Find the equation of the hyperbolic line through $A=(3,4)$, perpendicular to hyperbolic line $x^2+y^2=25$, $...
0
$begingroup$
I don't know how to solve this exercise in Hyperbolic Geometry.
Find the linear equation of hyperbolic line which passes through point $A=(3,4)$ and it is perpendicular to hyperbolic line with linear equation $x^2+y^2=25$ , $ y>0$.
hyperbolic-geometry
edited Dec 15 '18 at 17:05
Blue
48.5k870154
asked Dec 15 '18 at 16:28
Sotiris Z.Sotiris Z.
125
$endgroup$
add a comment |
0
$begingroup$
I don't know how to solve this exercise in Hyperbolic Geometry.
Find the linear equation of hyperbolic line which passes through point $A=(3,4)$ and it is perpendicular to hyperbolic line with linear equation $x^2+y^2=25$ , $ y>0$.
hyperbolic-geometry
edited Dec 15 '18 at 17:05
Blue
48.5k870154
asked Dec 15 '18 at 16:28
Sotiris Z.Sotiris Z.
125
$endgroup$
$begingroup$
To be clear: Is this an exercise in the Upper Half-Plane Model?
$endgroup$
– Blue
Dec 15 '18 at 17:07
$begingroup$
@Blue yes it is
$endgroup$
– Sotiris Z.
Dec 15 '18 at 17:10
$begingroup$
@Blue Can you help me?
$endgroup$
– Sotiris Z.
Dec 18 '18 at 20:30
$begingroup$
Since $A$ is on the given circle, the perpendicular "line" needed is the orthogonal circle through $A$. A neat thing about orthogonal circles: Where they meet, the tangent line to one circle passes through the center of the other. And, here, you know that the center of the target circle is on the $x$-axis. So, figure out where the tangent at $A$ meets the $x$-axis (recall that the tangent at a point is perpendicular to the radius to the point), and you'll have the center of the target circle; the distance from $A$ to that center gives you the radius; from there you can get the equation.
$endgroup$
– Blue
Dec 19 '18 at 2:53
add a comment |
0
0
0
$begingroup$
I don't know how to solve this exercise in Hyperbolic Geometry.
Find the linear equation of hyperbolic line which passes through point $A=(3,4)$ and it is perpendicular to hyperbolic line with linear equation $x^2+y^2=25$ , $ y>0$.
hyperbolic-geometry
edited Dec 15 '18 at 17:05
Blue
48.5k870154
asked Dec 15 '18 at 16:28
Sotiris Z.Sotiris Z.
125
$endgroup$
I don't know how to solve this exercise in Hyperbolic Geometry.
Find the linear equation of hyperbolic line which passes through point $A=(3,4)$ and it is perpendicular to hyperbolic line with linear equation $x^2+y^2=25$ , $ y>0$.
hyperbolic-geometry
hyperbolic-geometry
edited Dec 15 '18 at 17:05
Blue
48.5k870154
asked Dec 15 '18 at 16:28
Sotiris Z.Sotiris Z.
125
edited Dec 15 '18 at 17:05
Blue
48.5k870154
asked Dec 15 '18 at 16:28
Sotiris Z.Sotiris Z.
125
edited Dec 15 '18 at 17:05
Blue
48.5k870154
edited Dec 15 '18 at 17:05
Blue
48.5k870154
edited Dec 15 '18 at 17:05
Blue
48.5k870154
48.5k870154
asked Dec 15 '18 at 16:28
Sotiris Z.Sotiris Z.
125
asked Dec 15 '18 at 16:28
Sotiris Z.Sotiris Z.
125
asked Dec 15 '18 at 16:28
Sotiris Z.Sotiris Z.
125
125
$begingroup$
To be clear: Is this an exercise in the Upper Half-Plane Model?
$endgroup$
– Blue
Dec 15 '18 at 17:07
$begingroup$
@Blue yes it is
$endgroup$
– Sotiris Z.
Dec 15 '18 at 17:10
$begingroup$
@Blue Can you help me?
$endgroup$
– Sotiris Z.
Dec 18 '18 at 20:30
$begingroup$
Since $A$ is on the given circle, the perpendicular "line" needed is the orthogonal circle through $A$. A neat thing about orthogonal circles: Where they meet, the tangent line to one circle passes through the center of the other. And, here, you know that the center of the target circle is on the $x$-axis. So, figure out where the tangent at $A$ meets the $x$-axis (recall that the tangent at a point is perpendicular to the radius to the point), and you'll have the center of the target circle; the distance from $A$ to that center gives you the radius; from there you can get the equation.
