SOLUTION: Determine the coordinates of the foci of the original conic in exact form for the equation: 13x^2-6sqroot(3)xy+7y^2-16=0

The general conic equation is:



, , , , , 

This is a rotated ellipse because  and 

To eliminate the x'y'-term when we rotate through an angle
of  we use the formula



to find the angle of rotation  required to eliminate it:







When we replace  by 

and

 by 

into



using the proper angle of rotation, there will be no x'y' term
and we will get:



And since D and E are both 0, that simplifies to



The coefficient of  is



The coefficient of  is



So the equation of the ellipse when rotated counter-clockwise 60° is



Getting it in standard form we divide through by 16 to get
1 on the right:



or



Here is the graph.  The green axes are the x' and y' axes:



Comparing the equation to the standard equation for an ellipse
with center at the origin



We see that  or  
and that  or 

To find the foci with respect to the x'-y' coordinate system, we

use the Pythagorean relation for ellipses:







Thus the foci with respect to the rotated x'y'-coordinate
system are  (x',y') = (,0) and (,0)   

However we must find their coordinates with respect to the 
non-rotated original xy-coordinate system.

Let's plot the two foci, which are both  units from the
origin out on the green x' axis.  The x' axis makes a 60° angle with 
respect to the x-axis.



Now we draw a perpendicular from the focus to the x-axis forming
a right triangle with hypotenuse .  The horizontal
leg of that right triangle is the x-coordinate of the focus, and
the y-coordinate of that right triangle is the y-coordinate of
the focus.



From the right triangle,











So the upper focus is (,)

and by symmetry the lower focus is (,)

Edwin