**Problem : **
Find the coordinates of the foci of the ellipse
6*x*
^{2} +
*xy* + 7*y*
^{2} - 36 = 0
.

This ellipse has an
*xy*
-term, so we'll have to rotate the axes to eliminate
that term and find the standard form of the ellipse in the
*x'y'*
coordinate
system. Then we'll find the foci and convert back to
(*x*, *y*)
for the answer.

The axes must be rotated through an angle
*θ*
such that
cot(2*θ*) =
.
= -
. Therefore,
*θ* =
.

Next we must convert the
*x*
and
*y*
coordinates to
*x'*
and
*y'*
coordinates in
the new coordinate system which is a rotation of the coordinate axes by
*θ* =
radians. These conversions are as follows:
*x* = *x'*cos(*θ*) - *y'*sin(*θ*)
, and
*y* = *x'*sin(*θ*) + *y'*cos(*θ*)
. Substituting
*θ* =
, we find that
*x* =
, and
*y* =
. Then these values for
*x*
and
*y*
are substituted
in the equation
6*x*
^{2} +
*xy* + 7*y*
^{2} - 36 = 0
. After a lot of messy
algebra, the equation simplifies to
30*x'*
^{2} +22*y'*
^{2} = 144
. This equation in
standard form is
+ = 1
.

*a* > *b*
, so we know that
*a* 2.5584
and
*b* 2.1909
. Therefore
*c* 1.3211
. The major axis is vertical (based on the form of the
equation in which the
*y*
^{2}
term is the numerator of the fraction whose
denominator is
*a*
^{2}
). Therefore the foci are located at
(0,±1.3211)
.

Keep in mind that these are
(*x'*, *y'*)
coordinates, and not yet
(*x*, *y*)
coordinates. The
*x'*
and
*y'*
axes are rotated
radians in a
counterclockwise direction from the
*x*
and
*y*
axes. To find the
*x*
and
*y*
coordinates of the foci, we must convert
*x'*
and
*y'*
back to
*x*
and
*y*
. We
use the same equations as before, and eventually find out that the foci are
located at
(*x*, *y*) (- 1.144,.6605)
and
(1.144, - .6605)
. The
approximations were a result of square roots taken. This is how to rotate the
axes to eliminate the
*xy*
-term of a conic to get in into standard form.

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