How do I rotate the axes of and then graph y^2-6x^2-4x+2y=0y2−6x2−4x+2y=0?

1 Answer
bp
Mar 25, 2016

It is explained below.

Explanation:

In the 2nd degree equation Ax^2 +Cy^2 +Dx +Ey=0Ax2+Cy2+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(y+1)2−6x2−4x−1=0
Or, (y+1)^2 -6(x^2 +2/3 x)-1=0(y+1)2−6(x2+23x)−1=0

Or, (y+1)^2 -6(x^2 +2/3 x +1/9) +6/9 -1 =0(y+1)2−6(x2+23x+19)+69−1=0

Or, (y+1)^2 -6(x+1/3)^2=1/3(y+1)2−6(x+13)2=13

Or, 3(y+1)^2 - 18 (x+1/3)^2 =13(y+1)2−18(x+13)2=1

Or (y+1)^2 /(1/sqrt3)^2 - (x+1/3)^2 /(1/sqrt18)^2 =1(y+1)2(1√3)2−(x+13)2(1√18)2=1

Which clearly shows it is a hyperbola. Now if the angle of rotation of axes is thetaθ 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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