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.}
Equation of the ellipse
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
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}}.}
The directrices
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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
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
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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)} .
General Ellipse
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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
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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.}
K and "Standard Form"
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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
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Reflectivity of ellipse
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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}.}