Question

How do I rotate the axes of and then graph `y^2-6x^2-4x+2y=0`?

Answer 1

It is explained below.

Explanation:

In the 2nd degree equation `Ax^2 +Cy^2 +Dx +Ey=0`
if AC<0, the it represents a hyperbola. In the present case it is indeed so as AC=-6 which is < 0.

The equation can be rewritten as `(y+1)^2 -6x^2 -4x-1=0`
Or, `(y+1)^2 -6(x^2 +2/3 x)-1=0`

Or, `(y+1)^2 -6(x^2 +2/3 x +1/9) +6/9 -1 =0`

Or, `(y+1)^2 -6(x+1/3)^2=1/3`

Or, `3(y+1)^2 - 18 (x+1/3)^2 =1`

Or `(y+1)^2 /(1/sqrt3)^2 - (x+1/3)^2 /(1/sqrt18)^2 =1`

Which clearly shows it is a hyperbola. Now if the angle of rotation of axes is `theta` anticlockwise then new coordinates would be given by `x'= x cos theta +y sin theta` and `y'= x sin theta +y cos theta`