How do you use rotation of axes to identify and sketch the curve of #x^2+xy+y^2=1#?

1 Answer
Nov 14, 2016

Answer:

An ellipse
#1/2(x_(pi/4))^2+3/2(y_(pi/4))^2=1#

Explanation:

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.