Example

Example: Finding the Equation of an Elliptical Orbit Given Distances of 20 AU and 50 AU

Consider a planet moving in an elliptical orbit with the sun at one focus. The closest the planet gets to the sun (perihelion) is approximately 20 AU and the furthest (aphelion) is approximately 50 AU. Centering the elliptical orbit at the origin (0, 0) along the x-axis, the total length of the major axis is the sum of the closest and furthest distances: 20 + 50 = 70 AU. This gives a semi-major axis of a=702=35a = \frac{70}{2} = 35 AU, placing the vertices at (35,0)(-35, 0) and (35, 0) and meaning a2=1225a^2 = 1225. Since the closest distance is ac=20a - c = 20, the focal distance is c=3520=15c = 35 - 20 = 15 AU, placing the sun at the focus (15, 0). The value of b2b^2 is determined using the relation b2=a2c2b^2 = a^2 - c^2. Substituting the known values yields b2=352152=1225225=1000b^2 = 35^2 - 15^2 = 1225 - 225 = 1000. Substituting a2=1225a^2 = 1225 and b2=1000b^2 = 1000 into the standard equation x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 produces the equation of the orbit: x21225+y21000=1\frac{x^2}{1225} + \frac{y^2}{1000} = 1.

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Updated 2026-06-29

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