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

Mar 25, 2016

It is explained below.

#### Explanation:

In the 2nd degree equation $A {x}^{2} + C {y}^{2} + D x + E y = 0$
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 ${\left(y + 1\right)}^{2} - 6 {x}^{2} - 4 x - 1 = 0$
Or, ${\left(y + 1\right)}^{2} - 6 \left({x}^{2} + \frac{2}{3} x\right) - 1 = 0$

Or, ${\left(y + 1\right)}^{2} - 6 \left({x}^{2} + \frac{2}{3} x + \frac{1}{9}\right) + \frac{6}{9} - 1 = 0$

Or, ${\left(y + 1\right)}^{2} - 6 {\left(x + \frac{1}{3}\right)}^{2} = \frac{1}{3}$

Or, $3 {\left(y + 1\right)}^{2} - 18 {\left(x + \frac{1}{3}\right)}^{2} = 1$

Or ${\left(y + 1\right)}^{2} / {\left(\frac{1}{\sqrt{3}}\right)}^{2} - {\left(x + \frac{1}{3}\right)}^{2} / {\left(\frac{1}{\sqrt{18}}\right)}^{2} = 1$

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 $y ' = x \sin \theta + y \cos \theta$