Transforming a circle to get a parabola
$begingroup$
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.
transformation computational-geometry projective-geometry
edited Jan 9 at 13:12
user376343
3,9584829
asked Feb 24 '14 at 17:47
Mike GrahamMike Graham
1448
$endgroup$
add a comment |
$begingroup$
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.
transformation computational-geometry projective-geometry
edited Jan 9 at 13:12
user376343
3,9584829
asked Feb 24 '14 at 17:47
Mike GrahamMike Graham
1448
$endgroup$
add a comment |
$begingroup$
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.
transformation computational-geometry projective-geometry
edited Jan 9 at 13:12
user376343
3,9584829
asked Feb 24 '14 at 17:47
Mike GrahamMike Graham
1448
$endgroup$
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.
transformation computational-geometry projective-geometry
transformation computational-geometry projective-geometry
edited Jan 9 at 13:12
user376343
3,9584829
asked Feb 24 '14 at 17:47
Mike GrahamMike Graham
1448
edited Jan 9 at 13:12
user376343
3,9584829
asked Feb 24 '14 at 17:47
Mike GrahamMike Graham
1448
edited Jan 9 at 13:12
user376343
3,9584829
edited Jan 9 at 13:12
user376343
3,9584829
edited Jan 9 at 13:12
user376343
3,9584829
3,9584829
asked Feb 24 '14 at 17:47
Mike GrahamMike Graham
1448
asked Feb 24 '14 at 17:47
Mike GrahamMike Graham
1448
asked Feb 24 '14 at 17:47
Mike GrahamMike Graham
1448
1448
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
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$$ so you have a parabola rotated by $frac pi 4$
answered Feb 24 '14 at 17:58
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
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$begingroup$
The equation is $(x+y)^2-4x-2y-1=0$ so will be a parabola with an opening parallel to $x=-y$ Continuing, we have $$(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$$ so you have a parabola rotated by $frac pi 4$
answered Feb 24 '14 at 17:58
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
The equation is $(x+y)^2-4x-2y-1=0$ so will be a parabola with an opening parallel to $x=-y$ Continuing, we have $$(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$$ so you have a parabola rotated by $frac pi 4$
answered Feb 24 '14 at 17:58
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
The equation is $(x+y)^2-4x-2y-1=0$ so will be a parabola with an opening parallel to $x=-y$ Continuing, we have $$(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$$ so you have a parabola rotated by $frac pi 4$
answered Feb 24 '14 at 17:58
Ross MillikanRoss Millikan
301k24200375
$endgroup$
The equation is $(x+y)^2-4x-2y-1=0$ so will be a parabola with an opening parallel to $x=-y$ Continuing, we have $$(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 $frac pi 4$
answered Feb 24 '14 at 17:58
Ross MillikanRoss Millikan
301k24200375
edited Feb 25 '14 at 4:13
answered Feb 24 '14 at 17:58
Ross MillikanRoss Millikan
301k24200375
answered Feb 24 '14 at 17:58
Ross MillikanRoss Millikan
301k24200375
answered Feb 24 '14 at 17:58
Ross MillikanRoss Millikan
301k24200375
301k24200375
add a comment |
add a comment |
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