Ellipse

Ellipse

Figure 1: Ellipse (red curve) at origin with major axis horizontal.

Origin at point ${\displaystyle O}$${\displaystyle$ :(0,0)} .
Foci are points ${\displaystyle F_{1}(-c,0),\ F_{2}(c,0).OF_{1}=OF_{2}=c.}$
Line segment ${\displaystyle V_{1}OV_{2}}$ is the ${\displaystyle major\ axis.}$
${\displaystyle \ \ \ \ \ OV_{1}=OV_{2}=B_{1}F_{1}=B_{1}F_{2}=a.}$
Line segment ${\displaystyle B_{1}OB_{2}}$ is the ${\displaystyle minor\ axis.\ OB_{1}=OB_{2}=b.}$
Each line ${\displaystyle L_{1}F_{1}R_{1},L_{2}F_{2}R_{2}}$ is a ${\displaystyle latus\ rectum.}$
Each line ${\displaystyle x=D_{1},x=D_{2}}$ is a ${\displaystyle directrix.}$
${\displaystyle PF_{1}+PF_{2}=2a.}$

In cartesian geometry in two dimensions the ellipse is the locus of a point ${\displaystyle P}$ that moves relative to two fixed points called ${\displaystyle foci}$${\displaystyle :F_{1},F_{2}.}$ The distance ${\displaystyle F_{1}F_{2}}$ from one ${\displaystyle focus\ (F_{1})}$ to the other ${\displaystyle focus\ (F_{2})}$ is non-zero. The sum of the distances ${\displaystyle (PF_{1},PF_{2})}$ from point to foci is constant.

${\displaystyle PF_{1}+PF_{2}=K.}$ See figure 1.

The center of the ellipse is located at the origin ${\displaystyle O(0,0)}$ and the foci ${\displaystyle (F_{1},F_{2})}$ are on the ${\displaystyle X\ axis}$ at distance ${\displaystyle c}$ from ${\displaystyle O.}$

${\displaystyle F_{1}}$ has coordinates ${\displaystyle (-c,0).F_{2}}$ has coordinates ${\displaystyle (c,0)}$. Line segments ${\displaystyle OF_{1}=OF_{2}=c.}$

By definition ${\displaystyle PF_{1}+PF_{2}=B_{1}F_{1}+B_{1}F_{2}=V_{1}F_{1}+V_{1}F_{2}=K.}$

${\displaystyle V_{1}F_{1}=V_{2}F_{2}.\ \therefore V_{1}F_{1}+V_{1}F_{2}=V_{1}F_{2}+V_{2}F_{2}=V_{1}V_{2}=K=2a,}$ the length of the ${\displaystyle major\ axis\ (V_{1}V_{2}).\ OV_{1}=OV_{2}=a.}$

Each point ${\displaystyle (V_{1},V_{2})}$ where the curve intersects the major axis is called a ${\displaystyle vertex.\ V_{1},V_{2}}$ are the vertices of the ellipse.

Line segment ${\displaystyle B_{1}B_{2}}$ perpendicular to the major axis at the midpoint of the major axis is the ${\displaystyle minor\ axis}$ with length ${\displaystyle 2b.\ OB_{1}=OB_{2}=b.}$

${\displaystyle B_{1}F_{1}+B_{1}F_{2}=2a;\ B_{1}F_{1}=B_{1}F_{2}=a.\ \therefore a^{2}=b^{2}+c^{2}.}$

Any line segment that intersects the curve in two places is a ${\displaystyle chord.}$ A chord through the focus is a ${\displaystyle focal\ chord.}$ Each focal chord ${\displaystyle L_{1}F_{1}R_{1},\ L_{2}F_{2}R_{2}}$ perpendicular to the major axis is a ${\displaystyle latus\ rectum.}$

