SOLUTION: The moon's orbit is an ellipse with earth as one focus. If the maximum diatance from the moon to earth is 405 500 km and the minimum distance is 363 300 km, find the equation of th

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: The moon's orbit is an ellipse with earth as one focus. If the maximum diatance from the moon to earth is 405 500 km and the minimum distance is 363 300 km, find the equation of th      Log On


   



Question 1119858: The moon's orbit is an ellipse with earth as one focus. If the maximum diatance from the moon to earth is 405 500 km and the minimum distance is 363 300 km, find the equation of the ellipse in a cartesian coordinate system where earth is at the origin. Assume that the ellipse has horizontal major axis and that the minimum distance is achieved when the moon is to the right of the earth. Use 100 km as one unit.
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


According to the description, we are to view the ellipse as having a horizontal major axis. The standard form of the equation for such an ellipse is

%28x-h%29%5E2%2Fa%5E2%2B%28y-k%29%5E2%2Fb%5E2+=+1

In that form, the center of the ellipse as (h,k); a and b are the semi-major and semi-minor axes, and c is the distance from the center to each focus, where c^2 = a^2-b^2.

The minimum distance between earth and the moon is 3633 units to the right of earth; the maximum distance between earth and the moon is 4055 units to the left of earth. That means the length of the major axis is 3633+4055 = 7688 units; then the semi-major axis, a, is 7688/2 = 3844 units.

c, the distance from the center of the ellipse to a focus (the earth), is then 3844-3633 = 211 units.

Then, since the problem specifies that the earth be at the origin, the center of the ellipse is (-211,0).

We have h=-211 and k=0; and we know a=3844, so we can calculate a^2 = 14776336. Now we need to find b^2 using c^2 = a^2-b^2.

211%5E2+=+3844%5E2-b%5E2
b%5E2+=+3844%5E2-211%5E2+=+14731815

Now we have all the numbers we need to write the equation:

%28x-%28-211%29%29%5E2%2F14776336%2B%28y-0%29%5E2%2F14731815+=+1

or

%28x%2B211%29%5E2%2F14776336%2By%5E2%2F14731815+=+1

Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
.
The problem asks for the equation of the ellipse in a Cartesian coordinate system, where Earth is at the origin, 

in other words, the focus coinciding with the Earth is the origin.


Such an equation is "the polar form relative to focus" ( see this Wikipedia article

    https://en.wikipedia.org/wiki/Ellipse ,  the section "the polar form relative to focus" )


Such an equation has the form


    r%28theta%29 = %28a%2A%281-e%5E2%29%29%2F%281%2Be%2Acos%28theta%29%29,


where  "r" is the distance from the focus to the point at the ellipse  (= the current distance from Earth to the moon);  
theta is the angle with the horizontal axis (= with the major semi-axis);  "a"  is the major semi-axis  and  
"e"  is  the eccentricity, i.e. the ratio of the half the distance between focuses to the major semi-axis.


In given problem,  the major semi-axis  


    a = %284055%2B3633%29%2F2 = 3844  of the assigned units (each unit = 100 km),


while the half of the distance between focuses is


    e = %284055-3633%29%2F2 = 211  of the assigned units.


So the eccentricity is 


    e = 211%2F3844 = 0.0549.


Thus the requested equation is


    r%28theta%29 = %283844%2A%281-e%5E2%29%29%2F%281+%2B+e%2Acos%28theta%29%29 = 3844%2A%28%281-0.0549%5E2%29%2F%281%2B0.0549%2Acos%28theta%29%29%29 = 3844%2A%280.997%2F%281+%2B+0.0549%2Acos%28theta%29%29%29.