Question

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

Answer 1

`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.