document.write( "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. \n" ); document.write( "
Algebra.Com's Answer #735493 by ikleyn(52790)\"\" \"About 
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document.write( "The problem asks for the equation of the ellipse in a Cartesian coordinate system, where Earth is at the origin, \r\n" );
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document.write( "in other words, the focus coinciding with the Earth is the origin.\r\n" );
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document.write( "Such an equation is \"the polar form relative to focus\" ( see this Wikipedia article\r\n" );
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document.write( "    https://en.wikipedia.org/wiki/Ellipse ,  the section \"the polar form relative to focus\" )\r\n" );
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document.write( "Such an equation has the form\r\n" );
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document.write( "    \"r%28theta%29\" = \"%28a%2A%281-e%5E2%29%29%2F%281%2Be%2Acos%28theta%29%29\",\r\n" );
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document.write( "where  \"r\" is the distance from the focus to the point at the ellipse  (= the current distance from Earth to the moon);  \r\n" );
document.write( "\"theta\" is the angle with the horizontal axis (= with the major semi-axis);  \"a\"  is the major semi-axis  and  \r\n" );
document.write( "\"e\"  is  the eccentricity, i.e. the ratio of the half the distance between focuses to the major semi-axis.\r\n" );
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document.write( "In given problem,  the major semi-axis  \r\n" );
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document.write( "    a = \"%284055%2B3633%29%2F2\" = 3844  of the assigned units (each unit = 100 km),\r\n" );
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document.write( "while the half of the distance between focuses is\r\n" );
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document.write( "    e = \"%284055-3633%29%2F2\" = 211  of the assigned units.\r\n" );
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document.write( "So the eccentricity is \r\n" );
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document.write( "    e = \"211%2F3844\" = 0.0549.\r\n" );
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document.write( "Thus the requested equation is\r\n" );
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document.write( "    \"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\".\r\n" );
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