Second definition of ellipse

Graph of ellipse ${\displaystyle {\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=1}$ where ${\displaystyle a,b=20,12}$. At point ${\displaystyle B,\ {\frac {u}{v}}=e.}$ At point ${\displaystyle A,\ {\frac {a}{t}}=e.}$ Ellipse is path of point that moves so that ratio of distance to fixed point and distance to fixed line is constant. Let ${\displaystyle {\frac {p}{a}}=e}$ where: ${\displaystyle p}$ is non-zero, ${\displaystyle a>p,}$ ${\displaystyle a=p+u.}$ Therefore, ${\displaystyle 1>e>0.}$ Let directrix have equation ${\displaystyle x=t}$ where ${\displaystyle {\frac {a}{t}}=e.}$ At point ${\displaystyle B:}$ ${\displaystyle {\frac {p}{p+u}}={\frac {p+u}{p+u+v}}=e}$ ${\displaystyle (p+u)^{2}=p(p+u+v)}$ ${\displaystyle pp+pu+pu+uu=pp+pu+pv}$ ${\displaystyle pu+uu=pv}$ ${\displaystyle u(p+u)=pv}$ ${\displaystyle {\frac {u}{v}}={\frac {p}{p+u}}=e}$ ${\displaystyle {\frac {\text{distance to focus}}{\text{distance to directrix}}}=e\ \dots \ (3)}$ Statement ${\displaystyle (3)}$ is true at point ${\displaystyle A}$ also. Section under "Proof" below proves that statement (3) is true for any point ${\displaystyle P}$ on ellipse.

Proof

Proving that ${\displaystyle {\frac {\text{distance from point to focus}}{\text{distance from point to directrix}}}=e}$. Graph is part of curve ${\displaystyle {\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=1.}$ distance to Directrix1 ${\displaystyle =t-x={\frac {a}{e}}-x={\frac {a-ex}{e}}.}$ base = ${\displaystyle x-p=x-ae}$ ${\displaystyle {\text{(distance to Focus1)}}^{2}={\text{base}}^{2}+y^{2}}$ As expressed above in statement ${\displaystyle 3,}$ second definition of ellipse states that ellipse is path of point that moves so that ratio of distance to fixed point and distance to fixed line is constant. This section proves that this definition is true for any point ${\displaystyle P}$ on the ellipse. At point ${\displaystyle P:}$ ${\displaystyle (a^{2}-p^{2})x^{2}+a^{2}y^{2}-a^{2}(a^{2}-p^{2})=0}$ ${\displaystyle y^{2}={\frac {-(a^{2}-p^{2})x^{2}+a^{2}(a^{2}-p^{2})}{a^{2}}}}$ ${\displaystyle ={\frac {a^{2}e^{2}x^{2}-a^{2}x^{2}+a^{2}a^{2}-a^{2}a^{2}e^{2}}{a^{2}}}}$ ${\displaystyle =e^{2}x^{2}-x^{2}+a^{2}-a^{2}e^{2}}$ base ${\displaystyle =x-p=x-ae}$ ${\displaystyle ({\text{distance}}\ F_{1}P)^{2}=y^{2}+{\text{base}}^{2}=y^{2}+(x-ae)^{2}}$ ${\displaystyle =a^{2}-2aex+e^{2}x^{2}}$ ${\displaystyle =(a-ex)^{2}}$ ${\displaystyle {\text{distance to Focus1}}={\text{distance}}\ F_{1}P=a-ex}$ ${\displaystyle {\text{distance to Directrix1}}=t-x={\frac {a}{e}}-x={\frac {a-ex}{e}}}$ ${\displaystyle {\frac {\text{distance to Focus1}}{\text{distance to Directrix1}}}}$ ${\displaystyle =(a-ex){\frac {e}{(a-ex)}}}$ ${\displaystyle =e}$ Similar calculations can be used to prove the case for Focus2 ${\displaystyle (-p,0)}$ and Directrix2 ${\displaystyle (x=-t)}$ in which case: ${\displaystyle {\frac {\text{distance to Focus2}}{\text{distance to Directrix2}}}}$ ${\displaystyle =(a+ex){\frac {e}{(a+ex)}}}$ ${\displaystyle =e}$ Therefore: ${\displaystyle {\frac {\text{distance to focus}}{\text{distance to directrix}}}=e}$ where ${\displaystyle 1>e>0.}$ Ellipse is path of point that moves so that ratio of distance to fixed point and distance to fixed line is constant, called eccentricity ${\displaystyle e.}$

The directrices

Figure 1: Ellipse (red curve) at origin with major axis horizontal.

