Mixing
Focus,Latus Rectum and Directrix of Rectangular Hyperbola
$begingroup$
There are two forms of rectangular hyperbola $x^2-y^2 = a^2$ and xy = $c^2$. How do we find the focus, latus Rectum and Directrix of xy = $c^2$. For the former there is a set of well defined formulas.
geometry conic-sections
edited Jan 30 at 12:45
Parth Chauhan
636
asked Aug 30 '17 at 2:23
Samar Imam ZaidiSamar Imam Zaidi
1,6191520
$endgroup$
add a comment |
$begingroup$
There are two forms of rectangular hyperbola $x^2-y^2 = a^2$ and xy = $c^2$. How do we find the focus, latus Rectum and Directrix of xy = $c^2$. For the former there is a set of well defined formulas.
geometry conic-sections
edited Jan 30 at 12:45
Parth Chauhan
636
asked Aug 30 '17 at 2:23
Samar Imam ZaidiSamar Imam Zaidi
1,6191520
$endgroup$
$begingroup$
Use en.m.wikipedia.org/wiki/Rotation_of_axes to eliminate $xy$
$endgroup$
– lab bhattacharjee
Aug 30 '17 at 2:40
$begingroup$
For $xy=c^2$, the center is $(0,0)$, not the vertex. One vertex is $(c,c)$, the other is $(-c,-c)$.
$endgroup$
– Blue
Aug 30 '17 at 3:23
$begingroup$
Rotate the formulas.
$endgroup$
– amd
Aug 30 '17 at 4:47
add a comment |
$begingroup$
There are two forms of rectangular hyperbola $x^2-y^2 = a^2$ and xy = $c^2$. How do we find the focus, latus Rectum and Directrix of xy = $c^2$. For the former there is a set of well defined formulas.
geometry conic-sections
edited Jan 30 at 12:45
Parth Chauhan
636
asked Aug 30 '17 at 2:23
Samar Imam ZaidiSamar Imam Zaidi
1,6191520
$endgroup$
There are two forms of rectangular hyperbola $x^2-y^2 = a^2$ and xy = $c^2$. How do we find the focus, latus Rectum and Directrix of xy = $c^2$. For the former there is a set of well defined formulas.
geometry conic-sections
geometry conic-sections
edited Jan 30 at 12:45
Parth Chauhan
636
asked Aug 30 '17 at 2:23
Samar Imam ZaidiSamar Imam Zaidi
1,6191520
edited Jan 30 at 12:45
Parth Chauhan
636
asked Aug 30 '17 at 2:23
Samar Imam ZaidiSamar Imam Zaidi
1,6191520
edited Jan 30 at 12:45
Parth Chauhan
636
edited Jan 30 at 12:45
Parth Chauhan
636
edited Jan 30 at 12:45
Parth Chauhan
636
636
asked Aug 30 '17 at 2:23
Samar Imam ZaidiSamar Imam Zaidi
1,6191520
asked Aug 30 '17 at 2:23
Samar Imam ZaidiSamar Imam Zaidi
1,6191520
asked Aug 30 '17 at 2:23
Samar Imam ZaidiSamar Imam Zaidi
1,6191520
1,6191520
$begingroup$
Use en.m.wikipedia.org/wiki/Rotation_of_axes to eliminate $xy$
$endgroup$
– lab bhattacharjee
Aug 30 '17 at 2:40
$begingroup$
For $xy=c^2$, the center is $(0,0)$, not the vertex. One vertex is $(c,c)$, the other is $(-c,-c)$.
$endgroup$
– Blue
Aug 30 '17 at 3:23
$begingroup$
Rotate the formulas.
$endgroup$
– amd
Aug 30 '17 at 4:47
add a comment |
$begingroup$
Use en.m.wikipedia.org/wiki/Rotation_of_axes to eliminate $xy$
$endgroup$
– lab bhattacharjee
Aug 30 '17 at 2:40
$begingroup$
For $xy=c^2$, the center is $(0,0)$, not the vertex. One vertex is $(c,c)$, the other is $(-c,-c)$.
$endgroup$
– Blue
Aug 30 '17 at 3:23
$begingroup$
Rotate the formulas.
