**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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