×

Hello! Socratic's Terms of Service and Privacy Policy have been updated, which will be automatically effective on October 6, 2018. Please contact hello@socratic.com with any questions.

How do you rotate the axes to transform the equation #x^2+xy=3# into a new equation with no xy term and then find the angle of rotation?

1 Answer
Feb 1, 2017

Answer:

#(x'^2)/4(2+2sqrt2)+(y'^2)/4(2-2sqrt2)-3=0#

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+xy-3=0#, we have #A=1# and #B=1# and #C=0#, hence #cot2theta=1# i.e. #theta=pi/8#

Hence relation is give by #x=x'cos(pi/8)-y'sin(pi/8)# and #y=x'sin(pi/8)+y'cos(pi/8)# i.e.

#x=(x'sqrt(2+sqrt2))/2-(y'sqrt(2-sqrt2))/2# and #y=(x'sqrt(2-sqrt2))/2+(y'sqrt(2+sqrt2))/2#

Hence, we get #((x'sqrt(2+sqrt2))/2-(y'sqrt(2-sqrt2))/2)^2+((x'sqrt(2+sqrt2))/2-(y'sqrt(2-sqrt2))/2)((x'sqrt(2-sqrt2))/2+(y'sqrt(2+sqrt2))/2)-3=0#

or #((x'^2(2+sqrt2))/4+(y'^2(2-sqrt2))/4-(2x'y'sqrt2)/4)+((x'^2sqrt2)/4-(y'^2sqrt2)/4+(x'y'(2+sqrt2))/4-(x'y'(2-sqrt2))/4)-3=0#

or #(x'^2)/4(2+2sqrt2)+(y'^2)/4(2-2sqrt2)-3=0#

The two graphs are as follows:
graph{x^2+xy-3=0 [-10, 10, -5, 5]}
and
graph{(x^2)/4(2+2sqrt2)+(y^2)/4(2-2sqrt2)-3=0 [-10, 10, -5, 5]}