Mixing
Why do rotating lines intersect to form a circle or a hyperbola?
$begingroup$
It is possible to construct an ellipse or a hyperbola by tracing the intersections of offset lines that rotate at the same rate.
Firstly, here's a graph that shows the rotating lines, as well as the tracing of their intersections: https://www.desmos.com/calculator/bjpuuy1i4m. Play the slider $t$ to see the tracing.
Two lines rotate in opposite directions around a single point. At another point, two other lines do the same, but with a 90° offset to the first two lines.
These lines are represented by these four equations:
$y=tan(t)(x-u)$
$y=-tan(t)(x-u)$
$y=cot(t)(x+u)$
$y=-cot(t)(x+u)$
Where $t$ is the angle of rotation and $u$ is the x-intercept of the line.
For any value of $t$, all intersections between every line is a either a point on the circle $x^2+y^2=u^2$ or a point on the hyperbola $x^2-y^2=u^2$, as long as $u≠0$ .
Why do all of the intersections always fall on a circle or a hyperbola, and how might this construction connect to some of the other constructions of conics?
conic-sections
asked Jan 5 at 4:04
StymStym
132
$endgroup$
add a comment |
$begingroup$
It is possible to construct an ellipse or a hyperbola by tracing the intersections of offset lines that rotate at the same rate.
Firstly, here's a graph that shows the rotating lines, as well as the tracing of their intersections: https://www.desmos.com/calculator/bjpuuy1i4m. Play the slider $t$ to see the tracing.
Two lines rotate in opposite directions around a single point. At another point, two other lines do the same, but with a 90° offset to the first two lines.
These lines are represented by these four equations:
$y=tan(t)(x-u)$
$y=-tan(t)(x-u)$
$y=cot(t)(x+u)$
$y=-cot(t)(x+u)$
Where $t$ is the angle of rotation and $u$ is the x-intercept of the line.
For any value of $t$, all intersections between every line is a either a point on the circle $x^2+y^2=u^2$ or a point on the hyperbola $x^2-y^2=u^2$, as long as $u≠0$ .
Why do all of the intersections always fall on a circle or a hyperbola, and how might this construction connect to some of the other constructions of conics?
conic-sections
asked Jan 5 at 4:04
StymStym
132
$endgroup$
$begingroup$
Sounds like the circular directrix construction of an ellipse. See the last paragraph of this section of the Wikipedia ellipse article.
$endgroup$
– amd
Jan 5 at 4:57
$begingroup$
If you compute the coordinates of the intersection points of two lines, you end up with the parametric equation of either a circle or a hyperbola.
$endgroup$
– Aretino
Jan 6 at 10:40
add a comment |
$begingroup$
It is possible to construct an ellipse or a hyperbola by tracing the intersections of offset lines that rotate at the same rate.
Firstly, here's a graph that shows the rotating lines, as well as the tracing of their intersections: https://www.desmos.com/calculator/bjpuuy1i4m. Play the slider $t$ to see the tracing.
Two lines rotate in opposite directions around a single point. At another point, two other lines do the same, but with a 90° offset to the first two lines.
These lines are represented by these four equations:
$y=tan(t)(x-u)$
$y=-tan(t)(x-u)$
$y=cot(t)(x+u)$
$y=-cot(t)(x+u)$
Where $t$ is the angle of rotation and $u$ is the x-intercept of the line.
For any value of $t$, all intersections between every line is a either a point on the circle $x^2+y^2=u^2$ or a point on the hyperbola $x^2-y^2=u^2$, as long as $u≠0$ .
Why do all of the intersections always fall on a circle or a hyperbola, and how might this construction connect to some of the other constructions of conics?
conic-sections
asked Jan 5 at 4:04
StymStym
132
$endgroup$
It is possible to construct an ellipse or a hyperbola by tracing the intersections of offset lines that rotate at the same rate.
Firstly, here's a graph that shows the rotating lines, as well as the tracing of their intersections: https://www.desmos.com/calculator/bjpuuy1i4m. Play the slider $t$ to see the tracing.
Two lines rotate in opposite directions around a single point. At another point, two other lines do the same, but with a 90° offset to the first two lines.
These lines are represented by these four equations:
$y=tan(t)(x-u)$
$y=-tan(t)(x-u)$
$y=cot(t)(x+u)$
$y=-cot(t)(x+u)$
Where $t$ is the angle of rotation and $u$ is the x-intercept of the line.
For any value of $t$, all intersections between every line is a either a point on the circle $x^2+y^2=u^2$ or a point on the hyperbola $x^2-y^2=u^2$, as long as $u≠0$ .
