Hyperbola asymptotes from conic general equation [duplicate]

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• 
  Finding the asymptotes of a general hyperbola
  
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If I have the coefficients of the following equation:

$$AX^2 + BXY + CY^2 + DX + EY + F = 0$$

And I know it's a hyperbola, how can I get the equations for the asymptotes with respect to the coefficients A, B, C, D, E, and F?

i.e. similar to this question, except that question deals with properties of ellipses from the general equation instead of hyperbolas.

conic-sections

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asked Jul 24 at 5:37

Justin

1351112

marked as duplicate by Ng Chung Tak, Arnaud Mortier, Brahadeesh, Rhys Steele, Xander Henderson Aug 6 at 1:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What sort of equation(s) are you looking for? Deriving the general conic that represents the two asymptotes, for instance, is quite straightforward.
  – amd
  Jul 24 at 5:59
  

  
  
  
  

add a comment | 

up vote
3
down vote

favorite

1

This question already has an answer here:

• 
  Finding the asymptotes of a general hyperbola
  
  2 answers
  
  

If I have the coefficients of the following equation:

$$AX^2 + BXY + CY^2 + DX + EY + F = 0$$

And I know it's a hyperbola, how can I get the equations for the asymptotes with respect to the coefficients A, B, C, D, E, and F?

i.e. similar to this question, except that question deals with properties of ellipses from the general equation instead of hyperbolas.

conic-sections

share|cite|improve this question

asked Jul 24 at 5:37

Justin

1351112

marked as duplicate by Ng Chung Tak, Arnaud Mortier, Brahadeesh, Rhys Steele, Xander Henderson Aug 6 at 1:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What sort of equation(s) are you looking for? Deriving the general conic that represents the two asymptotes, for instance, is quite straightforward.
  – amd
  Jul 24 at 5:59
  

  
  
  
  

add a comment | 

up vote
3
down vote

favorite

1

up vote
3
down vote

favorite

1

1

This question already has an answer here:

• 
  Finding the asymptotes of a general hyperbola
  
  2 answers
  
  

If I have the coefficients of the following equation:

$$AX^2 + BXY + CY^2 + DX + EY + F = 0$$

And I know it's a hyperbola, how can I get the equations for the asymptotes with respect to the coefficients A, B, C, D, E, and F?

i.e. similar to this question, except that question deals with properties of ellipses from the general equation instead of hyperbolas.

conic-sections

share|cite|improve this question

asked Jul 24 at 5:37

Justin

1351112

This question already has an answer here:

• 
  Finding the asymptotes of a general hyperbola
  
  2 answers
  
  

If I have the coefficients of the following equation:

$$AX^2 + BXY + CY^2 + DX + EY + F = 0$$

And I know it's a hyperbola, how can I get the equations for the asymptotes with respect to the coefficients A, B, C, D, E, and F?

i.e. similar to this question, except that question deals with properties of ellipses from the general equation instead of hyperbolas.

This question already has an answer here:

• 
  Finding the asymptotes of a general hyperbola
  
  2 answers
  
  

conic-sections

share|cite|improve this question

asked Jul 24 at 5:37

Justin

1351112

share|cite|improve this question

share|cite|improve this question

asked Jul 24 at 5:37

Justin

1351112

asked Jul 24 at 5:37

Justin

1351112

asked Jul 24 at 5:37

Justin

1351112

1351112

marked as duplicate by Ng Chung Tak, Arnaud Mortier, Brahadeesh, Rhys Steele, Xander Henderson Aug 6 at 1:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

marked as duplicate by Ng Chung Tak, Arnaud Mortier, Brahadeesh, Rhys Steele, Xander Henderson Aug 6 at 1:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What sort of equation(s) are you looking for? Deriving the general conic that represents the two asymptotes, for instance, is quite straightforward.
  – amd
  Jul 24 at 5:59
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What sort of equation(s) are you looking for? Deriving the general conic that represents the two asymptotes, for instance, is quite straightforward.
  – amd
  Jul 24 at 5:59
  

  
  
  
  

What sort of equation(s) are you looking for? Deriving the general conic that represents the two asymptotes, for instance, is quite straightforward.
– amd
Jul 24 at 5:59

