Variable chord of hyperbola
$begingroup$
If a variable chord of hyperbola $x^2$$/4$ - $y^2$$/8$ $=$ $1$ subtends a right angle at the centre of hyperbola . If the chord touches a fixed concentric circle with hyperbola then we have to find the radius of the circle .
I thought of doing it by homozenizing , but not able to do how ?
conic-sections
asked Sep 7 '16 at 13:29
KoolmanKoolman
1,478821
$endgroup$
add a comment |
$begingroup$
If a variable chord of hyperbola $x^2$$/4$ - $y^2$$/8$ $=$ $1$ subtends a right angle at the centre of hyperbola . If the chord touches a fixed concentric circle with hyperbola then we have to find the radius of the circle .
I thought of doing it by homozenizing , but not able to do how ?
conic-sections
asked Sep 7 '16 at 13:29
KoolmanKoolman
1,478821
$endgroup$
add a comment |
$begingroup$
If a variable chord of hyperbola $x^2$$/4$ - $y^2$$/8$ $=$ $1$ subtends a right angle at the centre of hyperbola . If the chord touches a fixed concentric circle with hyperbola then we have to find the radius of the circle .
I thought of doing it by homozenizing , but not able to do how ?
conic-sections
asked Sep 7 '16 at 13:29
KoolmanKoolman
1,478821
$endgroup$
If a variable chord of hyperbola $x^2$$/4$ - $y^2$$/8$ $=$ $1$ subtends a right angle at the centre of hyperbola . If the chord touches a fixed concentric circle with hyperbola then we have to find the radius of the circle .
I thought of doing it by homozenizing , but not able to do how ?
conic-sections
conic-sections
asked Sep 7 '16 at 13:29
KoolmanKoolman
1,478821
asked Sep 7 '16 at 13:29
KoolmanKoolman
1,478821
edited Sep 7 '16 at 16:00
Koolman
asked Sep 7 '16 at 13:29
KoolmanKoolman
1,478821
asked Sep 7 '16 at 13:29
KoolmanKoolman
1,478821
asked Sep 7 '16 at 13:29
KoolmanKoolman
1,478821
1,478821
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Homogenise it and then
coeff.[x^2+y^2]=0 you will get constant term and since it is tangent to the circle x^2+y^2=r^2 then equate and u will get it.
answered Dec 7 '18 at 16:30
Anuj TanwarAnuj Tanwar
1
$endgroup$
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%2f1917905%2fvariable-chord-of-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
$begingroup$
Homogenise it and then
coeff.[x^2+y^2]=0 you will get constant term and since it is tangent to the circle x^2+y^2=r^2 then equate and u will get it.
answered Dec 7 '18 at 16:30
Anuj TanwarAnuj Tanwar
1
$endgroup$
add a comment |
$begingroup$
Homogenise it and then
coeff.[x^2+y^2]=0 you will get constant term and since it is tangent to the circle x^2+y^2=r^2 then equate and u will get it.
answered Dec 7 '18 at 16:30
Anuj TanwarAnuj Tanwar
1
$endgroup$
add a comment |
$begingroup$
Homogenise it and then
coeff.[x^2+y^2]=0 you will get constant term and since it is tangent to the circle x^2+y^2=r^2 then equate and u will get it.
answered Dec 7 '18 at 16:30
Anuj TanwarAnuj Tanwar
1
$endgroup$
Homogenise it and then
coeff.[x^2+y^2]=0 you will get constant term and since it is tangent to the circle x^2+y^2=r^2 then equate and u will get it.
answered Dec 7 '18 at 16:30
Anuj TanwarAnuj Tanwar
1
answered Dec 7 '18 at 16:30
Anuj TanwarAnuj Tanwar
1
answered Dec 7 '18 at 16:30
Anuj TanwarAnuj Tanwar
1
answered Dec 7 '18 at 16:30
Anuj TanwarAnuj Tanwar
1
1
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%2f1917905%2fvariable-chord-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
