Confusion about polar equation of hyperbola
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
add a comment |
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
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
2 days ago
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
2 days ago
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
2 days ago
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
2 days ago
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
2 days ago
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
2 days ago
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
2 days ago
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
2 days ago
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
2 days ago
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
2 days ago
"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
2 days ago
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
2 days ago
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
2 days ago
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
2 days ago
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
2 days ago
add a comment |
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3000749%2fconfusion-about-polar-equation-of-hyperbola%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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
2 days ago
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
2 days ago
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
2 days ago