$endgroup$
– Blue
Dec 19 '18 at 2:53
add a comment |
$begingroup$
To be clear: Is this an exercise in the Upper Half-Plane Model?
$endgroup$
– Blue
Dec 15 '18 at 17:07
$begingroup$
@Blue yes it is
$endgroup$
– Sotiris Z.
Dec 15 '18 at 17:10
$begingroup$
@Blue Can you help me?
$endgroup$
– Sotiris Z.
Dec 18 '18 at 20:30
$begingroup$
Since $A$ is on the given circle, the perpendicular "line" needed is the orthogonal circle through $A$. A neat thing about orthogonal circles: Where they meet, the tangent line to one circle passes through the center of the other. And, here, you know that the center of the target circle is on the $x$-axis. So, figure out where the tangent at $A$ meets the $x$-axis (recall that the tangent at a point is perpendicular to the radius to the point), and you'll have the center of the target circle; the distance from $A$ to that center gives you the radius; from there you can get the equation.
$endgroup$
– Blue
Dec 19 '18 at 2:53
$begingroup$
To be clear: Is this an exercise in the Upper Half-Plane Model?
$endgroup$
– Blue
Dec 15 '18 at 17:07
$begingroup$
To be clear: Is this an exercise in the Upper Half-Plane Model?
$endgroup$
– Blue
Dec 15 '18 at 17:07
$begingroup$
@Blue yes it is
$endgroup$
– Sotiris Z.
Dec 15 '18 at 17:10
$begingroup$
@Blue yes it is
$endgroup$
– Sotiris Z.
Dec 15 '18 at 17:10
$begingroup$
@Blue Can you help me?
$endgroup$
– Sotiris Z.
Dec 18 '18 at 20:30
$begingroup$
@Blue Can you help me?
$endgroup$
– Sotiris Z.
Dec 18 '18 at 20:30
$begingroup$
Since $A$ is on the given circle, the perpendicular "line" needed is the orthogonal circle through $A$. A neat thing about orthogonal circles: Where they meet, the tangent line to one circle passes through the center of the other. And, here, you know that the center of the target circle is on the $x$-axis. So, figure out where the tangent at $A$ meets the $x$-axis (recall that the tangent at a point is perpendicular to the radius to the point), and you'll have the center of the target circle; the distance from $A$ to that center gives you the radius; from there you can get the equation.
$endgroup$
– Blue
Dec 19 '18 at 2:53
$begingroup$
Since $A$ is on the given circle, the perpendicular "line" needed is the orthogonal circle through $A$. A neat thing about orthogonal circles: Where they meet, the tangent line to one circle passes through the center of the other. And, here, you know that the center of the target circle is on the $x$-axis. So, figure out where the tangent at $A$ meets the $x$-axis (recall that the tangent at a point is perpendicular to the radius to the point), and you'll have the center of the target circle; the distance from $A$ to that center gives you the radius; from there you can get the equation.
$endgroup$
– Blue
Dec 19 '18 at 2:53
add a comment |
0
active
oldest
votes
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
});
}
});
draft saved
draft discarded
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%2f3041678%2ffind-the-equation-of-the-hyperbolic-line-through-a-3-4-perpendicular-to-hyp%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f3041678%2ffind-the-equation-of-the-hyperbolic-line-through-a-3-4-perpendicular-to-hyp%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
$begingroup$
To be clear: Is this an exercise in the Upper Half-Plane Model?
$endgroup$
– Blue
Dec 15 '18 at 17:07
$begingroup$
@Blue yes it is
$endgroup$
– Sotiris Z.
Dec 15 '18 at 17:10
$begingroup$
@Blue Can you help me?
$endgroup$
– Sotiris Z.
Dec 18 '18 at 20:30
$begingroup$
Since $A$ is on the given circle, the perpendicular "line" needed is the orthogonal circle through $A$. A neat thing about orthogonal circles: Where they meet, the tangent line to one circle passes through the center of the other. And, here, you know that the center of the target circle is on the $x$-axis. So, figure out where the tangent at $A$ meets the $x$-axis (recall that the tangent at a point is perpendicular to the radius to the point), and you'll have the center of the target circle; the distance from $A$ to that center gives you the radius; from there you can get the equation.
$endgroup$
– Blue
Dec 19 '18 at 2:53