Question 1182836
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So, the major semi-axis length is a = 500/2 = 250 millions kilometers.


The focal distance is 400 millions kilometers; so the linear eccentricity "c" is half of it, i.e. c = 200 millions kilometers.


Then for the minor semi-axis of the ellipse "b" you have


    b = {{{sqrt(a^2-c^2)}}} = {{{sqrt(250^2-200^2)}}} = 150 millions kilometers.


The canonical equation of the ellipse is


    {{{x^2/250^2}}} + {{{y^2/150^2}}} = 1,


where x a y are coordinates of the orbit in millions of kilometers.
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Solved.


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On canonical equation of an ellipse read from the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Quadratic-relations-and-conic-sections/Ellipse-definition--canonical-equation--characteristic-points-and-elements.lesson>Ellipse definition, canonical equation, characteristic points and elements</A> 

in this site.