Is there a parabola which is similar to a branch of hyperbola?
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Parabola and a branch of hyperbola, visually looks similar.
The only difference I find is that, when x tends to infinity, hyperbola approaches a straight line (asymptote). Whereas if I draw an arbitrary line the parabola will rush past that line and keep going away from it.
But still, is there any parabola which is exactly what a branch of hyperbola looks like?
asymptotics conic-sections coordinate-systems
asked Dec 6 '18 at 5:21
Loop BackLoop Back
27711
$endgroup$
add a comment |
$begingroup$
Parabola and a branch of hyperbola, visually looks similar.
The only difference I find is that, when x tends to infinity, hyperbola approaches a straight line (asymptote). Whereas if I draw an arbitrary line the parabola will rush past that line and keep going away from it.
But still, is there any parabola which is exactly what a branch of hyperbola looks like?
asymptotics conic-sections coordinate-systems
asked Dec 6 '18 at 5:21
Loop BackLoop Back
27711
$endgroup$
3
$begingroup$
Since all parabolas have the same shape, one parabola isn't going to look more hyperbolic than anothesr.
$endgroup$
– bof
Dec 6 '18 at 5:28
add a comment |
$begingroup$
Parabola and a branch of hyperbola, visually looks similar.
The only difference I find is that, when x tends to infinity, hyperbola approaches a straight line (asymptote). Whereas if I draw an arbitrary line the parabola will rush past that line and keep going away from it.
But still, is there any parabola which is exactly what a branch of hyperbola looks like?
asymptotics conic-sections coordinate-systems
asked Dec 6 '18 at 5:21
Loop BackLoop Back
27711
$endgroup$
Parabola and a branch of hyperbola, visually looks similar.
The only difference I find is that, when x tends to infinity, hyperbola approaches a straight line (asymptote). Whereas if I draw an arbitrary line the parabola will rush past that line and keep going away from it.
But still, is there any parabola which is exactly what a branch of hyperbola looks like?
asymptotics conic-sections coordinate-systems
asymptotics conic-sections coordinate-systems
asked Dec 6 '18 at 5:21
Loop BackLoop Back
27711
asked Dec 6 '18 at 5:21
Loop BackLoop Back
27711
edited Dec 6 '18 at 11:32
Loop Back
asked Dec 6 '18 at 5:21
Loop BackLoop Back
27711
asked Dec 6 '18 at 5:21
Loop BackLoop Back
27711
asked Dec 6 '18 at 5:21
Loop BackLoop Back
27711
27711
3
$begingroup$
Since all parabolas have the same shape, one parabola isn't going to look more hyperbolic than anothesr.
$endgroup$
– bof
Dec 6 '18 at 5:28
add a comment |
3
$begingroup$
Since all parabolas have the same shape, one parabola isn't going to look more hyperbolic than anothesr.
$endgroup$
– bof
Dec 6 '18 at 5:28
3
3
$begingroup$
Since all parabolas have the same shape, one parabola isn't going to look more hyperbolic than anothesr.
$endgroup$
– bof
Dec 6 '18 at 5:28
$begingroup$
Since all parabolas have the same shape, one parabola isn't going to look more hyperbolic than anothesr.
$endgroup$
– bof
Dec 6 '18 at 5:28
add a comment |
3 Answers
3
active
oldest
votes
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The only differences between two parabolas are location, orientation, and scaling factor. As noted in a comment, they all have the same shape.
Hyperbolas, however, come in many different shapes. Some are asymptotic to a pair of perpendicular lines. Others live inside a much larger or much smaller angle between their asymptotic lines.
Now consider a sequence of hyperbolas constructed as follows. We put one vertex of the hyperbola at a fixed point and move the other vertex away, allowing the angle between the asymptotic lines to approach zero as the other vertex goes off to infinity. If we cleverly balance the rates at which the angle gets smaller and the other vertex gets farther, the hyperbolas will approach the shape of a parabola.
So no, you cannot make a parabola look like a branch of a typical hyperbola.
But you can make a branch of a hyperbola look almost like a parabola.
The match will still not be exact. You might as well ask for a positive number that is exactly zero.
answered Dec 6 '18 at 5:38
David KDavid K
53.1k341115
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No parabola has an asymptote, while every branch of a hyperbola has two asymptotes. Therefore, there can never be a parabola that looks exactly like a branch of a hyperbola.
answered Dec 6 '18 at 5:35
Robert HowardRobert Howard
1,9161822
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add a comment |
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Actually parabola is a special case of hyperbola, where eccentricity tends to 1.
Consider this equation of parabola. $$(x-2)^2+y^2=1•dfrac{|x|^2}{1}$$ This is a parabola with focus at $(2,0)$ and equation of directrix $x=0$. This looks something like this.
