How to determine what kind of conic section in the affine plane?
$begingroup$
So, I've been struggling a bit with understanding this problem.
Let $P^2$ be the real projective plane with homogenous coordinates $(x_0:x_1:x_2)$
Let $cal{C}$ be the line given by $$x_0^2 + 2x_0x_1+3x_1x_2-3(x_2^2)=0$$
Determine the conic section so that $cal{C}$ is in the affine plane, where $x_1neq x_2.$
I found that there are two points where $x_1-x_2=0$ and $cal{C}$ coincide, namely $x_0=x_1 ; text{and}; x_0=-x_1,$ and that when $x_2neq 0,$ $cal{C}$ is not degenerated.
What I dont understand is how we can know that the conic section in the affine plane will be a hyperbola.
The way this was explained to me is because we know that there are two points that are in the intersection with $x_2=x_1,$ although I dont see why it follows that it is a hyperbola. Similarly, how would it look if it were say, a circle.
geometry analytic-geometry conic-sections
edited Dec 11 '18 at 12:46
user376343
3,9584829
asked Dec 10 '18 at 19:40
AhmedMAhmedM
61
$endgroup$
add a comment |
$begingroup$
So, I've been struggling a bit with understanding this problem.
Let $P^2$ be the real projective plane with homogenous coordinates $(x_0:x_1:x_2)$
Let $cal{C}$ be the line given by $$x_0^2 + 2x_0x_1+3x_1x_2-3(x_2^2)=0$$
Determine the conic section so that $cal{C}$ is in the affine plane, where $x_1neq x_2.$
I found that there are two points where $x_1-x_2=0$ and $cal{C}$ coincide, namely $x_0=x_1 ; text{and}; x_0=-x_1,$ and that when $x_2neq 0,$ $cal{C}$ is not degenerated.
What I dont understand is how we can know that the conic section in the affine plane will be a hyperbola.
The way this was explained to me is because we know that there are two points that are in the intersection with $x_2=x_1,$ although I dont see why it follows that it is a hyperbola. Similarly, how would it look if it were say, a circle.
geometry analytic-geometry conic-sections
edited Dec 11 '18 at 12:46
user376343
3,9584829
asked Dec 10 '18 at 19:40
AhmedMAhmedM
61
$endgroup$
2
$begingroup$
You can also dehomogenize by setting $x_1-x_2=1$ or $x_1=x_2+1$ in $x_0^2 + 2x_0x_1+3x_1x_2-3(x_2^2)=0$ to get $2x_0x_2+3x_2+x_0^2+2x_0=0$, now an equation in two variables easily checked to be a hyperbola. Also, a quick google search gave this link to address your real question.
$endgroup$
– Jan-Magnus Økland
Dec 10 '18 at 21:21
$begingroup$
Note that circles can’t be distinguished from other ellipses until you impose a Euclidean geometry on the affine plane.
$endgroup$
– amd
Dec 10 '18 at 23:27
add a comment |
$begingroup$
So, I've been struggling a bit with understanding this problem.
Let $P^2$ be the real projective plane with homogenous coordinates $(x_0:x_1:x_2)$
Let $cal{C}$ be the line given by $$x_0^2 + 2x_0x_1+3x_1x_2-3(x_2^2)=0$$
Determine the conic section so that $cal{C}$ is in the affine plane, where $x_1neq x_2.$
I found that there are two points where $x_1-x_2=0$ and $cal{C}$ coincide, namely $x_0=x_1 ; text{and}; x_0=-x_1,$ and that when $x_2neq 0,$ $cal{C}$ is not degenerated.
What I dont understand is how we can know that the conic section in the affine plane will be a hyperbola.
The way this was explained to me is because we know that there are two points that are in the intersection with $x_2=x_1,$ although I dont see why it follows that it is a hyperbola. Similarly, how would it look if it were say, a circle.
geometry analytic-geometry conic-sections
edited Dec 11 '18 at 12:46
user376343
3,9584829
asked Dec 10 '18 at 19:40
AhmedMAhmedM
61
$endgroup$
So, I've been struggling a bit with understanding this problem.
Let $P^2$ be the real projective plane with homogenous coordinates $(x_0:x_1:x_2)$
Let $cal{C}$ be the line given by $$x_0^2 + 2x_0x_1+3x_1x_2-3(x_2^2)=0$$
Determine the conic section so that $cal{C}$ is in the affine plane, where $x_1neq x_2.$
I found that there are two points where $x_1-x_2=0$ and $cal{C}$ coincide, namely $x_0=x_1 ; text{and}; x_0=-x_1,$ and that when $x_2neq 0,$ $cal{C}$ is not degenerated.