What sort of equation(s) are you looking for? Deriving the general conic that represents the two asymptotes, for instance, is quite straightforward.
– amd
Jul 24 at 5:59

add a comment | 

4 Answers
4

active

oldest

votes

up vote
3
down vote



accepted

First find the centre of the conic. This is the point $(u,v)$
such that the equation of the conic can be rewritten as
$$A(X-u)^2+B(X-u)(Y-v)+C(Y-v)^2+F'=0.$$
For the conic to be a hyperbola, the quadratic part has to factor
into distinct linear factors over $Bbb R$:
$$AX^2+BXY+CY^2=(R_1X+S_1Y)(R_2X+S_2Y).$$
Then the asymptotes are
$$R_i (X-u)+S_i(Y-v)=0$$
($i=1$, $2$).

share|cite|improve this answer

answered Jul 24 at 6:04

Lord Shark the Unknown

85.2k950111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Looks good. I just need to figure out the degenerate cases and how to handle them...
  – Justin
  Jul 24 at 18:52
  

  
  
  
  

add a comment | 

up vote
1
down vote

The equation of a hyperbola whose asymptotes have equations $ax+by+c=0$ and $dx+ey+f=0$ can be written as:
$$
(ax+by+c)(dx+ey+f)=g.
$$
You only need then to expand the above equation and compare the coefficient of the resulting polynomial with $Adots F$ to obtain $adots g$.

In practice, to avoid redundancies, it's convenient to divide the equation of the hyperbola by $A$ (if $Ane0$) and thus set $a=d=1$, to get a system of five equations in five unknowns.

share|cite|improve this answer

answered Jul 24 at 8:51

Aretino

21.7k21342

add a comment | 

up vote
1
down vote

In matrix form the equation is $$x^TAx+2b^Tx+c=0$$ where $A$ is $2times2$ symmetric.

It can be centered with

$$(x+A^-1b)^TA(x+A^-1b)-b^TA^-TA^Tb+c=y^TAy+c'=0.$$

Then, by diagonalizing $A$,

$$y^TPLambda P^-1y+c'=z^TLambda z+c'=0.$$

For an hyperbola, the Eigenvalues have opposite signs and the reduced equation is

$$lambda_uu^2-lambda_vv^2=(sqrtlambda_uu+sqrtlambda_vv)(sqrtlambda_uu-sqrtlambda_vv)=-c'.$$

The two factors are the asymptotes, and in the original coordinates

$$x=P^-1z-A^-1b.$$

share|cite|improve this answer

answered Jul 24 at 9:18

Yves Daoust

111k665203

add a comment | 

up vote
0
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The asymptotes of a hyperbola are the tangents at its intersections with the line at infinity. The matrix of this general equation is $$Q = beginbmatrixA&frac B2&frac D2\frac B2&C&frac E2\frac D2&frac E2&Fendbmatrix$$ and the line at infinity is $mathscr l = [0,0,1]^T$. Let $mathscr l_times$ be the “cross product matrix” of $mathscr l$. We can compute the hyperbola-line intersection by finding a value of $alpha$ that makes $$mathscr l_times^TQmathscr l_times+alphamathscr l_times = beginbmatrixC&-alpha-frac B2&0\alpha-frac B2&A&0\0&0&0endbmatrix$$ a rank-one matrix. This occurs when the principal $2times2$ minor vanishes, which leads to a quadratic equation in $alpha$ with solutions $pmfrac12sqrtB^2-4AC$. Taking the positive root gives the matrix $$beginbmatrixC&-frac12left(B+sqrtB^2-4ACright)&0 \ -frac12left(B^2-sqrtB^2-4ACright)&A&0\0&0&0endbmatrix.$$ The two intersection points are a row-column pair of this matrix that has a nonzero diagonal element. For example, if $Cne0$, then the two points are $mathbf p_1 = left[C,-frac12left(B+sqrtB^2-4ACright),0right]^T$ and $mathbf p_2 = left[C,-frac12left(B-sqrtB^2-4ACright),0right]^T$, the projective equivalents of the asymptotes’ direction vectors, with corresponding tangents $mathscr m_i = Qmathbf p_i$. The implicit Cartesian equations of the lines that these vectors represent are $[x,y,1]mathscr m_i=0$, or $[x,y,1]Qmathbf p_i=0$.

share|cite|improve this answer

edited Jul 24 at 7:17

answered Jul 24 at 7:05

amd

25.8k2943

add a comment | 

4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

First find the centre of the conic. This is the point $(u,v)$
such that the equation of the conic can be rewritten as
$$A(X-u)^2+B(X-u)(Y-v)+C(Y-v)^2+F'=0.$$
For the conic to be a hyperbola, the quadratic part has to factor
into distinct linear factors over $Bbb R$:
$$AX^2+BXY+CY^2=(R_1X+S_1Y)(R_2X+S_2Y).$$
Then the asymptotes are
$$R_i (X-u)+S_i(Y-v)=0$$
($i=1$, $2$).

