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How do you use rotation of axes to identify and sketch the curve of #sqrt3xy+y^2=1#?

1 Answer
Aug 27, 2016

Answer:

#3/2X^2-Y^2 = 1#

Explanation:

This quadratic can be written as

#p.M.p^T = 1#

where

#p = (x,y)# and #M = ((0,sqrt(3)/2),(sqrt(3)/2,1))#

Choosing a rotation given by

#R(theta) = ((Costheta, -Sintheta),(Sintheta, Costheta))#

and a new set of coordinates

#P = (X,Y)=p.R(theta)# then

#P.R^(-1)(theta).M.R(theta).P^T = 1#

with

#R^(-1)(theta).M.R(theta) = ((Sintheta (-sqrt[3] Costheta + Sintheta), 1/2 (sqrt[3] Cos(2theta) - Sin(2 theta))),(1/ 2 (sqrt[3] Cos(2theta) - Sin(2 theta)), Cos theta (Costheta + sqrt[3] Sintheta)))#

Choosing #theta# such that

#1/2 (sqrt[3] Cos(2theta) - Sin(2 theta))=0# for

#theta =-pi/3# or #theta = pi/6#

we have

#3/2X^2-Y^2 = 1#

which is a hyperbola.