Origin at point ${\displaystyle O}$${\displaystyle$ :(0,0)} .
Foci are points ${\displaystyle F_{1}(-c,0),\ F_{2}(c,0).OF_{1}=OF_{2}=c.}$
Directrices are lines ${\displaystyle x=D_{1},\ x=D_{2},\ D_{1}=-D_{2}.}$ ${\displaystyle OV_{1}=OV_{2}=a.}$
${\displaystyle eccentricity\ e={\frac {c}{a}}={\frac {F_{2}V_{2}}{V_{2}D_{2}}}={\frac {V_{1}F_{2}}{V_{1}D_{2}}}={\frac {OV_{2}}{OD_{2}}}}$
${\displaystyle e={\frac {PF_{1}}{PE_{1}}}}$ (yellow lines); ${\displaystyle e={\frac {PF_{2}}{PE_{2}}}}$ (purple lines)
To define ellipse specify ${\displaystyle focus,\ directrix,\ e.}$

See Figure 1. The vertical line through ${\displaystyle D_{2}}$ with equation ${\displaystyle x=D_{2}}$ is a ${\displaystyle directrix}$ of the ellipse. Likewise for the vertical line ${\displaystyle x=D_{1};\ D_{1}=-D_{2}.}$

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 ${\displaystyle e=eccentricity,\ 0<e<1.}$

Consider the point ${\displaystyle V_{2}}$ in the figure. ${\displaystyle {\frac {F_{2}V_{2}}{V_{2}D_{2}}}=e.}$

Let the length ${\displaystyle V_{2}D_{2}=d.\ {\frac {a-c}{d}}=e.\ d={\frac {a-c}{e}}}$

Consider the point ${\displaystyle V_{1}}$:

${\displaystyle {\frac {V_{1}F_{2}}{V_{1}D_{2}}}={\frac {a+c}{2a+d}}=e=(a+c)/(2a+{\frac {a-c}{e}})=(a+c)/({\frac {2ae+a-c}{e}})={\frac {e(a+c)}{2ae+a-c}}.}$

${\displaystyle 2ae+a-c=a+c;\ 2ae=2c.}$

${\displaystyle e={\frac {c}{a}}.}$

Length ${\displaystyle OD_{2}=a+d=a+{\frac {a-c}{e}}=a+{\frac {a-c}{c/a}}=a+{\frac {a(a-c)}{c}}={\frac {ac+aa-ac}{c}}={\frac {a^{2}}{c}}}$.

Distance from center to directrix ${\displaystyle =OD_{2}={\frac {a^{2}}{c}}={\frac {a^{2}}{ea}}={\frac {a}{e}}={\frac {c}{e^{2}}}.\ e={\frac {OV_{2}}{OD_{2}}}.}$

Distance from center to directrix = ${\displaystyle OF_{2}+F_{2}D_{2}=c+d_{0}={\frac {c}{e^{2}}}}$.

${\displaystyle e^{2}c+e^{2}d_{0}=c;\ c-e^{2}c=e^{2}d_{0};\ c={\frac {e^{2}d_{0}}{1-e^{2}}}}$.

${\displaystyle 1-e^{2}=1-{\frac {c^{2}}{a^{2}}}={\frac {a^{2}-c^{2}}{a^{2}}}={\frac {b^{2}}{a^{2}}}}$.

${\displaystyle {\frac {e^{2}}{1-e^{2}}}={\frac {c^{2}}{a^{2}}}/{\frac {b^{2}}{a^{2}}}={\frac {c^{2}}{b^{2}}}}$.

Distance from focus to directrix ${\displaystyle =F_{2}D_{2}={\frac {b^{2}}{c}}}$.

Ellipse using focus and directrix

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Figure 2: Ellipse (red curve) at origin with major axis vertical.

Origin at point ${\displaystyle O}$${\displaystyle$ :(0,0)} .
Focus at point ${\displaystyle F}$${\displaystyle$ :(0,-c).}
Directrix is line ${\displaystyle DE}$${\displaystyle$ :\ y={\frac {-c}{e^{2}}}.}
Point ${\displaystyle P}$ has coordinates: ${\displaystyle (x,y).}$
${\displaystyle eccentricity\ e={\frac {PF}{PE}}}$ (yellow lines)

See Figure 2. Let the point ${\displaystyle P}$ be ${\displaystyle (x,y)}$, the focus ${\displaystyle F}$ be ${\displaystyle (0,-c)}$ and the directrix ${\displaystyle DE}$ have equation ${\displaystyle y={\frac {-c}{e^{2}}}}$ where ${\displaystyle 1>e>0.}$ The directrix is horizontal and the major axis vertical through the origin ${\displaystyle (0,0)}$.

