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

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1 Answer
bp
Mar 25, 2016

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#

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