What I dont understand is how we can know that the conic section in the affine plane will be a hyperbola.
The way this was explained to me is because we know that there are two points that are in the intersection with $x_2=x_1,$ although I dont see why it follows that it is a hyperbola. Similarly, how would it look if it were say, a circle.
geometry analytic-geometry conic-sections
geometry analytic-geometry conic-sections
edited Dec 11 '18 at 12:46
user376343
3,9584829
asked Dec 10 '18 at 19:40
AhmedMAhmedM
61
edited Dec 11 '18 at 12:46
user376343
3,9584829
asked Dec 10 '18 at 19:40
AhmedMAhmedM
61
edited Dec 11 '18 at 12:46
user376343
3,9584829
edited Dec 11 '18 at 12:46
user376343
3,9584829
edited Dec 11 '18 at 12:46
user376343
3,9584829
3,9584829
asked Dec 10 '18 at 19:40
AhmedMAhmedM
61
asked Dec 10 '18 at 19:40
AhmedMAhmedM
61
asked Dec 10 '18 at 19:40
AhmedMAhmedM
61
61
2
$begingroup$
You can also dehomogenize by setting $x_1-x_2=1$ or $x_1=x_2+1$ in $x_0^2 + 2x_0x_1+3x_1x_2-3(x_2^2)=0$ to get $2x_0x_2+3x_2+x_0^2+2x_0=0$, now an equation in two variables easily checked to be a hyperbola. Also, a quick google search gave this link to address your real question.
$endgroup$
– Jan-Magnus Økland
Dec 10 '18 at 21:21
$begingroup$
Note that circles can’t be distinguished from other ellipses until you impose a Euclidean geometry on the affine plane.
$endgroup$
– amd
Dec 10 '18 at 23:27
add a comment |
2
$begingroup$
You can also dehomogenize by setting $x_1-x_2=1$ or $x_1=x_2+1$ in $x_0^2 + 2x_0x_1+3x_1x_2-3(x_2^2)=0$ to get $2x_0x_2+3x_2+x_0^2+2x_0=0$, now an equation in two variables easily checked to be a hyperbola. Also, a quick google search gave this link to address your real question.
$endgroup$
– Jan-Magnus Økland
Dec 10 '18 at 21:21
$begingroup$
Note that circles can’t be distinguished from other ellipses until you impose a Euclidean geometry on the affine plane.
$endgroup$
– amd
Dec 10 '18 at 23:27
2
2
$begingroup$
You can also dehomogenize by setting $x_1-x_2=1$ or $x_1=x_2+1$ in $x_0^2 + 2x_0x_1+3x_1x_2-3(x_2^2)=0$ to get $2x_0x_2+3x_2+x_0^2+2x_0=0$, now an equation in two variables easily checked to be a hyperbola. Also, a quick google search gave this link to address your real question.
$endgroup$
– Jan-Magnus Økland
Dec 10 '18 at 21:21
$begingroup$
You can also dehomogenize by setting $x_1-x_2=1$ or $x_1=x_2+1$ in $x_0^2 + 2x_0x_1+3x_1x_2-3(x_2^2)=0$ to get $2x_0x_2+3x_2+x_0^2+2x_0=0$, now an equation in two variables easily checked to be a hyperbola. Also, a quick google search gave this link to address your real question.
$endgroup$
– Jan-Magnus Økland
Dec 10 '18 at 21:21
$begingroup$
Note that circles can’t be distinguished from other ellipses until you impose a Euclidean geometry on the affine plane.
$endgroup$
– amd
Dec 10 '18 at 23:27
$begingroup$
Note that circles can’t be distinguished from other ellipses until you impose a Euclidean geometry on the affine plane.
$endgroup$
– amd
Dec 10 '18 at 23:27
add a comment |
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$begingroup$
You can also dehomogenize by setting $x_1-x_2=1$ or $x_1=x_2+1$ in $x_0^2 + 2x_0x_1+3x_1x_2-3(x_2^2)=0$ to get $2x_0x_2+3x_2+x_0^2+2x_0=0$, now an equation in two variables easily checked to be a hyperbola. Also, a quick google search gave this link to address your real question.
$endgroup$
– Jan-Magnus Økland
Dec 10 '18 at 21:21
$begingroup$
Note that circles can’t be distinguished from other ellipses until you impose a Euclidean geometry on the affine plane.
$endgroup$
– amd
Dec 10 '18 at 23:27