
B
=
0
;
K
=
A
G
2
+
C
H
2
−
F
A
C
=
9
(
50
)
2
+
25
(
40
2
)
−
54400
9
(
25
)
=
36
{\displaystyle B=0;\ K={\frac {AG^{2}+CH^{2}-F}{AC}}={\frac {9(50)^{2}+25(40^{2})-54400}{9(25)}}=36}
and the equation in "standard form" is:
324
x
2
+
900
y
2
−
32400
x
−
72000
y
+
1958400
=
0
{\displaystyle 324x^{2}+900y^{2}-32400x-72000y+1958400=0}
.
Tangent at latus rectum
edit
Reflectivity of ellipse
edit
Figure 1: Ellipse (red curve) with major axis horizontal. Origin at point
O
{\displaystyle O}
:
(
0
,
0
)
{\displaystyle :(0,0)}
. Foci are points
F
1
(
−
c
,
0
)
,
F
2
(
c
,
0
)
.
{\displaystyle F_{1}(-c,0),\ F_{2}(c,0).}
Line
T
1
P
T
2
{\displaystyle T_{1}PT_{2}}
tangent to curve at
P
{\displaystyle P}
. Angle of incidence = angle of reflection:
∠
F
2
P
T
2
=
∠
F
1
P
T
1
.
{\displaystyle \angle {F_{2}PT_{2}}=\angle {F_{1}PT_{1}}.}
See Figure 1.
The curve (red line) is an ellipse with equation:
b
2
x
2
+
a
2
y
2
−
a
2
b
2
=
0
{\displaystyle b^{2}x^{2}+a^{2}y^{2}-a^{2}b^{2}=0}
where
A
=
b
2
,
C
=
a
2
.
{\displaystyle A=b^{2},\ C=a^{2}.}
a
2
y
2
=
a
2
b
2
−
b
2
x
2
y
2
=
a
2
b
2
−
b
2
x
2
a
2
y
=
a
2
b
2
−
b
2
x
2
a
2
=
a
2
b
2
−
b
2
x
2
a
{\displaystyle {\begin{aligned}a^{2}y^{2}=a^{2}b^{2}-b^{2}x^{2}\\\\y^{2}={\frac {a^{2}b^{2}-b^{2}x^{2}}{a^{2}}}\\\\y={\sqrt {\frac {a^{2}b^{2}-b^{2}x^{2}}{a^{2}}}}={\frac {\sqrt {a^{2}b^{2}-b^{2}x^{2}}}{a}}\end{aligned}}}
Foci
F
1
,
F
2
{\displaystyle F_{1},F_{2}}
have coordinates
(
−
c
,
0
)
,
(
c
,
0
)
.
{\displaystyle (-c,0),(c,0).}
Line
T
1
P
T
2
{\displaystyle T_{1}PT_{2}}
is tangent to the curve at point
P
.
{\displaystyle P.}
A ray of light emanating from focus
F
2
{\displaystyle F_{2}}
is reflected from the inside surface of the ellipse at point
P
{\displaystyle P}
and passes through the other focus
F
1
.
{\displaystyle F_{1}.}
The aim is to prove that
∠
F
1
P
T
1
=
∠
F
2
P
T
2
.
{\displaystyle \angle {F_{1}PT_{1}}=\angle {F_{2}PT_{2}}.}
Point
N
{\displaystyle N}
has coordinates
(
c
+
u
,
0
)
.
{\displaystyle (c+u,0).}
At point
P
,
x
=
c
+
u
,
y
=
a
2
b
2
−
b
2
x
2
a
=
a
2
b
2
−
b
2
(
c
+
u
)
2
a
=
R
a
{\displaystyle P,\ x=c+u,\ y={\frac {\sqrt {a^{2}b^{2}-b^{2}x^{2}}}{a}}={\frac {\sqrt {a^{2}b^{2}-b^{2}(c+u)^{2}}}{a}}={\frac {R}{a}}}
Slope of line
F
2
P
=
P
N
F
2
N
=
R
a
u
=
m
2
.
{\displaystyle F_{2}P={\frac {PN}{F_{2}N}}={\frac {R}{au}}=m_{2}.}
Slope of line
F
1
P
=
P
N
F
1
N
=
R
a
(
2
c
+
u
)
=
m
1
.
