On http://www.cs.mtu.edu/~shene/COURSES/cs3621/NOTES/geometry/geo-tran.html
I am unable to understand the following point
Obviously, this transformation sends $\;(x,y,w)=(1,0,1)\;$ to $\;(x',y',w') = > (1,-1,0).\;$ That is, this projective transformation sends $\;(1,0)\;$ on the $xy-$plane to the point at infinity in direction $\;<1,-1>.\;$ From the right-hand side of the matrix equation $\;x=Px'\;$ we have
x = 2x' + y' y = x' + y' (1) w = 2x' + y' + w'Let us consider a circle $\;x^2 + y^2 = 1.\;$ Plugging the above equations into the circle equation changes it to the following:
x^2 + 2xy + y^2 - 4xw - 2yw - w^2 = 0 (2)Dividing the above by $\;w^2\;$ to convert it back to conventional form yields
x^2 + 2xy + y^2 - 4x - 2y - 1 = 0 (3)This is a parabola! (Why?) Therefore, a circle that has no point at infinity is transformed to a parabola that does have point at infinity.
How is (2) the equation of a circle. I tried converting to homogenous coordinates, applying the transformation, and then comparing and I can't get the same answer.
How is (3) a parabola? It has an $\;x^2\;$ term and a $\;y^2\;$ term.
The equation is $(x+y)^2-4x-2y-1=0$ so will be a parabola with an opening parallel to $x=-y$ Continuing, we have $$(x+y)^2-4x-2y-1=0\\(x+y)^2-3(x+y)+\frac 94+(y-x)-\frac{13}4=0\\(x+y-\frac 32)^2=-(y-x)+\frac{13}4$$ so you have a parabola rotated by $\frac \pi 4$