Question 23284: Please help!
Like all planets Jupiter has an elliptical orbit, with the centre of the sun located at a focus. Write an equation of the ellipse that models Jupiter's orbit around the sun. Assume that the centre of the sun in on x-axis.
------------------jupiter
__740_____sun___810______ --------> assuming this is a ellipse and Jupiter is
around the orbit of the sun.
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! standard eqn. of an ellipse with foci on x axis,centre as origin and directrices parallel to y axis is
(X^2/A^2)+(Y^2/B^2)=1,where A and B are semi major/minor axes
foci are given by (A*E,0),(-A*E,0)
eccentricity =E =[(A^2-B^2)/A^2]^0.5
we are given the sun is at one focus and the 2 numbers you have given as 740 and 810 are NOT ELABORATED AS TO WHAT THESE NUMBERS ARE.
I back calculated the major / minor axes as per the answer given by you and came out
with following interpretation of the numbers given by you.
740 is the minimum distance of jupiter from sun
810 is the maximum distance of jupiter from sun
so total distance =major axis ,since sun is at one of the foci and is on x axis or major axis
major axis =2*A=740+810=1550..or A=1550/2= 775
as given above foci are given by (A*E,0) and (-A*E,0),with centre taken as origin.
since sun is at focus and its coordinate is (775-740,0) and (775-810,0) or (35,0),(-35,0) we have
A*E=35..that is.... 775*E=35...or....E=35/775=7/155
further we have E=[(A^2-B^2)/A^2]^0.5
so B=[A^2-(E^2)(A^2)]^0.5 =A*(1-E^2)^0.5=774.2093
hence eqn. Of ellipse is
(X^2/775^2)+(Y^2/774.2093^2)=1
(X^2/600625)+(Y^2/599400)=1
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