This conic can be represented as
#p^TMp = 1#
with #p=(x,y)^T# and #M = ((1,1/2),(1/2,1))#
Making a change of coordinates such that #q = (x_theta,y_theta)#
#q=R_(theta) p# with #R_(theta)=((costheta,-sintheta),(sintheta,costheta))# we have
#p = R_(theta)^Tq# and
#p^TMp=q^TR_(theta)MR_(theta)^Tq = q^TM_(theta)q=1#
with #M_(theta) = ((1-sinthetacostheta,1/2cos2theta),(1/2cos2theta,1-sinthetacostheta))#
Choosing #theta# such that #cos2theta=0# or #2theta=pi/2# or
#theta=pi/4# we have
#M_(pi/4)=((1/2,0),(0,3/2))# and in this new reference frame the conic looks as
#(x_(pi/4),y_(pi/4))M_(pi/4)((x_(pi/4)),(y_(pi/4))) = 1/2(x_(pi/4))^2+3/2(y_(pi/4))^2=1#
which is an ellipse.