Why do all of the intersections always fall on a circle or a hyperbola, and how might this construction connect to some of the other constructions of conics?
conic-sections
conic-sections
asked Jan 5 at 4:04
StymStym
132
asked Jan 5 at 4:04
StymStym
132
asked Jan 5 at 4:04
StymStym
132
asked Jan 5 at 4:04
StymStym
132
asked Jan 5 at 4:04
StymStym
132
132
$begingroup$
Sounds like the circular directrix construction of an ellipse. See the last paragraph of this section of the Wikipedia ellipse article.
$endgroup$
– amd
Jan 5 at 4:57
$begingroup$
If you compute the coordinates of the intersection points of two lines, you end up with the parametric equation of either a circle or a hyperbola.
$endgroup$
– Aretino
Jan 6 at 10:40
add a comment |
$begingroup$
Sounds like the circular directrix construction of an ellipse. See the last paragraph of this section of the Wikipedia ellipse article.
$endgroup$
– amd
Jan 5 at 4:57
$begingroup$
If you compute the coordinates of the intersection points of two lines, you end up with the parametric equation of either a circle or a hyperbola.
$endgroup$
– Aretino
Jan 6 at 10:40
$begingroup$
Sounds like the circular directrix construction of an ellipse. See the last paragraph of this section of the Wikipedia ellipse article.
$endgroup$
– amd
Jan 5 at 4:57
$begingroup$
Sounds like the circular directrix construction of an ellipse. See the last paragraph of this section of the Wikipedia ellipse article.
$endgroup$
– amd
Jan 5 at 4:57
$begingroup$
If you compute the coordinates of the intersection points of two lines, you end up with the parametric equation of either a circle or a hyperbola.
$endgroup$
– Aretino
Jan 6 at 10:40
$begingroup$
If you compute the coordinates of the intersection points of two lines, you end up with the parametric equation of either a circle or a hyperbola.
$endgroup$
– Aretino
Jan 6 at 10:40
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Consider a line $r$ of slope $m$ passing through $A=(-u,0)$: its equation is $y=m(x+u)$.
Line $s$, perpendicular to $r$ and passing through $B=(u,0)$,
has equation $y=(-1/m)(x-u)$.
Lines $r$ and $s$ intersect at $P=left({1-m^2over1+m^2}u,{2mover1+m^2}uright)$
and by Thale's theorem $P$ lies on the circle of diameter $AB$ (you can also check that $x_P^2+y_P^2=u^2$).
Line $t$, reflection of $s$ about line $x=u$, has equation $y=(1/m)(x-u)$ and intersects $r$ at $Q=left({1+m^2over1-m^2}u,{2mover1-m^2}uright)$. You can check that $x_Q^2-y_Q^2=u^2$, hence $Q$ lies on the equilateral hyperbola having $A$ and $B$ as vertices.
As $m$ varies in $(-infty,+infty)$, points $P$ and $Q$ thus describe a circle and a hyperbola, as explained above. In the equations you report, the convenient parameterization $m=tan t$ is used, but that is not necessary. In addition, a fourth line $r'$ is added: it does nothing new, but meets $s$ on the hyperbola and $t$ on the circle, as you can easily check.
answered Jan 7 at 12:46
AretinoAretino
22.9k21443
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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%2f3062388%2fwhy-do-rotating-lines-intersect-to-form-a-circle-or-a-hyperbola%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider a line $r$ of slope $m$ passing through $A=(-u,0)$: its equation is $y=m(x+u)$.
Line $s$, perpendicular to $r$ and passing through $B=(u,0)$,
has equation $y=(-1/m)(x-u)$.
Lines $r$ and $s$ intersect at $P=left({1-m^2over1+m^2}u,{2mover1+m^2}uright)$
and by Thale's theorem $P$ lies on the circle of diameter $AB$ (you can also check that $x_P^2+y_P^2=u^2$).
Line $t$, reflection of $s$ about line $x=u$, has equation $y=(1/m)(x-u)$ and intersects $r$ at $Q=left({1+m^2over1-m^2}u,{2mover1-m^2}uright)$. You can check that $x_Q^2-y_Q^2=u^2$, hence $Q$ lies on the equilateral hyperbola having $A$ and $B$ as vertices.
As $m$ varies in $(-infty,+infty)$, points $P$ and $Q$ thus describe a circle and a hyperbola, as explained above. In the equations you report, the convenient parameterization $m=tan t$ is used, but that is not necessary. In addition, a fourth line $r'$ is added: it does nothing new, but meets $s$ on the hyperbola and $t$ on the circle, as you can easily check.
answered Jan 7 at 12:46
AretinoAretino
22.9k21443
$endgroup$
add a comment |
$begingroup$
Consider a line $r$ of slope $m$ passing through $A=(-u,0)$: its equation is $y=m(x+u)$.
