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ellipse

Parameter Form

$x:=a⋅\cos t$

$y:=b⋅\sin t$

where:

$\sin x:=÷{exp(i⋅x)-\exp(-i⋅x)}{2⋅i}$

$\cos x:=÷{exp(i⋅x)+\exp(-i⋅x)}{2}$

where:

$\tan x:=÷{\sin x}{\cos x}$

Properties

$\sin(x+y)=\sin(x)⋅\cos(y)+\cos(x)⋅\sin(y)$
$\cos(x+y)=\cos(x)⋅\cos(y)-\sin(x)⋅\sin(y)$
$\sin(-x)=-\sin(x)$
$(\cos(x))²+(\sin(x))²=1$
$÷{d}{dx}(\sin x)=\cos x$
$÷{d}{dx}(\cos x)=-\sin x$
$\exp(i⋅x)=\cos(x)+i⋅\sin(x)$ , see Exponentiation for "exp".
$(\sin a)⋅(\sin b)=÷{1}{2}⋅\cos(a-b)-÷{1}{2}⋅\cos(a+b)$
$(\cos a)⋅(\cos b)=÷{1}{2}⋅\cos(a-b)+÷{1}{2}⋅\cos(a+b)$

Circumference

The Circumference s is:

$s:=∫\limits_{∂} ds$

where ds :: short_arc.

Since:

... it follows that:

$s:=∫\limits_{0}^{2⋅π} √{ẋ²+ẏ²}⋅dt$

$s=a⋅∫ \limits_{0}^{2⋅π} √{1-k²⋅(\sin s)²}⋅ds$

Numeric Eccentricity

Excentricity.

And thus:

Parameter Form in Polar Coordinates, center of the ellipse as center of coordinate system

$r(φ)=a⋅√{1-ε²⋅(\sin φ)²}$

Because:

$x=a⋅\cos φ$

$y=a⋅√{1-ε²}⋅\sin φ$

$r²=a²⋅((\cos φ)²+(1-ε²)⋅(\sin φ)²)$

$r²=a²⋅((\cos φ)²+(\sin φ)²-ε²⋅(\sin φ)²)$

$r²=a²⋅(1-ε²⋅(\sin φ)²)$

... after choosing the positive root:

$r=a⋅√{1-ε²⋅(\sin φ)²}$

Author: Danny (remove the ".nospam" to send)

Last modification on: Thu, 09 May 2013 .