Guess what this is
Looks similar to the above parabola, but it isn't. It's a hyperbola with same focus, same equation of directrix but with eccentricity 1.000001. Don't believe me see this zoomed out image.
How far the second focus has reached, near to $-1times 10^7$. As the eccentricity tends to 1, the second focus tends to $-infty$ and the hyperbola looks more like a parabola.
answered Dec 6 '18 at 10:08
Loop BackLoop Back
27711
$endgroup$
add a comment |
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3 Answers
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3 Answers
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active
oldest
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$begingroup$
The only differences between two parabolas are location, orientation, and scaling factor. As noted in a comment, they all have the same shape.
Hyperbolas, however, come in many different shapes. Some are asymptotic to a pair of perpendicular lines. Others live inside a much larger or much smaller angle between their asymptotic lines.
Now consider a sequence of hyperbolas constructed as follows. We put one vertex of the hyperbola at a fixed point and move the other vertex away, allowing the angle between the asymptotic lines to approach zero as the other vertex goes off to infinity. If we cleverly balance the rates at which the angle gets smaller and the other vertex gets farther, the hyperbolas will approach the shape of a parabola.
So no, you cannot make a parabola look like a branch of a typical hyperbola.
But you can make a branch of a hyperbola look almost like a parabola.
The match will still not be exact. You might as well ask for a positive number that is exactly zero.
answered Dec 6 '18 at 5:38
David KDavid K
53.1k341115
$endgroup$
add a comment |
$begingroup$
The only differences between two parabolas are location, orientation, and scaling factor. As noted in a comment, they all have the same shape.
Hyperbolas, however, come in many different shapes. Some are asymptotic to a pair of perpendicular lines. Others live inside a much larger or much smaller angle between their asymptotic lines.
Now consider a sequence of hyperbolas constructed as follows. We put one vertex of the hyperbola at a fixed point and move the other vertex away, allowing the angle between the asymptotic lines to approach zero as the other vertex goes off to infinity. If we cleverly balance the rates at which the angle gets smaller and the other vertex gets farther, the hyperbolas will approach the shape of a parabola.
So no, you cannot make a parabola look like a branch of a typical hyperbola.
But you can make a branch of a hyperbola look almost like a parabola.
The match will still not be exact. You might as well ask for a positive number that is exactly zero.
answered Dec 6 '18 at 5:38
David KDavid K
53.1k341115
$endgroup$
add a comment |
$begingroup$
The only differences between two parabolas are location, orientation, and scaling factor. As noted in a comment, they all have the same shape.
Hyperbolas, however, come in many different shapes. Some are asymptotic to a pair of perpendicular lines. Others live inside a much larger or much smaller angle between their asymptotic lines.
Now consider a sequence of hyperbolas constructed as follows. We put one vertex of the hyperbola at a fixed point and move the other vertex away, allowing the angle between the asymptotic lines to approach zero as the other vertex goes off to infinity. If we cleverly balance the rates at which the angle gets smaller and the other vertex gets farther, the hyperbolas will approach the shape of a parabola.
So no, you cannot make a parabola look like a branch of a typical hyperbola.
But you can make a branch of a hyperbola look almost like a parabola.
The match will still not be exact. You might as well ask for a positive number that is exactly zero.
answered Dec 6 '18 at 5:38
David KDavid K
53.1k341115
$endgroup$
The only differences between two parabolas are location, orientation, and scaling factor. As noted in a comment, they all have the same shape.
Hyperbolas, however, come in many different shapes. Some are asymptotic to a pair of perpendicular lines. Others live inside a much larger or much smaller angle between their asymptotic lines.
Now consider a sequence of hyperbolas constructed as follows. We put one vertex of the hyperbola at a fixed point and move the other vertex away, allowing the angle between the asymptotic lines to approach zero as the other vertex goes off to infinity. If we cleverly balance the rates at which the angle gets smaller and the other vertex gets farther, the hyperbolas will approach the shape of a parabola.
So no, you cannot make a parabola look like a branch of a typical hyperbola.
But you can make a branch of a hyperbola look almost like a parabola.