${\displaystyle PF={\sqrt {x^{2}+(y+c)^{2}}}}$

${\displaystyle PE=y+{\frac {c}{e^{2}}}}$

${\displaystyle e={\frac {PF}{PE}};\ PF=ePE.}$

${\displaystyle {\sqrt {x^{2}+(y+c)^{2}}}=e(y+{\frac {c}{e^{2}}})}$

${\displaystyle e{\sqrt {x^{2}+(y+c)^{2}}}=ee(y+{\frac {c}{e^{2}}})=eey+c}$

${\displaystyle ee(x^{2}+(y+c)^{2})=(eey+c)^{2}}$

${\displaystyle ee(x^{2}+(y+c)^{2})-(eey+c)^{2}=0}$

Expand and the result is:

${\displaystyle +eexx+eeyy-eeeeyy+ccee-cc=0}$

${\displaystyle xx+yy-eeyy+cc-{\frac {cc}{ee}}=0}$

${\displaystyle xx+(1-ee)yy+cc-aa=0}$

${\displaystyle xx+(1-ee)yy-(aa-cc)=0}$

${\displaystyle xx+(1-ee)yy-bb=0}$

${\displaystyle 1-ee=1-{\frac {cc}{aa}}={\frac {aa-cc}{aa}}={\frac {bb}{aa}}}$

${\displaystyle xx+({\frac {bb}{aa}})yy-bb=0}$

${\displaystyle a^{2}x^{2}+b^{2}y^{2}-a^{2}b^{2}=0}$

Compare this equation with the equation generated earlier: ${\displaystyle b^{2}x^{2}+a^{2}y^{2}-a^{2}b^{2}=0}$.

When the equation is ${\displaystyle b^{2}x^{2}+a^{2}y^{2}-a^{2}b^{2}=0,}$ the major axis is horizontal.

When the equation is ${\displaystyle a^{2}x^{2}+b^{2}y^{2}-a^{2}b^{2}=0,}$ the major axis is vertical.

General ellipse at origin

Figure 1: Ellipse (red curve) at origin with major axis oblique.

Origin at point ${\displaystyle O}$${\displaystyle$ :(0,0)} .
Foci are points ${\displaystyle F_{1}(-p,-q),\ F_{2}(p,q).}$
${\displaystyle OF_{1}=OF_{2}=c={\sqrt {p^{2}+q^{2}}}.}$
Line segment ${\displaystyle V_{1}OV_{2}}$ is the ${\displaystyle major\ axis.}$
${\displaystyle \ \ \ \ \ OV_{1}=OV_{2}=a.}$
Line ${\displaystyle D_{1}OD_{2}}$ is the major axis extended.
Lines perpendicular to ${\displaystyle D_{1}OD_{2}}$ at ${\displaystyle D_{1},F_{1},O,F_{2},D_{2}}$ are directrix, latus rectum. minor axis, latus rectum, directrix.
${\displaystyle OD_{1}=OD_{2}={\frac {a^{2}}{c}}.}$
${\displaystyle F_{1}D_{1}=F_{2}D_{2}={\frac {b^{2}}{c}}.}$
${\displaystyle PF_{1}+PF_{2}=2a.}$

The general ellipse allows for the major axis to have slope other than horizontal or vertical. See Figure 1.

The line ${\displaystyle V_{1}V_{2}}$ is the major axis of the ellipse shown in red. Points ${\displaystyle F_{1},F_{2}}$ are the foci with coordinates ${\displaystyle (-p,-q),(p,q)}$ respectively.

Point ${\displaystyle P(x,y)}$ is any point on the curve. By definition ${\displaystyle PF_{1}+PF_{2}=2a.}$

Length ${\displaystyle PF_{1}={\sqrt {(x-(-p))^{2}+(y-(-q))^{2}}}={\sqrt {(x+p)^{2}+(y+q)^{2}}}=M.}$

Length ${\displaystyle PF_{2}={\sqrt {(x-p)^{2}+(y-q)^{2}}}=N.}$

${\displaystyle {\sqrt {(x+p)^{2}+(y+q)^{2}}}+{\sqrt {(x-p)^{2}+(y-q)^{2}}}=2a.}$

${\displaystyle M+N=2a}$

${\displaystyle MM+2MN+NN=4aa}$

${\displaystyle 2MN=4aa-MM-NN}$

${\displaystyle 4MMNN=(4aa-MM-NN)^{2}}$

${\displaystyle 4MMNN-(4aa-MM-NN)^{2}=0.}$

Make appropriate substitutions, expand and the result is:

${\displaystyle (aa-pp)x^{2}-2pqxy+(aa-qq)y^{2}+aapp+aaqq-aaaa=0}$.

