Question 1166886
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As usual at this forum,  the condition is presented  INCORRECTLY   (it is just a bad tradition here).


The corrected version is below.


<pre>
    An orbit of a satellite around a planet is an ellipse, with the planet at one focus of this ellipse. 
    The distance of the satellite from this {{{highlight(cross(star))}}} <U>planet</U> varies from 300, 000 km to 500, 000 km, 
    attained when the satellite is at each of the two vertices. Find the equation of this ellipse, 
    if its center is at the origin, and the vertices are on the x-axis. Assume all units are in 100, 000 km
</pre>


<U>Solution</U>



<pre>
According to the condition,


    a + c = 500,000 km,  or 5 units     (1)     (   the largest distance from the focus to the point on the ellipse
                                                  = the largest distance from the planet's center to the satellite)

    a - c = 300,000 km,  or 3 units     (2)     (   the smallest distance from the focus to the point on the ellipse
                                                  = the smallest distance from the planet's center to the satellite)


In these equations, "a" is the major semi-axis and "c" is the linear eccentricity (half of the focus distance).


By adding equations (1) and (2),  we get


    2a = 5+3 units = 8 units;  hence  a = 4 units.


Then c = 1 unit.


From the other side,  c = {{{sqrt(a^2-b^2)}}};  c^2 = a^2 - b^2;  b^2 = a^2 - c^2 = 4^2 - 1^2 = 15,

where b is the minor semi-axis.


Thus the equation of the ellipse is


    {{{x^2/16}}} + {{{y^2/15}}} = 1.     <U>ANSWER</U>
</pre>

Solved.