Question 836757
<pre>
{{{drawing(5200/11,600,-13,13,-16.5,16.5, 
green(line(-12,0,12,0), line(0,-15.49,0,15.49) ),
arc(0,0,24,2*-15.49193338),
circle(0,-9.29516003,0.25),circle(0,-9.29516003,0.22),circle(0,-9.29516003,0.21),circle(0,-9.29516003,0.15),circle(0,-9.29516003,0.07),circle(0,-9.29516003,0.05),circle(0,-9.29516003,0.03),circle(0,-9.29516003,0.01),
red(circle(0,-9.29516003,4)),

circle(0,15.49,0.25),circle(0,15.49,0.22),circle(0,15.49,0.21),circle(0,15.49,0.15),circle(0,15.49,0.07),circle(0,15.49,0.05),circle(0,15.49,0.03),circle(0,15.49,0.01),


circle(0,-5.29516003,0.25),circle(0,-5.29516003,0.22),circle(0,-5.29516003,0.21),circle(0,-5.29516003,0.15),circle(0,-5.29516003,0.07),circle(0,-5.29516003,0.05),circle(0,-5.29516003,0.03),circle(0,-5.29516003,0.01),

circle(0,0,0.25),circle(0,0,0.22),circle(0,0,0.21),circle(0,0,0.15),circle(0,0,0.07),circle(0,0,0.05),circle(0,0,0.03),circle(0,0,0.01),

circle(12,0,0.25),circle(12,0,0.22),circle(12,0,0.21),circle(12,0,0.15),circle(12,0,0.07),circle(12,0,0.05),circle(12,0,0.03),circle(12,0,0.01),


locate(.4,-9,C), locate(.4,-4.5,S),locate(.4,.9,O),locate(.5,16.4,A),
locate(12.4,.9,B)




 )}}}

On the graph, we will say the scale 1 unit = 1000 miles.

The equation of the ellipse might be {{{x^2/b^2+y^2/a^2=1}}}, 
but we don't need that.

The center of the earth is at point C, one of the focal points of the
elliptical orbit of the satellite.  The red circle represents a cross section
of the earth.  We assume the radius of the earth is 4000 mile = 4 units.
The point S is the surface of the earth where the satellite is at its
highest distance above the surface of the earth.

We want to find the length of SA.

OB=b is the semi-minor axis and it is given as 12 units (12000 miles).

The Pythagorean relation for every ellipse is 

c²=a²-b²

We are given the eccentricity 0.6 = {{{c/a}}}

So c = OC = 0.6*OA = 0.6a, and we are given b = OB = 12 (12000 miles)

Substituting

{{{c^2}}}{{{""=""}}}{{{a^2-b^2}}}

{{{(0.6a)^2}}}{{{""=""}}}{{{a^2-12^2}}}

{{{0.36a^2}}}{{{""=""}}}{{{a^2-144}}}

{{{144}}}{{{""=""}}}{{{a^2-0.36a^2}}}

{{{144}}}{{{""=""}}}{{{0.64a^2}}}

{{{144/.64}}}{{{""=""}}}{{{a^2}}}

{{{225}}}{{{""=""}}}{{{a^2}}}

{{{15}}}{{{""=""}}}{{{a}}}

So OA = a = 15

and OC = c = 0.6a = .6(15) = 9 = 

SC = 4 (radius =4000 mi.)

OS = OC-SC = 9-4 = 5

So SA = OS+OA = 5+15 = 20

So the maximum altitude of the satellite is 20000 miles.

Edwin</pre>