$endgroup$
– amd
Aug 30 '17 at 4:47
$begingroup$
Use en.m.wikipedia.org/wiki/Rotation_of_axes to eliminate $xy$
$endgroup$
– lab bhattacharjee
Aug 30 '17 at 2:40
$begingroup$
Use en.m.wikipedia.org/wiki/Rotation_of_axes to eliminate $xy$
$endgroup$
– lab bhattacharjee
Aug 30 '17 at 2:40
$begingroup$
For $xy=c^2$, the center is $(0,0)$, not the vertex. One vertex is $(c,c)$, the other is $(-c,-c)$.
$endgroup$
– Blue
Aug 30 '17 at 3:23
$begingroup$
For $xy=c^2$, the center is $(0,0)$, not the vertex. One vertex is $(c,c)$, the other is $(-c,-c)$.
$endgroup$
– Blue
Aug 30 '17 at 3:23
$begingroup$
Rotate the formulas.
$endgroup$
– amd
Aug 30 '17 at 4:47
$begingroup$
Rotate the formulas.
$endgroup$
– amd
Aug 30 '17 at 4:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You can find what you need without rotations, just remembering the geometric properties of foci $F$ and $F'$. First of all, they lie on the transverse (or major) axis of the hyperbola (line $y=x$ in our case) and are symmetric with respect to center $(0,0)$. We can then conveniently set $F=(f,f)$ and $F'=(-f,-f)$, with $f>0$.
The hyperbola is the locus of points $P$ such that
$$|F'P-FP|=k,$$
where $k$ is a constant. $A=(c,c)$ and $B=(-c,-c)$ are the vertices of the hyperbola (that is its intersections with the transverse axis): as $|F'A-FA|=AB=2sqrt2c$ we then get $k=2sqrt2c$. Apply now the equation to any convenient point on the hyperbola, for instance $P=(f,c^2/f)$:
you have $FA=f-c^2/f$, $F'A=sqrt{4f^2+(f+c^2/f)^2}$ and from the above equation, after some algebra, you can get $f=sqrt2c$.
To find the latus rectum $FL$ you just need to find $L$ as one of the intersections between the hyperbola and a line through $F$ perpendicular to the transverse axis. Similarly, from the definition of directrix you can find its equation.
answered Aug 30 '17 at 8:44
AretinoAretino
25.8k31545
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
You can find what you need without rotations, just remembering the geometric properties of foci $F$ and $F'$. First of all, they lie on the transverse (or major) axis of the hyperbola (line $y=x$ in our case) and are symmetric with respect to center $(0,0)$. We can then conveniently set $F=(f,f)$ and $F'=(-f,-f)$, with $f>0$.
The hyperbola is the locus of points $P$ such that
$$|F'P-FP|=k,$$
where $k$ is a constant. $A=(c,c)$ and $B=(-c,-c)$ are the vertices of the hyperbola (that is its intersections with the transverse axis): as $|F'A-FA|=AB=2sqrt2c$ we then get $k=2sqrt2c$. Apply now the equation to any convenient point on the hyperbola, for instance $P=(f,c^2/f)$:
you have $FA=f-c^2/f$, $F'A=sqrt{4f^2+(f+c^2/f)^2}$ and from the above equation, after some algebra, you can get $f=sqrt2c$.
To find the latus rectum $FL$ you just need to find $L$ as one of the intersections between the hyperbola and a line through $F$ perpendicular to the transverse axis. Similarly, from the definition of directrix you can find its equation.
answered Aug 30 '17 at 8:44
AretinoAretino
25.8k31545
$endgroup$
add a comment |
$begingroup$
You can find what you need without rotations, just remembering the geometric properties of foci $F$ and $F'$. First of all, they lie on the transverse (or major) axis of the hyperbola (line $y=x$ in our case) and are symmetric with respect to center $(0,0)$. We can then conveniently set $F=(f,f)$ and $F'=(-f,-f)$, with $f>0$.
