Confusion about polar equation of hyperbola
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Suppose I have a hyperbola $frac{x^2}{a^2}-frac{y^2}{b^2}=1$. I know that when choosing the right-hand side focus as the pole and the polar axis has the same direction as x-axis, the equation of the hyperbola in polar coordinates is $r=dfrac{p}{1-varepsiloncosvarphi}$, where $p=frac{b^2}{a}$, $varepsilon$ is the eccentricity of the hyperbola and $r,varphi$ are the polar coordinates of a point on hyperbola.
I understand that when using this equation, I only get the right-hand side branch of the hyperbola. However when I graph this for example in Geogebra, it gives me the whole hyperbola. What am I missing here?
geometry conic-sections polar-coordinates
asked Nov 16 at 5:12
user570271
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Suppose I have a hyperbola $frac{x^2}{a^2}-frac{y^2}{b^2}=1$. I know that when choosing the right-hand side focus as the pole and the polar axis has the same direction as x-axis, the equation of the hyperbola in polar coordinates is $r=dfrac{p}{1-varepsiloncosvarphi}$, where $p=frac{b^2}{a}$, $varepsilon$ is the eccentricity of the hyperbola and $r,varphi$ are the polar coordinates of a point on hyperbola.
I understand that when using this equation, I only get the right-hand side branch of the hyperbola. However when I graph this for example in Geogebra, it gives me the whole hyperbola. What am I missing here?
geometry conic-sections polar-coordinates
asked Nov 16 at 5:12
user570271
334
2
"I understand that when using this equation, I only get the right-hand side branch of the hyperbola." Don't understand that; it isn't so! :) In your example, consider $theta=0$, which gives $$a = frac{9/4}{1-frac54cdot 1}=-9$$ Negative values are plotted in the opposite direction, so $theta=0$ and $a=-9$ gives the vertex on the left branch of the hyperbola. Likewise, all negative $a$ values give points on the left branch. The transition from left to right (that is, negative to positive) happens when $1-epsilon costheta = 0$, which corresponds to the angle of the asymptote.
– Blue
Nov 16 at 12:09
Oh, I see it now, thank you so much! I have never used polar coordinates so it was super confusing to me. Can you please also explain why is there a formula with both plus and minus sign given in Wikipedia en.wikipedia.org/wiki/Hyperbola#Polar_coordinates? Does it have to do with the choice of direction of the polar axis?
– user570271
Nov 16 at 12:41
The $pm$ determines which focus appears at the pole. (In the parabola case (where $epsilon=1$), you can think of it as "which direction the parabola opens".) I personally favor having the left-hand focus there; that way, $theta=0$ corresponds to the closer branch. In any case, to help wrap your mind around what's happening, check some test points; in particular, $theta=0$ and $theta=180^circ$. (Again, in the parabola case, you'll know which way the curve opens, because one of those values gives a denominator of zero, so that the corresponding point is "at infinity".)
– Blue
Nov 16 at 12:50
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose I have a hyperbola $frac{x^2}{a^2}-frac{y^2}{b^2}=1$. I know that when choosing the right-hand side focus as the pole and the polar axis has the same direction as x-axis, the equation of the hyperbola in polar coordinates is $r=dfrac{p}{1-varepsiloncosvarphi}$, where $p=frac{b^2}{a}$, $varepsilon$ is the eccentricity of the hyperbola and $r,varphi$ are the polar coordinates of a point on hyperbola.
I understand that when using this equation, I only get the right-hand side branch of the hyperbola. However when I graph this for example in Geogebra, it gives me the whole hyperbola. What am I missing here?
geometry conic-sections polar-coordinates
asked Nov 16 at 5:12
user570271
334
Suppose I have a hyperbola $frac{x^2}{a^2}-frac{y^2}{b^2}=1$. I know that when choosing the right-hand side focus as the pole and the polar axis has the same direction as x-axis, the equation of the hyperbola in polar coordinates is $r=dfrac{p}{1-varepsiloncosvarphi}$, where $p=frac{b^2}{a}$, $varepsilon$ is the eccentricity of the hyperbola and $r,varphi$ are the polar coordinates of a point on hyperbola.
