#13x^2+6sqrt3xy+7y^2-16=0# is equivalent to
#((x),(y))^T((13,3sqrt(3)),(3sqrt(3),7))((x),(y)) - 16=0#
Introducing a change of variables
#((X),(Y)) = ((costheta, -Sintheta),(Sintheta, Costheta))((x),(y))#
we get at
#((X),(Y))^T((10 + 3 Cos2theta - 3 sqrt[3] Sin2theta,
3 (sqrt[3] Cos2theta + Sin2theta)),(3 (sqrt[3] Cos2theta + Sin2theta),
10 - 3 Cos2theta + 3 sqrt[3] Sin2theta))((X),(Y))-16=0#
Choosing #theta# such that
# 3 (sqrt[3] Cos2theta + Sin2theta)=0#
we have #theta = pi/3# or #theta = -pi/6#
and the quadratic in this new set of coordinates reads
#16X^2+4Y^2-16=0#
or
#4X^2+16Y^2-16=0#
In both cases the quadratic is characterized as an ellipse.
Note. The quadratic kind can be determined computing the characteristic polynomial roots. In this case
#M=((13,3sqrt(3)),(3sqrt(3),7))#
has the characteristic polynomial
#lambda^2-"trace"(M)lambda + det(M)=0#
or
#lambda^2-20lambda+64=0# with roots
#lambda = {4, 16}# characteririzing an ellipse.