share|cite|improve this answer

answered Jul 24 at 6:04

Lord Shark the Unknown

85.2k950111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Looks good. I just need to figure out the degenerate cases and how to handle them...
  – Justin
  Jul 24 at 18:52
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

First find the centre of the conic. This is the point $(u,v)$
such that the equation of the conic can be rewritten as
$$A(X-u)^2+B(X-u)(Y-v)+C(Y-v)^2+F'=0.$$
For the conic to be a hyperbola, the quadratic part has to factor
into distinct linear factors over $Bbb R$:
$$AX^2+BXY+CY^2=(R_1X+S_1Y)(R_2X+S_2Y).$$
Then the asymptotes are
$$R_i (X-u)+S_i(Y-v)=0$$
($i=1$, $2$).

share|cite|improve this answer

answered Jul 24 at 6:04

Lord Shark the Unknown

85.2k950111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Looks good. I just need to figure out the degenerate cases and how to handle them...
  – Justin
  Jul 24 at 18:52
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

First find the centre of the conic. This is the point $(u,v)$
such that the equation of the conic can be rewritten as
$$A(X-u)^2+B(X-u)(Y-v)+C(Y-v)^2+F'=0.$$
For the conic to be a hyperbola, the quadratic part has to factor
into distinct linear factors over $Bbb R$:
$$AX^2+BXY+CY^2=(R_1X+S_1Y)(R_2X+S_2Y).$$
Then the asymptotes are
$$R_i (X-u)+S_i(Y-v)=0$$
($i=1$, $2$).

share|cite|improve this answer

answered Jul 24 at 6:04

Lord Shark the Unknown

85.2k950111

First find the centre of the conic. This is the point $(u,v)$
such that the equation of the conic can be rewritten as
$$A(X-u)^2+B(X-u)(Y-v)+C(Y-v)^2+F'=0.$$
For the conic to be a hyperbola, the quadratic part has to factor
into distinct linear factors over $Bbb R$:
$$AX^2+BXY+CY^2=(R_1X+S_1Y)(R_2X+S_2Y).$$
Then the asymptotes are
$$R_i (X-u)+S_i(Y-v)=0$$
($i=1$, $2$).

share|cite|improve this answer

answered Jul 24 at 6:04

Lord Shark the Unknown

85.2k950111

share|cite|improve this answer

share|cite|improve this answer

answered Jul 24 at 6:04

Lord Shark the Unknown

85.2k950111

answered Jul 24 at 6:04

Lord Shark the Unknown

85.2k950111

answered Jul 24 at 6:04

Lord Shark the Unknown

85.2k950111

85.2k950111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Looks good. I just need to figure out the degenerate cases and how to handle them...
  – Justin
  Jul 24 at 18:52
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Looks good. I just need to figure out the degenerate cases and how to handle them...
  – Justin
  Jul 24 at 18:52
  

  
  
  
  

Looks good. I just need to figure out the degenerate cases and how to handle them...
– Justin
Jul 24 at 18:52

Looks good. I just need to figure out the degenerate cases and how to handle them...
– Justin
Jul 24 at 18:52

add a comment | 

up vote
1
down vote

The equation of a hyperbola whose asymptotes have equations $ax+by+c=0$ and $dx+ey+f=0$ can be written as:
$$
(ax+by+c)(dx+ey+f)=g.
$$
You only need then to expand the above equation and compare the coefficient of the resulting polynomial with $Adots F$ to obtain $adots g$.

In practice, to avoid redundancies, it's convenient to divide the equation of the hyperbola by $A$ (if $Ane0$) and thus set $a=d=1$, to get a system of five equations in five unknowns.

share|cite|improve this answer

answered Jul 24 at 8:51

Aretino

21.7k21342

add a comment | 

up vote
1
down vote

The equation of a hyperbola whose asymptotes have equations $ax+by+c=0$ and $dx+ey+f=0$ can be written as:
$$
(ax+by+c)(dx+ey+f)=g.
$$
You only need then to expand the above equation and compare the coefficient of the resulting polynomial with $Adots F$ to obtain $adots g$.

