Can you construct a hyperbola with only the eccentricity, two axes of symmetry and semimajor axis length...
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I am trying to construct a hyperbola for a project I'm doing and I have the two axes of symmetry, the length of the semimajor axis and the eccentricity. Is it possible? If so, how?
geometry conic-sections
asked Jan 10 at 23:51
user140323user140323
1214
$endgroup$
add a comment |
0
$begingroup$
I am trying to construct a hyperbola for a project I'm doing and I have the two axes of symmetry, the length of the semimajor axis and the eccentricity. Is it possible? If so, how?
geometry conic-sections
asked Jan 10 at 23:51
user140323user140323
1214
$endgroup$
1
$begingroup$
Semi-major axis and eccentricity are enough to determine the shape and size of the hyperbola; if I understand "centerline" correctly as one of the lines of symmetry (specifically the one through the vertices?), then we have orientation and part of the location. However, the exact location of the center of the hyperbola remains undetermined.
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– Blue
Jan 11 at 0:01
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My bad, by centerline I mean the line of symmetry not passing through the vertices. And, now that I think of it, I have both lines given. I'll edit my question.
$endgroup$
– user140323
Jan 11 at 2:14
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If you know both axes of symmetry, semimajor axis and eccentricity, then you know the position of both vertices and can easily find the position of the foci. Hence the hyperbola is uniquely determined.
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– Aretino
Jan 12 at 11:16
add a comment |
0
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0
$begingroup$
I am trying to construct a hyperbola for a project I'm doing and I have the two axes of symmetry, the length of the semimajor axis and the eccentricity. Is it possible? If so, how?
geometry conic-sections
asked Jan 10 at 23:51
user140323user140323
1214
$endgroup$
I am trying to construct a hyperbola for a project I'm doing and I have the two axes of symmetry, the length of the semimajor axis and the eccentricity. Is it possible? If so, how?
geometry conic-sections
geometry conic-sections
asked Jan 10 at 23:51
user140323user140323
1214
asked Jan 10 at 23:51
user140323user140323
1214
edited Jan 11 at 2:15
user140323
asked Jan 10 at 23:51
user140323user140323
1214
asked Jan 10 at 23:51
user140323user140323
1214
asked Jan 10 at 23:51
user140323user140323
1214
1214
1
$begingroup$
Semi-major axis and eccentricity are enough to determine the shape and size of the hyperbola; if I understand "centerline" correctly as one of the lines of symmetry (specifically the one through the vertices?), then we have orientation and part of the location. However, the exact location of the center of the hyperbola remains undetermined.
$endgroup$
– Blue
Jan 11 at 0:01
$begingroup$
My bad, by centerline I mean the line of symmetry not passing through the vertices. And, now that I think of it, I have both lines given. I'll edit my question.
$endgroup$
– user140323
Jan 11 at 2:14
$begingroup$
If you know both axes of symmetry, semimajor axis and eccentricity, then you know the position of both vertices and can easily find the position of the foci. Hence the hyperbola is uniquely determined.
$endgroup$
– Aretino
Jan 12 at 11:16
add a comment |
1
$begingroup$
Semi-major axis and eccentricity are enough to determine the shape and size of the hyperbola; if I understand "centerline" correctly as one of the lines of symmetry (specifically the one through the vertices?), then we have orientation and part of the location. However, the exact location of the center of the hyperbola remains undetermined.
$endgroup$
– Blue
Jan 11 at 0:01
$begingroup$
My bad, by centerline I mean the line of symmetry not passing through the vertices. And, now that I think of it, I have both lines given. I'll edit my question.
$endgroup$
– user140323
Jan 11 at 2:14
$begingroup$
If you know both axes of symmetry, semimajor axis and eccentricity, then you know the position of both vertices and can easily find the position of the foci. Hence the hyperbola is uniquely determined.
$endgroup$
– Aretino
Jan 12 at 11:16
1
1
$begingroup$
Semi-major axis and eccentricity are enough to determine the shape and size of the hyperbola; if I understand "centerline" correctly as one of the lines of symmetry (specifically the one through the vertices?), then we have orientation and part of the location. However, the exact location of the center of the hyperbola remains undetermined.
$endgroup$
– Blue
Jan 11 at 0:01
$begingroup$
Semi-major axis and eccentricity are enough to determine the shape and size of the hyperbola; if I understand "centerline" correctly as one of the lines of symmetry (specifically the one through the vertices?), then we have orientation and part of the location. However, the exact location of the center of the hyperbola remains undetermined.
$endgroup$
– Blue
Jan 11 at 0:01
$begingroup$
My bad, by centerline I mean the line of symmetry not passing through the vertices. And, now that I think of it, I have both lines given. I'll edit my question.
$endgroup$
– user140323
Jan 11 at 2:14
$begingroup$
My bad, by centerline I mean the line of symmetry not passing through the vertices. And, now that I think of it, I have both lines given. I'll edit my question.
$endgroup$
– user140323
Jan 11 at 2:14
$begingroup$
If you know both axes of symmetry, semimajor axis and eccentricity, then you know the position of both vertices and can easily find the position of the foci. Hence the hyperbola is uniquely determined.
$endgroup$
– Aretino
Jan 12 at 11:16
$begingroup$
If you know both axes of symmetry, semimajor axis and eccentricity, then you know the position of both vertices and can easily find the position of the foci. Hence the hyperbola is uniquely determined.
$endgroup$
– Aretino
Jan 12 at 11:16
add a comment |
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1
$begingroup$
Semi-major axis and eccentricity are enough to determine the shape and size of the hyperbola; if I understand "centerline" correctly as one of the lines of symmetry (specifically the one through the vertices?), then we have orientation and part of the location. However, the exact location of the center of the hyperbola remains undetermined.
$endgroup$
– Blue
Jan 11 at 0:01
$begingroup$
My bad, by centerline I mean the line of symmetry not passing through the vertices. And, now that I think of it, I have both lines given. I'll edit my question.
$endgroup$
– user140323
Jan 11 at 2:14
$begingroup$
If you know both axes of symmetry, semimajor axis and eccentricity, then you know the position of both vertices and can easily find the position of the foci. Hence the hyperbola is uniquely determined.
$endgroup$
– Aretino
Jan 12 at 11:16