The match will still not be exact. You might as well ask for a positive number that is exactly zero.
answered Dec 6 '18 at 5:38
David KDavid K
53.1k341115
answered Dec 6 '18 at 5:38
David KDavid K
53.1k341115
answered Dec 6 '18 at 5:38
David KDavid K
53.1k341115
answered Dec 6 '18 at 5:38
David KDavid K
53.1k341115
53.1k341115
add a comment |
add a comment |
$begingroup$
No parabola has an asymptote, while every branch of a hyperbola has two asymptotes. Therefore, there can never be a parabola that looks exactly like a branch of a hyperbola.
answered Dec 6 '18 at 5:35
Robert HowardRobert Howard
1,9161822
$endgroup$
add a comment |
$begingroup$
No parabola has an asymptote, while every branch of a hyperbola has two asymptotes. Therefore, there can never be a parabola that looks exactly like a branch of a hyperbola.
answered Dec 6 '18 at 5:35
Robert HowardRobert Howard
1,9161822
$endgroup$
add a comment |
$begingroup$
No parabola has an asymptote, while every branch of a hyperbola has two asymptotes. Therefore, there can never be a parabola that looks exactly like a branch of a hyperbola.
answered Dec 6 '18 at 5:35
Robert HowardRobert Howard
1,9161822
$endgroup$
No parabola has an asymptote, while every branch of a hyperbola has two asymptotes. Therefore, there can never be a parabola that looks exactly like a branch of a hyperbola.
answered Dec 6 '18 at 5:35
Robert HowardRobert Howard
1,9161822
answered Dec 6 '18 at 5:35
Robert HowardRobert Howard
1,9161822
answered Dec 6 '18 at 5:35
Robert HowardRobert Howard
1,9161822
answered Dec 6 '18 at 5:35
Robert HowardRobert Howard
1,9161822
1,9161822
add a comment |
add a comment |
$begingroup$
Actually parabola is a special case of hyperbola, where eccentricity tends to 1.
Consider this equation of parabola. $$(x-2)^2+y^2=1•dfrac{|x|^2}{1}$$ This is a parabola with focus at $(2,0)$ and equation of directrix $x=0$. This looks something like this.
Guess what this is
Looks similar to the above parabola, but it isn't. It's a hyperbola with same focus, same equation of directrix but with eccentricity 1.000001. Don't believe me see this zoomed out image.
How far the second focus has reached, near to $-1times 10^7$. As the eccentricity tends to 1, the second focus tends to $-infty$ and the hyperbola looks more like a parabola.
answered Dec 6 '18 at 10:08
Loop BackLoop Back
27711
$endgroup$
add a comment |
$begingroup$
Actually parabola is a special case of hyperbola, where eccentricity tends to 1.
Consider this equation of parabola. $$(x-2)^2+y^2=1•dfrac{|x|^2}{1}$$ This is a parabola with focus at $(2,0)$ and equation of directrix $x=0$. This looks something like this.
Guess what this is
Looks similar to the above parabola, but it isn't. It's a hyperbola with same focus, same equation of directrix but with eccentricity 1.000001. Don't believe me see this zoomed out image.
How far the second focus has reached, near to $-1times 10^7$. As the eccentricity tends to 1, the second focus tends to $-infty$ and the hyperbola looks more like a parabola.
answered Dec 6 '18 at 10:08
Loop BackLoop Back
27711
$endgroup$
add a comment |
$begingroup$
Actually parabola is a special case of hyperbola, where eccentricity tends to 1.
Consider this equation of parabola. $$(x-2)^2+y^2=1•dfrac{|x|^2}{1}$$ This is a parabola with focus at $(2,0)$ and equation of directrix $x=0$. This looks something like this.
Guess what this is
Looks similar to the above parabola, but it isn't. It's a hyperbola with same focus, same equation of directrix but with eccentricity 1.000001. Don't believe me see this zoomed out image.
How far the second focus has reached, near to $-1times 10^7$. As the eccentricity tends to 1, the second focus tends to $-infty$ and the hyperbola looks more like a parabola.
answered Dec 6 '18 at 10:08
Loop BackLoop Back
27711
$endgroup$
Actually parabola is a special case of hyperbola, where eccentricity tends to 1.
Consider this equation of parabola. $$(x-2)^2+y^2=1•dfrac{|x|^2}{1}$$ This is a parabola with focus at $(2,0)$ and equation of directrix $x=0$. This looks something like this.
Guess what this is
Looks similar to the above parabola, but it isn't. It's a hyperbola with same focus, same equation of directrix but with eccentricity 1.000001. Don't believe me see this zoomed out image.
How far the second focus has reached, near to $-1times 10^7$. As the eccentricity tends to 1, the second focus tends to $-infty$ and the hyperbola looks more like a parabola.
answered Dec 6 '18 at 10:08
Loop BackLoop Back
27711
edited Dec 6 '18 at 11:33
answered Dec 6 '18 at 10:08
Loop BackLoop Back
27711
answered Dec 6 '18 at 10:08
Loop BackLoop Back
27711
answered Dec 6 '18 at 10:08
Loop BackLoop Back
27711
27711
add a comment |
add a comment |
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$begingroup$
Since all parabolas have the same shape, one parabola isn't going to look more hyperbolic than anothesr.
$endgroup$
– bof
Dec 6 '18 at 5:28