In practice, to avoid redundancies, it's convenient to divide the equation of the hyperbola by $A$ (if $Ane0$) and thus set $a=d=1$, to get a system of five equations in five unknowns.

share|cite|improve this answer

answered Jul 24 at 8:51

Aretino

21.7k21342

add a comment | 

up vote
1
down vote

up vote
1
down vote

The equation of a hyperbola whose asymptotes have equations $ax+by+c=0$ and $dx+ey+f=0$ can be written as:
$$
(ax+by+c)(dx+ey+f)=g.
$$
You only need then to expand the above equation and compare the coefficient of the resulting polynomial with $Adots F$ to obtain $adots g$.

In practice, to avoid redundancies, it's convenient to divide the equation of the hyperbola by $A$ (if $Ane0$) and thus set $a=d=1$, to get a system of five equations in five unknowns.

share|cite|improve this answer

answered Jul 24 at 8:51

Aretino

21.7k21342

The equation of a hyperbola whose asymptotes have equations $ax+by+c=0$ and $dx+ey+f=0$ can be written as:
$$
(ax+by+c)(dx+ey+f)=g.
$$
You only need then to expand the above equation and compare the coefficient of the resulting polynomial with $Adots F$ to obtain $adots g$.

In practice, to avoid redundancies, it's convenient to divide the equation of the hyperbola by $A$ (if $Ane0$) and thus set $a=d=1$, to get a system of five equations in five unknowns.

share|cite|improve this answer

answered Jul 24 at 8:51

Aretino

21.7k21342

share|cite|improve this answer

share|cite|improve this answer

answered Jul 24 at 8:51

Aretino

21.7k21342

answered Jul 24 at 8:51

Aretino

21.7k21342

answered Jul 24 at 8:51

Aretino

21.7k21342

21.7k21342

add a comment | 

add a comment | 

up vote
1
down vote

In matrix form the equation is $$x^TAx+2b^Tx+c=0$$ where $A$ is $2times2$ symmetric.

It can be centered with

$$(x+A^-1b)^TA(x+A^-1b)-b^TA^-TA^Tb+c=y^TAy+c'=0.$$

Then, by diagonalizing $A$,

$$y^TPLambda P^-1y+c'=z^TLambda z+c'=0.$$

For an hyperbola, the Eigenvalues have opposite signs and the reduced equation is

$$lambda_uu^2-lambda_vv^2=(sqrtlambda_uu+sqrtlambda_vv)(sqrtlambda_uu-sqrtlambda_vv)=-c'.$$

The two factors are the asymptotes, and in the original coordinates

$$x=P^-1z-A^-1b.$$

share|cite|improve this answer

answered Jul 24 at 9:18

Yves Daoust

111k665203

add a comment | 

up vote
1
down vote

In matrix form the equation is $$x^TAx+2b^Tx+c=0$$ where $A$ is $2times2$ symmetric.

It can be centered with

$$(x+A^-1b)^TA(x+A^-1b)-b^TA^-TA^Tb+c=y^TAy+c'=0.$$

Then, by diagonalizing $A$,

$$y^TPLambda P^-1y+c'=z^TLambda z+c'=0.$$

For an hyperbola, the Eigenvalues have opposite signs and the reduced equation is

$$lambda_uu^2-lambda_vv^2=(sqrtlambda_uu+sqrtlambda_vv)(sqrtlambda_uu-sqrtlambda_vv)=-c'.$$

The two factors are the asymptotes, and in the original coordinates

$$x=P^-1z-A^-1b.$$

share|cite|improve this answer

answered Jul 24 at 9:18

Yves Daoust

111k665203

add a comment | 

up vote
1
down vote

up vote
1
down vote

In matrix form the equation is $$x^TAx+2b^Tx+c=0$$ where $A$ is $2times2$ symmetric.

It can be centered with

$$(x+A^-1b)^TA(x+A^-1b)-b^TA^-TA^Tb+c=y^TAy+c'=0.$$

Then, by diagonalizing $A$,

$$y^TPLambda P^-1y+c'=z^TLambda z+c'=0.$$

For an hyperbola, the Eigenvalues have opposite signs and the reduced equation is

$$lambda_uu^2-lambda_vv^2=(sqrtlambda_uu+sqrtlambda_vv)(sqrtlambda_uu-sqrtlambda_vv)=-c'.$$

The two factors are the asymptotes, and in the original coordinates

$$x=P^-1z-A^-1b.$$

share|cite|improve this answer

answered Jul 24 at 9:18

Yves Daoust

111k665203

In matrix form the equation is $$x^TAx+2b^Tx+c=0$$ where $A$ is $2times2$ symmetric.