Line $s$, perpendicular to $r$ and passing through $B=(u,0)$,
has equation $y=(-1/m)(x-u)$.
Lines $r$ and $s$ intersect at $P=left({1-m^2over1+m^2}u,{2mover1+m^2}uright)$
and by Thale's theorem $P$ lies on the circle of diameter $AB$ (you can also check that $x_P^2+y_P^2=u^2$).
Line $t$, reflection of $s$ about line $x=u$, has equation $y=(1/m)(x-u)$ and intersects $r$ at $Q=left({1+m^2over1-m^2}u,{2mover1-m^2}uright)$. You can check that $x_Q^2-y_Q^2=u^2$, hence $Q$ lies on the equilateral hyperbola having $A$ and $B$ as vertices.
As $m$ varies in $(-infty,+infty)$, points $P$ and $Q$ thus describe a circle and a hyperbola, as explained above. In the equations you report, the convenient parameterization $m=tan t$ is used, but that is not necessary. In addition, a fourth line $r'$ is added: it does nothing new, but meets $s$ on the hyperbola and $t$ on the circle, as you can easily check.
answered Jan 7 at 12:46
AretinoAretino
22.9k21443
$endgroup$
add a comment |
$begingroup$
Consider a line $r$ of slope $m$ passing through $A=(-u,0)$: its equation is $y=m(x+u)$.
Line $s$, perpendicular to $r$ and passing through $B=(u,0)$,
has equation $y=(-1/m)(x-u)$.
Lines $r$ and $s$ intersect at $P=left({1-m^2over1+m^2}u,{2mover1+m^2}uright)$
and by Thale's theorem $P$ lies on the circle of diameter $AB$ (you can also check that $x_P^2+y_P^2=u^2$).
Line $t$, reflection of $s$ about line $x=u$, has equation $y=(1/m)(x-u)$ and intersects $r$ at $Q=left({1+m^2over1-m^2}u,{2mover1-m^2}uright)$. You can check that $x_Q^2-y_Q^2=u^2$, hence $Q$ lies on the equilateral hyperbola having $A$ and $B$ as vertices.
As $m$ varies in $(-infty,+infty)$, points $P$ and $Q$ thus describe a circle and a hyperbola, as explained above. In the equations you report, the convenient parameterization $m=tan t$ is used, but that is not necessary. In addition, a fourth line $r'$ is added: it does nothing new, but meets $s$ on the hyperbola and $t$ on the circle, as you can easily check.
answered Jan 7 at 12:46
AretinoAretino
22.9k21443
$endgroup$
Consider a line $r$ of slope $m$ passing through $A=(-u,0)$: its equation is $y=m(x+u)$.
Line $s$, perpendicular to $r$ and passing through $B=(u,0)$,
has equation $y=(-1/m)(x-u)$.
Lines $r$ and $s$ intersect at $P=left({1-m^2over1+m^2}u,{2mover1+m^2}uright)$
and by Thale's theorem $P$ lies on the circle of diameter $AB$ (you can also check that $x_P^2+y_P^2=u^2$).
Line $t$, reflection of $s$ about line $x=u$, has equation $y=(1/m)(x-u)$ and intersects $r$ at $Q=left({1+m^2over1-m^2}u,{2mover1-m^2}uright)$. You can check that $x_Q^2-y_Q^2=u^2$, hence $Q$ lies on the equilateral hyperbola having $A$ and $B$ as vertices.
As $m$ varies in $(-infty,+infty)$, points $P$ and $Q$ thus describe a circle and a hyperbola, as explained above. In the equations you report, the convenient parameterization $m=tan t$ is used, but that is not necessary. In addition, a fourth line $r'$ is added: it does nothing new, but meets $s$ on the hyperbola and $t$ on the circle, as you can easily check.
answered Jan 7 at 12:46
AretinoAretino
22.9k21443
edited Jan 7 at 14:14
answered Jan 7 at 12:46
AretinoAretino
22.9k21443
answered Jan 7 at 12:46
AretinoAretino
22.9k21443
answered Jan 7 at 12:46
AretinoAretino
22.9k21443
22.9k21443
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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%2f3062388%2fwhy-do-rotating-lines-intersect-to-form-a-circle-or-a-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
$begingroup$
Sounds like the circular directrix construction of an ellipse. See the last paragraph of this section of the Wikipedia ellipse article.
$endgroup$
– amd
Jan 5 at 4:57
$begingroup$
If you compute the coordinates of the intersection points of two lines, you end up with the parametric equation of either a circle or a hyperbola.
$endgroup$
– Aretino
Jan 6 at 10:40