Question

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

Answer 1

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]}