I understand that when using this equation, I only get the right-hand side branch of the hyperbola. However when I graph this for example in Geogebra, it gives me the whole hyperbola. What am I missing here?
geometry conic-sections polar-coordinates
geometry conic-sections polar-coordinates
asked Nov 16 at 5:12
user570271
334
asked Nov 16 at 5:12
user570271
334
edited Nov 16 at 5:40
asked Nov 16 at 5:12
user570271
334
asked Nov 16 at 5:12
user570271
334
asked Nov 16 at 5:12
user570271
334
334
2
"I understand that when using this equation, I only get the right-hand side branch of the hyperbola." Don't understand that; it isn't so! :) In your example, consider $theta=0$, which gives $$a = frac{9/4}{1-frac54cdot 1}=-9$$ Negative values are plotted in the opposite direction, so $theta=0$ and $a=-9$ gives the vertex on the left branch of the hyperbola. Likewise, all negative $a$ values give points on the left branch. The transition from left to right (that is, negative to positive) happens when $1-epsilon costheta = 0$, which corresponds to the angle of the asymptote.
– Blue
Nov 16 at 12:09
Oh, I see it now, thank you so much! I have never used polar coordinates so it was super confusing to me. Can you please also explain why is there a formula with both plus and minus sign given in Wikipedia en.wikipedia.org/wiki/Hyperbola#Polar_coordinates? Does it have to do with the choice of direction of the polar axis?
– user570271
Nov 16 at 12:41
The $pm$ determines which focus appears at the pole. (In the parabola case (where $epsilon=1$), you can think of it as "which direction the parabola opens".) I personally favor having the left-hand focus there; that way, $theta=0$ corresponds to the closer branch. In any case, to help wrap your mind around what's happening, check some test points; in particular, $theta=0$ and $theta=180^circ$. (Again, in the parabola case, you'll know which way the curve opens, because one of those values gives a denominator of zero, so that the corresponding point is "at infinity".)
– Blue
Nov 16 at 12:50
add a comment |
2
"I understand that when using this equation, I only get the right-hand side branch of the hyperbola." Don't understand that; it isn't so! :) In your example, consider $theta=0$, which gives $$a = frac{9/4}{1-frac54cdot 1}=-9$$ Negative values are plotted in the opposite direction, so $theta=0$ and $a=-9$ gives the vertex on the left branch of the hyperbola. Likewise, all negative $a$ values give points on the left branch. The transition from left to right (that is, negative to positive) happens when $1-epsilon costheta = 0$, which corresponds to the angle of the asymptote.
– Blue
Nov 16 at 12:09
Oh, I see it now, thank you so much! I have never used polar coordinates so it was super confusing to me. Can you please also explain why is there a formula with both plus and minus sign given in Wikipedia en.wikipedia.org/wiki/Hyperbola#Polar_coordinates? Does it have to do with the choice of direction of the polar axis?
– user570271
Nov 16 at 12:41
The $pm$ determines which focus appears at the pole. (In the parabola case (where $epsilon=1$), you can think of it as "which direction the parabola opens".) I personally favor having the left-hand focus there; that way, $theta=0$ corresponds to the closer branch. In any case, to help wrap your mind around what's happening, check some test points; in particular, $theta=0$ and $theta=180^circ$. (Again, in the parabola case, you'll know which way the curve opens, because one of those values gives a denominator of zero, so that the corresponding point is "at infinity".)
– Blue
Nov 16 at 12:50
2
2
"I understand that when using this equation, I only get the right-hand side branch of the hyperbola." Don't understand that; it isn't so! :) In your example, consider $theta=0$, which gives $$a = frac{9/4}{1-frac54cdot 1}=-9$$ Negative values are plotted in the opposite direction, so $theta=0$ and $a=-9$ gives the vertex on the left branch of the hyperbola. Likewise, all negative $a$ values give points on the left branch. The transition from left to right (that is, negative to positive) happens when $1-epsilon costheta = 0$, which corresponds to the angle of the asymptote.