{\displaystyle F_{1}P={\frac {PN}{F_{1}N}}={\frac {R}{a(2c+u)}}=m_{1}.}
Slope of curve at
P
=
−
A
x
C
y
=
−
A
(
c
+
u
)
C
(
R
/
a
)
=
−
A
(
c
+
u
)
a
C
R
=
m
.
{\displaystyle P={\frac {-Ax}{Cy}}={\frac {-A(c+u)}{C(R/a)}}={\frac {-A(c+u)a}{CR}}=m.}
Using
tan
(
A
−
B
)
=
tan
(
A
)
−
tan
(
B
)
1
+
tan
(
A
)
tan
(
B
)
{\displaystyle \tan(A-B)={\frac {\tan(A)-\tan(B)}{1+\tan(A)\tan(B)}}}
,
tan
(
∠
T
1
P
F
1
)
=
m
1
−
m
1
+
m
1
m
=
a
A
(
C
+
c
c
+
c
u
)
R
(
C
(
2
c
+
u
)
−
A
(
c
+
u
)
)
{\displaystyle \tan(\angle {T_{1}PF_{1}})={\frac {m_{1}-m}{1+m_{1}m}}={\frac {aA(C+cc+cu)}{R(C(2c+u)-A(c+u))}}}
tan
(
∠
T
2
P
F
2
)
=
m
−
m
2
1
+
m
m
2
=
a
A
(
−
C
+
c
c
+
c
u
)
R
(
C
u
−
A
(
c
+
u
)
)
{\displaystyle \tan(\angle {T_{2}PF_{2}})={\frac {m-m_{2}}{1+mm_{2}}}={\frac {aA(-C+cc+cu)}{R(Cu-A(c+u))}}}
if
∠
T
1
P
F
1
==
∠
T
2
P
F
2
{\displaystyle \angle {T_{1}PF_{1}}==\angle {T_{2}PF_{2}}}
then:
tan
(
∠
T
1
P
F
1
)
=
tan
(
∠
T
2
P
F
2
)
{\displaystyle \tan(\angle {T_{1}PF_{1}})=\tan(\angle {T_{2}PF_{2}})}
,
a
A
(
C
+
c
c
+
c
u
)
R
(
C
(
2
c
+
u
)
−
A
(
c
+
u
)
)
=
a
A
(
−
C
+
c
c
+
c
u
)
R
(
C
u
−
A
(
c
+
u
)
)
{\displaystyle {\frac {aA(C+cc+cu)}{R(C(2c+u)-A(c+u))}}={\frac {aA(-C+cc+cu)}{R(Cu-A(c+u))}}}
,
(
C
+
c
c
+
c
u
)
(
C
u
−
A
(
c
+
u
)
)
=
(
C
(
2
c
+
u
)
−
A
(
c
+
u
)
)
(
−
C
+
c
c
+
c
u
)
{\displaystyle (C+cc+cu)(Cu-A(c+u))=(C(2c+u)-A(c+u))(-C+cc+cu)}
and
(
C
+
c
c
+
c
u
)
(
C
u
−
A
(
c
+
u
)
)
−
(
C
(
2
c
+
u
)
−
A
(
c
+
u
)
)
(
−
C
+
c
c
+
c
u
)
=
0
{\displaystyle (C+cc+cu)(Cu-A(c+u))-(C(2c+u)-A(c+u))(-C+cc+cu)=0}
where
A
=
b
b
,
C
=
a
a
,
c
c
=
a
a
−
b
b
.
{\displaystyle A=bb,\ C=aa,\ cc=aa-bb.}
If you make the substitutions and expand, the result is
0
{\displaystyle 0}
.
Therefore, angle of reflection
∠
F
1
P
T
1
=
{\displaystyle \angle {F_{1}PT_{1}}=}
angle of incidence
∠
F
2
P
T
2
{\displaystyle \angle {F_{2}PT_{2}}}
and the reflected ray
P
F
1
{\displaystyle PF_{1}}
passes through the other focus
F
1
.
{\displaystyle F_{1}.}