```Question 300296
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
<pre><font size = 4 color = "indigo"><b>

{{{13x^2-6sqrt(3)xy+7y^2-16=0}}}

The general conic equation is:

{{{Ax^2+Bxy+Cy^2+Dx+Ey+F=0}}}

{{{A=13}}}, {{{B=-6sqrt(3)}}}, {{{C=7}}}, {{{D=0}}}, {{{E=0}}}, {{{F=-16}}}

This is a rotated ellipse because {{{B^2-4AC=(-6sqrt(3))^2-4(13)(7)=36*3-364=-256}}} and {{{-256<0}}}

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

{{{tan(2theta)=B/(A-C)}}}

to find the angle of rotation {{{theta}}} required to eliminate it:

{{{tan(2theta)=(-6sqrt(3))/(13-7)}}}
{{{tan(2theta)=(-6sqrt(3))/6}}}
{{{tan(2theta)=-sqrt(3)}}}
{{{2theta="120°"}}}
{{{theta="60°"}}}

When we replace {{{x}}} by {{{"x'"*cos(theta)-"y'"*sin(theta)}}}

and

{{{y}}} by {{{"x'"*sin(theta)+"y'"*cos(theta)}}}

into

{{{Ax^2+Bxy+Cy^2+Dx+Ey+F=0}}}

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

{{{(A*Cos^2theta+B*Sin(theta)Cos(theta)+C*Sin^2theta)*"x'"^2+

(A*Sin^2theta-B*Sin(theta)Cos(theta)+C*Cos^2theta)*"y'"^2

+ (D*Cos(theta)+E*Sin(theta))*"x'" +

(-D*Sin(theta)+E*Cos(theta))*"x'" +  F=0

}}}

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

{{{(A*Cos^2theta+B*Sin(theta)Cos(theta)+C*Sin^2theta)*"x'"^2+

(A*Sin^2theta-B*Sin(theta)Cos(theta)+C*Cos^2theta)*"y'"^2
+  F=0
}}}

The coefficient of {{{"x'"^2}}} is

{{{13*Cos^2("60°")+(-6sqrt(3))*Sin("60°")Cos("60°")+7*Sin^2("60°") =

13*(1/2)^2-6sqrt(3)*(sqrt(3)/2)(1/2)+7*(sqrt(3)/2)^2 =

13(1/4)-(6*3)/4+7*(3/4)=13/4-18/4+21/4=16/4=4}}}

The coefficient of {{{"y'"^2}}} is

{{{13*Sin^2("60°")-(-6sqrt(3))*Sin("60°")Cos("60°")+7*Cos^2("60°") =

13*(sqrt(3)/2)^2+6sqrt(3)*(sqrt(3)/2)(1/2)+7*(1/2)^2 =

13(3/4)+(6*3)/4+7*(1/4)=39/4+18/4+7/4=64/4=16}}}

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

{{{4*"x'"^2+16*"y'"^2=16}}}

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

{{{4*"x'"^2/16+16*"y'"^2/16=16/16}}}

or

{{{"x'"^2/4+"y'"^2/1=1}}}

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

{{{drawing(400,400,-3,3,-3,3,

graph(400,400,-3,3,-3,3,(3sqrt(3)x+4sqrt(7-4x^2))/7) ,

graph(400,400,-3,3,-3,3,(3sqrt(3)x-4sqrt(7-4x^2))/7),

green(line(5,5*sqrt(3),-5,-5*sqrt(3))),

green(line(5,-5*sqrt(3)/3,-5,5*sqrt(3)/3)), locate(1.3,2.9,"x'"),
locate(-2.8,1.9,"y'")

)}}}

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

{{{x^2/a^2+y^2/b^2=1}}}

We see that {{{a^2=4}}} or {{{a=2}}}
and that {{{b^2=1}}} or {{{b=1}}}

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

use the Pythagorean relation for ellipses:

{{{c^2=a^2-b^2}}}
{{{c^2=2^2-1^2}}}
{{{c^2=4-1}}}
{{{c^2=3}}}
{{{c=sqrt(3)}}}

Thus the foci with respect to the rotated x'y'-coordinate
system are  (x',y') = ({{{sqrt(3)}}},0) and ({{{-sqrt(3)}}},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 {{{sqrt(3)}}} units from the
origin out on the green x' axis.  The x' axis makes a 60° angle with
respect to the x-axis.

