Class 38: Geometry of Conic Sections Class 38: Geometry of Conic - PowerPoint PPT Presentation

Class 38: Geometry of Conic Sections Class 38: Geometry of Conic Sections

Orbital equation       2 Radial : ( r - r ) f(r)     r    2 Tangential Tangential : : L L r 1 u  r r 2 2 2 2 3 L u d u L u Solving this will give the     f f    2 2 shape of the orbit d k f  f  Special case Special case 2 r R  r Solution S l i    1 cos

Conic Sections R R  r    1 cos We will assume   0:  = 0 Circle R>0  <1 Ellipse R>0  =1 Parabola R>0  >1 Hyperbola – attractive R>0 Hyperbola – repulsive R<0

Inversion Convention I will adopt the “inversion convention” in allowing r to be I will adopt the inversion convention in allowing r to be negative. If you see a negative r, you just treat it like a vector, and do an inversion through the origin (focus). For ellipse, parabola, and circle, ‐ r the r for the whole curve have the same sign, but not for hyperbola: For R>0: r  We will use dotted line to represent the section with represent the section with negative r. Positive r Negative r

Ellipse, parabola, and circle The ellipse (0<  <1) will always look like this: Th lli (0< <1) ill l l k lik thi r  Circle (  =0) is a special case of ellipse when the two Circle (  0) is a special case of ellipse when the two foci coincide at the same point (the center). Parabola (  =1) is a special case when the left end of the above ellipse goes to infinite far left.

Hyperbola Hyperbola (  >1) has two branches Hyperbola (  >1) has two branches. Following our inversion Following our inversion convention, one has positive r (solid line) and the other has negative r (dotted line) r>0 r<0 r<0 r>0 r r   Case for R>0 Case for R<0 (attractive potential) (repulsive potential) ( p p )

r a and r p 2 1 L k            1. r a and r p are roots of 1 r and r are roots of 2 2 E E 2 2 E E r r 2 2 k k r r - L L 0 0  2 2 r r 2. For ellipse, parabola, and circle, r a and r p are positive. For ellipse, r a > r p > 0 For parabola, r a =  > r p > 0 r a r p For circle, r a = r p > 0 , a p 3. For Hyperbola r>0 0 r<0 0 r<0 0 r>0 r a <0 , |r a | > | r p | r p <0 , |r a | > | r p | (attractive potential) (repulsive potential)

Relationship between (R,  ) and (r a , r p ) R R   r         p o 1 1 cos cos 0 0 1 1 R R   r        a o 1 1 cos cos 180 180 1 1   r r   a p  r r a a p p 2r r R  a p r  r a p

Conic Sections in Cartesian Coordinates Ellipse Ellipse x    With the exception of parabola the With the exception of parabola, the x a a y origin is now at the center of the conics and the focus will have the x ‐ coordinate f   f a ` ` x There is a line call directrix at (a  ,0) a x  x   y Distance between curve and focus   Di t Distance b t between curve and d di directrix t i (Directrix of a circle is at infinite far away) x (a  ,0) (a  0) General Equation (not for parabola): G l E ti ( t f b l ) 2 2 x y 2    1 1 x  x    a a 2 2 2 a b Note that a<0 for this picture Hyperbola B is imaginary for hyperbola.

Parabola The Cartesian Coordinate equation for parabola is y 2 = ‐ 4a’x This is not in our general form. The a’ here is not even the same “a “ as in the general equation. We can consider a parabola as part of a big ellipse as  1. However , as the ellipse becomes bigger and bigger, it center (or the origin of the Cartesian coordinates) shifts towards the left at infinity far away and that’s the problem We coordinates) shifts towards the left at infinity far away and that s the problem. We are forced to translate the coordinates by “a” to the right back to the vertex of the parabola. For this reason, the equation of parabola is not in pour general form:  2 2 (x a) y   Lim 1 2 2   a b a  2 2 2ax a y    Lim 1 2 2     a a b b a a 2 2 y    Lim x 0 2   a b a 2 b    2 2 y -4a' 4 ' x wit it h h a' ' Li Lim 2a   a

Parabola y x (a’,0) x  a x  a ' 2 b  a' Lim   2 a a, b

Relationship between (a,b) and (r a , r p )   r r  a p a 2 b  r a r p   r and r are roots of a p p    2 2 r 2 ar b 0 Note: 1. a can be positive or negative for hyperbola, N t 1 b iti ti f h b l depending on the values of r a and r p . 2 b is imaginary for hyperbola because one of r and 2. b is imaginary for hyperbola because one of r a and r p (but not both) is negative.

Relationship between (a,b) and (R and  ) R R   r r     R 1 1       a p a a   2 2 2 1 R R R     b r r ( ( )( )( ) )     a a p p   1 1 2 1   2  2 a b   a 2 b R  R a

The Triangular Relationship R,  2 b  a' Lim for parabola   2 a a, b  r r R   a p  a      2 r r r r 1 1 a p R  b 2r r    a p 2 2  R 1 r   2 a b r r r   R R a p  r a   p 1 2 b R   R r   a a a 1  r r   a p a b r r a a p p 2 2 r a ,r p a,b    2 2 Roots of r 2 ar b 0

Some famous triangles Ellipse The follow the same relationship: y    2 2 2 2 a b a If we allow b to be imaginary. a b x a  a  Hyperbola y y |a| |b| x |a| 

The Triangular Relationship R,  E,L 2 b  a' Lim for parabola   2 a a, b r and r are roots of a p  r r 2 1 L k   R E   a p   2 2 r r a           2 2 2 r r r r 2 E r 2 k r - L 0 1 1 a p R  b 2r r    a p 2 2  R 1 r   2 a b r r r   R R a p  r a   p 1 2 b R   R r   a a a 1  r r   a p a b r r a a p p 2 2 r a ,r p a,b    2 2 Roots of r 2 ar b 0

A Simplified Relationship That’s all you need to remember, other things are simple algebra. E,L r and r are roots of a p 2 1 L k   E  2 2 r r      2 2 2 2 2 2 E E r 2 2 k k r - L L 0 0 R R   r p r r    a p 1 a 2 R   1 r r a b  b r a r   R,  r a ,r p p a,b