This equation has the form ${\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0}$ where:

${\displaystyle A=a^{2}-p^{2}}$

${\displaystyle B=-2pq}$

${\displaystyle C=a^{2}-q^{2}}$

${\displaystyle D=E=0}$ because center is at origin,

${\displaystyle F=a^{2}p^{2}+a^{2}q^{2}-a^{4}=a^{2}(p^{2}+q^{2}-a^{2})=a^{2}(c^{2}-a^{2})=-a^{2}b^{2}.}$

An example

Let ${\displaystyle (p,q)}$ be (3,4) and ${\displaystyle a=8.}$

The ellipse is: ${\displaystyle 880x^{2}-384xy+768y^{2}+(0)x+(0)y-39936=0,}$ or ${\displaystyle 55x^{2}-24xy+48y^{2}+(0)x+(0)y-2496=0.}$

Reverse-engineering ellipse at origin

Given an ellipse in format ${\displaystyle Ax^{2}+Bxy+Cy^{2}+(0)x+(0)y+F=0,}$ calculate ${\displaystyle p,q,a.\ (B}$ is non-zero.${\displaystyle )}$

${\displaystyle A=a^{2}-p^{2}}$

${\displaystyle B=-2pq}$

${\displaystyle C=a^{2}-q^{2}}$

${\displaystyle F=a^{2}p^{2}+a^{2}q^{2}-a^{4}.}$

Coefficients provided could be, for example, ${\displaystyle (55,-24,48,0,0,-2496)}$ or ${\displaystyle (110,-48,96,0,0,-4992)}$

or ${\displaystyle (55k,-24k,48k,0,0,-2496k)}$ where ${\displaystyle k}$ is an arbitrary constant and all groups of coefficients define the same ellipse.

To produce consistent, correct values for ${\displaystyle p,q,a}$ the equations become:

${\displaystyle KA=a^{2}-p^{2}}$

${\displaystyle KB=-2pq}$

${\displaystyle KC=a^{2}-q^{2}}$

${\displaystyle KF=a^{2}p^{2}+a^{2}q^{2}-a^{4}.}$

or:

${\displaystyle KA-(a^{2}-p^{2})=0}$

${\displaystyle KB-(-2pq)=0}$

${\displaystyle KC-(a^{2}-q^{2})=0}$

${\displaystyle KF-(a^{2}p^{2}+a^{2}q^{2}-a^{4})=0.}$

Solutions are:

${\displaystyle K={\frac {4F}{B^{2}-4AC}}}$. This formula for ${\displaystyle K}$ is valid if both ${\displaystyle D,E}$ are ${\displaystyle 0}$.

${\displaystyle 4(pp)^{2}+4K(A-C)pp-B^{2}K^{2}=0}$ where ${\displaystyle pp=p^{2}.}$

You should see one positive value ${\displaystyle (p^{2})}$ and one negative value ${\displaystyle (-(q^{2}))}$ for ${\displaystyle pp.}$ Choose the positive value and ${\displaystyle p={\sqrt {pp}}}$.

${\displaystyle q={\frac {-KB}{2p}}}$

${\displaystyle a={\sqrt {AK+p^{2}}}}$

The solutions become simpler if K == 1. if ( K != 1 ) { A ← KA; B ← KB; C ← KC; } and the solutions for ${\displaystyle p,q,a}$ become:

${\displaystyle 4(pp)^{2}+4(A-C)pp-B^{2}=0}$ where ${\displaystyle pp=p^{2}.}$

${\displaystyle q={\frac {-B}{2p}}}$

${\displaystyle a={\sqrt {A+p^{2}}}}$

With values ${\displaystyle p,q,a}$ available all the familiar values of the ellipse may be calculated:

${\displaystyle c={\sqrt {p^{2}+q^{2}}}}$

Equation of major axis: ${\displaystyle y={\frac {q}{p}}x;\ qx-py+0=0;\ {\frac {q}{c}}x-{\frac {p}{c}}y+0=0}$ in normal form.

Equation of minor axis: ${\displaystyle {\frac {p}{c}}x+{\frac {q}{c}}y+0=0}$ in normal form.