It can be centered with

$$(x+A^-1b)^TA(x+A^-1b)-b^TA^-TA^Tb+c=y^TAy+c'=0.$$

Then, by diagonalizing $A$,

$$y^TPLambda P^-1y+c'=z^TLambda z+c'=0.$$

For an hyperbola, the Eigenvalues have opposite signs and the reduced equation is

$$lambda_uu^2-lambda_vv^2=(sqrtlambda_uu+sqrtlambda_vv)(sqrtlambda_uu-sqrtlambda_vv)=-c'.$$

The two factors are the asymptotes, and in the original coordinates

$$x=P^-1z-A^-1b.$$

share|cite|improve this answer

answered Jul 24 at 9:18

Yves Daoust

111k665203

share|cite|improve this answer

share|cite|improve this answer

answered Jul 24 at 9:18

Yves Daoust

111k665203

answered Jul 24 at 9:18

Yves Daoust

111k665203

answered Jul 24 at 9:18

Yves Daoust

111k665203

111k665203

add a comment | 

add a comment | 

up vote
0
down vote

The asymptotes of a hyperbola are the tangents at its intersections with the line at infinity. The matrix of this general equation is $$Q = beginbmatrixA&frac B2&frac D2\frac B2&C&frac E2\frac D2&frac E2&Fendbmatrix$$ and the line at infinity is $mathscr l = [0,0,1]^T$. Let $mathscr l_times$ be the “cross product matrix” of $mathscr l$. We can compute the hyperbola-line intersection by finding a value of $alpha$ that makes $$mathscr l_times^TQmathscr l_times+alphamathscr l_times = beginbmatrixC&-alpha-frac B2&0\alpha-frac B2&A&0\0&0&0endbmatrix$$ a rank-one matrix. This occurs when the principal $2times2$ minor vanishes, which leads to a quadratic equation in $alpha$ with solutions $pmfrac12sqrtB^2-4AC$. Taking the positive root gives the matrix $$beginbmatrixC&-frac12left(B+sqrtB^2-4ACright)&0 \ -frac12left(B^2-sqrtB^2-4ACright)&A&0\0&0&0endbmatrix.$$ The two intersection points are a row-column pair of this matrix that has a nonzero diagonal element. For example, if $Cne0$, then the two points are $mathbf p_1 = left[C,-frac12left(B+sqrtB^2-4ACright),0right]^T$ and $mathbf p_2 = left[C,-frac12left(B-sqrtB^2-4ACright),0right]^T$, the projective equivalents of the asymptotes’ direction vectors, with corresponding tangents $mathscr m_i = Qmathbf p_i$. The implicit Cartesian equations of the lines that these vectors represent are $[x,y,1]mathscr m_i=0$, or $[x,y,1]Qmathbf p_i=0$.

share|cite|improve this answer

edited Jul 24 at 7:17

answered Jul 24 at 7:05

amd

25.8k2943

add a comment | 

up vote
0
down vote

The asymptotes of a hyperbola are the tangents at its intersections with the line at infinity. The matrix of this general equation is $$Q = beginbmatrixA&frac B2&frac D2\frac B2&C&frac E2\frac D2&frac E2&Fendbmatrix$$ and the line at infinity is $mathscr l = [0,0,1]^T$. Let $mathscr l_times$ be the “cross product matrix” of $mathscr l$. We can compute the hyperbola-line intersection by finding a value of $alpha$ that makes $$mathscr l_times^TQmathscr l_times+alphamathscr l_times = beginbmatrixC&-alpha-frac B2&0\alpha-frac B2&A&0\0&0&0endbmatrix$$ a rank-one matrix. This occurs when the principal $2times2$ minor vanishes, which leads to a quadratic equation in $alpha$ with solutions $pmfrac12sqrtB^2-4AC$. Taking the positive root gives the matrix $$beginbmatrixC&-frac12left(B+sqrtB^2-4ACright)&0 \ -frac12left(B^2-sqrtB^2-4ACright)&A&0\0&0&0endbmatrix.$$ The two intersection points are a row-column pair of this matrix that has a nonzero diagonal element. For example, if $Cne0$, then the two points are $mathbf p_1 = left[C,-frac12left(B+sqrtB^2-4ACright),0right]^T$ and $mathbf p_2 = left[C,-frac12left(B-sqrtB^2-4ACright),0right]^T$, the projective equivalents of the asymptotes’ direction vectors, with corresponding tangents $mathscr m_i = Qmathbf p_i$. The implicit Cartesian equations of the lines that these vectors represent are $[x,y,1]mathscr m_i=0$, or $[x,y,1]Qmathbf p_i=0$.