– Blue
Nov 16 at 12:09
"I understand that when using this equation, I only get the right-hand side branch of the hyperbola." Don't understand that; it isn't so! :) In your example, consider $theta=0$, which gives $$a = frac{9/4}{1-frac54cdot 1}=-9$$ Negative values are plotted in the opposite direction, so $theta=0$ and $a=-9$ gives the vertex on the left branch of the hyperbola. Likewise, all negative $a$ values give points on the left branch. The transition from left to right (that is, negative to positive) happens when $1-epsilon costheta = 0$, which corresponds to the angle of the asymptote.
– Blue
Nov 16 at 12:09
Oh, I see it now, thank you so much! I have never used polar coordinates so it was super confusing to me. Can you please also explain why is there a formula with both plus and minus sign given in Wikipedia en.wikipedia.org/wiki/Hyperbola#Polar_coordinates? Does it have to do with the choice of direction of the polar axis?
– user570271
Nov 16 at 12:41
Oh, I see it now, thank you so much! I have never used polar coordinates so it was super confusing to me. Can you please also explain why is there a formula with both plus and minus sign given in Wikipedia en.wikipedia.org/wiki/Hyperbola#Polar_coordinates? Does it have to do with the choice of direction of the polar axis?
– user570271
Nov 16 at 12:41
The $pm$ determines which focus appears at the pole. (In the parabola case (where $epsilon=1$), you can think of it as "which direction the parabola opens".) I personally favor having the left-hand focus there; that way, $theta=0$ corresponds to the closer branch. In any case, to help wrap your mind around what's happening, check some test points; in particular, $theta=0$ and $theta=180^circ$. (Again, in the parabola case, you'll know which way the curve opens, because one of those values gives a denominator of zero, so that the corresponding point is "at infinity".)
– Blue
Nov 16 at 12:50
The $pm$ determines which focus appears at the pole. (In the parabola case (where $epsilon=1$), you can think of it as "which direction the parabola opens".) I personally favor having the left-hand focus there; that way, $theta=0$ corresponds to the closer branch. In any case, to help wrap your mind around what's happening, check some test points; in particular, $theta=0$ and $theta=180^circ$. (Again, in the parabola case, you'll know which way the curve opens, because one of those values gives a denominator of zero, so that the corresponding point is "at infinity".)
– Blue
Nov 16 at 12:50
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"I understand that when using this equation, I only get the right-hand side branch of the hyperbola." Don't understand that; it isn't so! :) In your example, consider $theta=0$, which gives $$a = frac{9/4}{1-frac54cdot 1}=-9$$ Negative values are plotted in the opposite direction, so $theta=0$ and $a=-9$ gives the vertex on the left branch of the hyperbola. Likewise, all negative $a$ values give points on the left branch. The transition from left to right (that is, negative to positive) happens when $1-epsilon costheta = 0$, which corresponds to the angle of the asymptote.
– Blue
Nov 16 at 12:09
Oh, I see it now, thank you so much! I have never used polar coordinates so it was super confusing to me. Can you please also explain why is there a formula with both plus and minus sign given in Wikipedia en.wikipedia.org/wiki/Hyperbola#Polar_coordinates? Does it have to do with the choice of direction of the polar axis?
– user570271
Nov 16 at 12:41
The $pm$ determines which focus appears at the pole. (In the parabola case (where $epsilon=1$), you can think of it as "which direction the parabola opens".) I personally favor having the left-hand focus there; that way, $theta=0$ corresponds to the closer branch. In any case, to help wrap your mind around what's happening, check some test points; in particular, $theta=0$ and $theta=180^circ$. (Again, in the parabola case, you'll know which way the curve opens, because one of those values gives a denominator of zero, so that the corresponding point is "at infinity".)
– Blue
Nov 16 at 12:50