Equations of directrices: ${\displaystyle {\frac {p}{c}}x+{\frac {q}{c}}y\pm {\frac {a^{2}}{c}}=0}$ in normal form.

An example using focus and directrix

Figure 2: Ellipse (red curve) at origin with major axis oblique.

Origin at point ${\displaystyle O}$${\displaystyle$ :(0,0)} .
Focus at point ${\displaystyle F}$${\displaystyle$ :(-20,15)}
Directrix is line ${\displaystyle DE}$${\displaystyle$ :{\frac {4}{5}}x-{\frac {3}{5}}y+36=0.}
eccentricity ${\displaystyle e={\frac {5}{6}}={\frac {P_{1}F}{P_{1}D}}={\frac {P_{2}F}{P_{2}E}}.}$
Ellipse has equation:
${\displaystyle 20x^{2}+24xy+27y^{2}-9900=0.}$

Given focus ${\displaystyle (-20,15),\ e={\frac {5}{6}}}$ and directrix with equation ${\displaystyle {\frac {4}{5}}x-{\frac {3}{5}}y+36=0}$, calculate equation of the ellipse in form ${\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0.}$

See Figure 2. Let point ${\displaystyle P\ (x,y)}$ be any point on the ellipse.

Distance from ${\displaystyle P}$ to focus ${\displaystyle ={\sqrt {(x+20)^{2}+(y-15)^{2}}}}$

Distance from ${\displaystyle P}$ to directrix ${\displaystyle =0.8x-0.6y+36}$.

${\displaystyle {\sqrt {(x+20)^{2}+(y-15)^{2}}}={\frac {5}{6}}(0.8x-0.6y+36)}$.

${\displaystyle 6{\sqrt {(x+20)^{2}+(y-15)^{2}}}=5(0.8x-0.6y+36)=4x-3y+180}$.

${\displaystyle 36((x+20)^{2}+(y-15)^{2})=(4x-3y+180)^{2}}$.

${\displaystyle 36((x+20)^{2}+(y-15)^{2})-(4x-3y+180)^{2}=0}$.

Expand and the result is: ${\displaystyle 20x^{2}+24xy+27y^{2}+(0)x+(0)y-9900=0.}$

Because ${\displaystyle D=E=0,}$ the center of the ellipse is at the origin and the various lines have equations as follows.

Minor axis: ${\displaystyle {\frac {4}{5}}x-{\frac {3}{5}}y+0=0}$

Other directrix: ${\displaystyle {\frac {4}{5}}x-{\frac {3}{5}}y-36=0}$

Major axis: ${\displaystyle {\frac {3}{5}}x+{\frac {4}{5}}y+0=0}$

If you reverse-engineer the ellipse using the method above, ${\displaystyle K=25}$,

the expression ${\displaystyle KA=a^{2}-p^{2}}$ becomes ${\displaystyle (25)(20)=30^{2}-20^{2}}$, and

the expression ${\displaystyle KF=-a^{2}b^{2}}$ becomes ${\displaystyle (25)(-9900)=-(30^{2})(275)}$.

General Ellipse

Figure 1: The general ellipse (red curve).
Foci at ${\displaystyle S_{1},S_{2},}$ center at ${\displaystyle C_{2}.}$
Point ${\displaystyle V}$ is a vertex. Length ${\displaystyle VC_{2}=a.}$
${\displaystyle PS_{1}+PS_{2}=2a.}$

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 ${\displaystyle foci:\ S_{1}(s,t),\ S_{2}(u,v),}$ ${\displaystyle major\ axis}$ of length ${\displaystyle 2a}$ and point ${\displaystyle P(x,y)}$

${\displaystyle PS_{1}+PS_{2}=2a}$

${\displaystyle {\sqrt {(x-s)^{2}+(y-t)^{2}}}+{\sqrt {(x-u)^{2}+(y-v)^{2}}}=2a}$

The expansion of this expression is somewhat complicated because it contains 5 variables ${\displaystyle s,t,u,v,a.}$

The expansion may be simplified by reducing the number of variables from ${\displaystyle 5}$ to ${\displaystyle 3}$, the familiar ${\displaystyle p,q,a.}$

Let ${\displaystyle C_{2}}$, the center of the ellipse, have coordinates ${\displaystyle (G,H)}$ where ${\displaystyle G={\frac {s+u}{2}};\ H={\frac {t+v}{2}}}$ and ${\displaystyle p=u-G;\ q=v-H.}$