How do you graph the conic #x^2-4xy+y^2+1=0# by first rotations the axis and eliminating the #xy# term?

1 Answer
Jan 28, 2017

Answer:

Please see below.

Explanation:

A conic equation of the type of #Ax^2+Bxy+Cy^2+Dx+Ey+F=0# is rotated by an angle #theta#, to form a new Cartesian plane with coordinates #(x',y')#, if #theta# is appropriately chosen, we can have a new equation without term #xy# i.e. of standard form.
http://philschatz.com/precalculus-book/contents/m49441.html
The relation between coordinates #(x,y)# and #(x'.y')# can be expressed as
#x=x'costheta-y'sintheta# and #y=x'sintheta+y'costheta#

or #x'=xcostheta+ysintheta# and #y=-xsintheta+ycostheta#

for this we need to have #theta# given by #cot2theta=(A-C)/B#

In the given case as equation is #x^2-4xy+y^2+1=0#, we have #A=C=1# and #B=-4# and hence #cot2theta=0# i.e. #theta=pi/4#

Hence relation is give by #x=x'cos(pi/4)-y'sin(pi/4)# and #y=x'sin(pi/4)+y'cos(pi/4)# i.e.

#x=(x')/sqrt2-(y')/sqrt2# and #y=(x')/sqrt2+(y')/sqrt2#

Hence, we get #((x')/sqrt2-(y')/sqrt2)^2-4((x')/sqrt2-(y')/sqrt2)((x')/sqrt2+(y')/sqrt2)+((x')/sqrt2+(y')/sqrt2)^2+1=0#

or #((x'^2)/2+(y'^2)/2-x'y')-4((x'^2)/2-(y'^2)/2)+((x'^2)/2+(y'^2)/2+x'y')+1=0#

or #-x'^2+3y'^2+1=0# or #x'^2-3y'^2=1#

The two graphs are as follows:
graph{x^2-4xy+y^2+1=0 [-10, 10, -5, 5]}
and
graph{x^2-3y^2=1 [-10, 10, -5, 5]}