{{{drawing(400,400,-3,3,-3,3,

graph(400,400,-3,3,-3,3,(3sqrt(3)x+4sqrt(7-4x^2))/7) ,

graph(400,400,-3,3,-3,3,(3sqrt(3)x-4sqrt(7-4x^2))/7),

green(line(5,5*sqrt(3),-5,-5*sqrt(3))),

green(line(5,-5*sqrt(3)/3,-5,5*sqrt(3)/3)), locate(1.3,2.9,"x'"),
locate(-2.8,1.9,"y'"),

line(sqrt(3)/2+.05,3/2,sqrt(3)/2-.05,3/2), line(sqrt(3)/2,3/2+.05,sqrt(3)/2,3/2-.05), line(sqrt(3)/2+.05,3/2+.05,sqrt(3)/2-.05,3/2-.05), line(sqrt(3)/2+.05,3/2-.05,sqrt(3)/2-.05,3/2+.05),

line(-sqrt(3)/2+.05,-3/2,-sqrt(3)/2-.05,-3/2), line(-sqrt(3)/2,-3/2+.05,-sqrt(3)/2,-3/2-.05), line(-sqrt(3)/2+.05,-3/2+.05,-sqrt(3)/2-.05,-3/2-.05), line(-sqrt(3)/2+.05,-3/2-.05,-sqrt(3)/2-.05,-3/2+.05), locate(.2,.3,"60°"),

red(arc(0,0,1.4,-1.4,0,60)), locate(.25,1,sqrt(3))

)}}}

Now we draw a perpendicular from the focus to the x-axis forming
a right triangle with hypotenuse {{{sqrt(3)}}}.  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.

{{{drawing(400,400,-3,3,-3,3,

graph(400,400,-3,3,-3,3,(3sqrt(3)x+4sqrt(7-4x^2))/7) ,

graph(400,400,-3,3,-3,3,(3sqrt(3)x-4sqrt(7-4x^2))/7),

green(line(5,5*sqrt(3),-5,-5*sqrt(3))),

green(line(5,-5*sqrt(3)/3,-5,5*sqrt(3)/3)), locate(1.3,2.9,"x'"),
locate(-2.8,1.9,"y'"),

line(sqrt(3)/2+.05,3/2,sqrt(3)/2-.05,3/2), line(sqrt(3)/2,3/2+.05,sqrt(3)/2,3/2-.05), line(sqrt(3)/2+.05,3/2+.05,sqrt(3)/2-.05,3/2-.05), line(sqrt(3)/2+.05,3/2-.05,sqrt(3)/2-.05,3/2+.05),

line(-sqrt(3)/2+.05,-3/2,-sqrt(3)/2-.05,-3/2), line(-sqrt(3)/2,-3/2+.05,-sqrt(3)/2,-3/2-.05), line(-sqrt(3)/2+.05,-3/2+.05,-sqrt(3)/2-.05,-3/2-.05), line(-sqrt(3)/2+.05,-3/2-.05,-sqrt(3)/2-.05,-3/2+.05), locate(.2,.3,"60°"),

red(arc(0,0,1.4,-1.4,0,60)), locate(.25,1,sqrt(3)), locate(.4,0.05,x),
locate(.9,.9,y),
blue(line(sqrt(3)/2,3/2,sqrt(3)/2,0)), locate(sqrt(3)/2+.1,3/2+.2,"(x,y)")

)}}}

From the right triangle,

{{{x/sqrt(3) = cos("60°")}}}
{{{x/sqrt(3) = 1/2}}}
{{{2x=sqrt(3)}}}
{{{x=sqrt(3)/2}}}

{{{y/sqrt(3) = sin("60°")}}}
{{{y/sqrt(3) = sqrt(3)/2}}}
{{{2y=3}}}
{{{y=3/2}}}

So the upper focus is ({{{sqrt(3)/2}}},{{{3/2}}})

and by symmetry the lower focus is ({{{-sqrt(3)/2}}},{{{-3/2}}})

Edwin</pre>```