The hyperbola is the locus of points $P$ such that
$$|F'P-FP|=k,$$
where $k$ is a constant. $A=(c,c)$ and $B=(-c,-c)$ are the vertices of the hyperbola (that is its intersections with the transverse axis): as $|F'A-FA|=AB=2sqrt2c$ we then get $k=2sqrt2c$. Apply now the equation to any convenient point on the hyperbola, for instance $P=(f,c^2/f)$:
you have $FA=f-c^2/f$, $F'A=sqrt{4f^2+(f+c^2/f)^2}$ and from the above equation, after some algebra, you can get $f=sqrt2c$.
To find the latus rectum $FL$ you just need to find $L$ as one of the intersections between the hyperbola and a line through $F$ perpendicular to the transverse axis. Similarly, from the definition of directrix you can find its equation.
answered Aug 30 '17 at 8:44
AretinoAretino
25.8k31545
$endgroup$
add a comment |
$begingroup$
You can find what you need without rotations, just remembering the geometric properties of foci $F$ and $F'$. First of all, they lie on the transverse (or major) axis of the hyperbola (line $y=x$ in our case) and are symmetric with respect to center $(0,0)$. We can then conveniently set $F=(f,f)$ and $F'=(-f,-f)$, with $f>0$.
The hyperbola is the locus of points $P$ such that
$$|F'P-FP|=k,$$
where $k$ is a constant. $A=(c,c)$ and $B=(-c,-c)$ are the vertices of the hyperbola (that is its intersections with the transverse axis): as $|F'A-FA|=AB=2sqrt2c$ we then get $k=2sqrt2c$. Apply now the equation to any convenient point on the hyperbola, for instance $P=(f,c^2/f)$:
you have $FA=f-c^2/f$, $F'A=sqrt{4f^2+(f+c^2/f)^2}$ and from the above equation, after some algebra, you can get $f=sqrt2c$.
To find the latus rectum $FL$ you just need to find $L$ as one of the intersections between the hyperbola and a line through $F$ perpendicular to the transverse axis. Similarly, from the definition of directrix you can find its equation.
answered Aug 30 '17 at 8:44
AretinoAretino
25.8k31545
$endgroup$
You can find what you need without rotations, just remembering the geometric properties of foci $F$ and $F'$. First of all, they lie on the transverse (or major) axis of the hyperbola (line $y=x$ in our case) and are symmetric with respect to center $(0,0)$. We can then conveniently set $F=(f,f)$ and $F'=(-f,-f)$, with $f>0$.
The hyperbola is the locus of points $P$ such that
$$|F'P-FP|=k,$$
where $k$ is a constant. $A=(c,c)$ and $B=(-c,-c)$ are the vertices of the hyperbola (that is its intersections with the transverse axis): as $|F'A-FA|=AB=2sqrt2c$ we then get $k=2sqrt2c$. Apply now the equation to any convenient point on the hyperbola, for instance $P=(f,c^2/f)$:
you have $FA=f-c^2/f$, $F'A=sqrt{4f^2+(f+c^2/f)^2}$ and from the above equation, after some algebra, you can get $f=sqrt2c$.
To find the latus rectum $FL$ you just need to find $L$ as one of the intersections between the hyperbola and a line through $F$ perpendicular to the transverse axis. Similarly, from the definition of directrix you can find its equation.
answered Aug 30 '17 at 8:44
AretinoAretino
25.8k31545
edited Aug 30 '17 at 13:02
answered Aug 30 '17 at 8:44
AretinoAretino
25.8k31545
answered Aug 30 '17 at 8:44
AretinoAretino
25.8k31545
answered Aug 30 '17 at 8:44
AretinoAretino
25.8k31545
25.8k31545
add a comment |
add a comment |
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$begingroup$
Use en.m.wikipedia.org/wiki/Rotation_of_axes to eliminate $xy$
$endgroup$
– lab bhattacharjee
Aug 30 '17 at 2:40
$begingroup$
For $xy=c^2$, the center is $(0,0)$, not the vertex. One vertex is $(c,c)$, the other is $(-c,-c)$.
$endgroup$
– Blue
Aug 30 '17 at 3:23
$begingroup$
Rotate the formulas.
$endgroup$
– amd
Aug 30 '17 at 4:47