share|cite|improve this answer

edited Jul 24 at 7:17

answered Jul 24 at 7:05

amd

25.8k2943

add a comment | 

up vote
0
down vote

up vote
0
down vote

The asymptotes of a hyperbola are the tangents at its intersections with the line at infinity. The matrix of this general equation is $$Q = beginbmatrixA&frac B2&frac D2\frac B2&C&frac E2\frac D2&frac E2&Fendbmatrix$$ and the line at infinity is $mathscr l = [0,0,1]^T$. Let $mathscr l_times$ be the “cross product matrix” of $mathscr l$. We can compute the hyperbola-line intersection by finding a value of $alpha$ that makes $$mathscr l_times^TQmathscr l_times+alphamathscr l_times = beginbmatrixC&-alpha-frac B2&0\alpha-frac B2&A&0\0&0&0endbmatrix$$ a rank-one matrix. This occurs when the principal $2times2$ minor vanishes, which leads to a quadratic equation in $alpha$ with solutions $pmfrac12sqrtB^2-4AC$. Taking the positive root gives the matrix $$beginbmatrixC&-frac12left(B+sqrtB^2-4ACright)&0 \ -frac12left(B^2-sqrtB^2-4ACright)&A&0\0&0&0endbmatrix.$$ The two intersection points are a row-column pair of this matrix that has a nonzero diagonal element. For example, if $Cne0$, then the two points are $mathbf p_1 = left[C,-frac12left(B+sqrtB^2-4ACright),0right]^T$ and $mathbf p_2 = left[C,-frac12left(B-sqrtB^2-4ACright),0right]^T$, the projective equivalents of the asymptotes’ direction vectors, with corresponding tangents $mathscr m_i = Qmathbf p_i$. The implicit Cartesian equations of the lines that these vectors represent are $[x,y,1]mathscr m_i=0$, or $[x,y,1]Qmathbf p_i=0$.

share|cite|improve this answer

edited Jul 24 at 7:17

answered Jul 24 at 7:05

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The asymptotes of a hyperbola are the tangents at its intersections with the line at infinity. The matrix of this general equation is $$Q = beginbmatrixA&frac B2&frac D2\frac B2&C&frac E2\frac D2&frac E2&Fendbmatrix$$ and the line at infinity is $mathscr l = [0,0,1]^T$. Let $mathscr l_times$ be the “cross product matrix” of $mathscr l$. We can compute the hyperbola-line intersection by finding a value of $alpha$ that makes $$mathscr l_times^TQmathscr l_times+alphamathscr l_times = beginbmatrixC&-alpha-frac B2&0\alpha-frac B2&A&0\0&0&0endbmatrix$$ a rank-one matrix. This occurs when the principal $2times2$ minor vanishes, which leads to a quadratic equation in $alpha$ with solutions $pmfrac12sqrtB^2-4AC$. Taking the positive root gives the matrix $$beginbmatrixC&-frac12left(B+sqrtB^2-4ACright)&0 \ -frac12left(B^2-sqrtB^2-4ACright)&A&0\0&0&0endbmatrix.$$ The two intersection points are a row-column pair of this matrix that has a nonzero diagonal element. For example, if $Cne0$, then the two points are $mathbf p_1 = left[C,-frac12left(B+sqrtB^2-4ACright),0right]^T$ and $mathbf p_2 = left[C,-frac12left(B-sqrtB^2-4ACright),0right]^T$, the projective equivalents of the asymptotes’ direction vectors, with corresponding tangents $mathscr m_i = Qmathbf p_i$. The implicit Cartesian equations of the lines that these vectors represent are $[x,y,1]mathscr m_i=0$, or $[x,y,1]Qmathbf p_i=0$.

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edited Jul 24 at 7:17

answered Jul 24 at 7:05

amd

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share|cite|improve this answer

share|cite|improve this answer

edited Jul 24 at 7:17

edited Jul 24 at 7:17

edited Jul 24 at 7:17

answered Jul 24 at 7:05

amd

25.8k2943

answered Jul 24 at 7:05

amd

25.8k2943

answered Jul 24 at 7:05

amd

25.8k2943

25.8k2943

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