Circle transformation
$begingroup$
Let $K$ be a circle $(x-1)^2+(y-4)^2=5$ and $P$ a line with the equation $y=-3x$.
We transform the circle $K$ with a transformation, defined with the next conditions:
points are tranformed, so that their distance from the line $P$ gets
multiplied by the factor $3$vector from the original point to the image is perpendicular to $P$
- the original points and their images are on the same side of $P$
Let $E$ be the new curve we get by tranformimg $K$ like this.
What is the equation of $E$?
circle conic-sections transformation
asked Dec 28 '18 at 14:12
nene123nene123
284
$endgroup$
add a comment |
$begingroup$
Let $K$ be a circle $(x-1)^2+(y-4)^2=5$ and $P$ a line with the equation $y=-3x$.
We transform the circle $K$ with a transformation, defined with the next conditions:
points are tranformed, so that their distance from the line $P$ gets
multiplied by the factor $3$vector from the original point to the image is perpendicular to $P$
- the original points and their images are on the same side of $P$
Let $E$ be the new curve we get by tranformimg $K$ like this.
What is the equation of $E$?
circle conic-sections transformation
asked Dec 28 '18 at 14:12
nene123nene123
284
$endgroup$
$begingroup$
The only idea I have is that an ellipse with major axis $x=3y-11$ forms and that $(-1.2,3.6)$ and $(-1,3)$ lie on that ellipse. Also, the minor axis is parallel to $y=-3x$. I am currently working on the solution.
$endgroup$
– Mohammad Zuhair Khan
Dec 28 '18 at 15:03
$begingroup$
Interesting question, but what work have you done on this yourself? Where are you getting stuck?
$endgroup$
– Théophile
Dec 28 '18 at 15:52
$begingroup$
A lot of effort later, I have found that $(5.2+4.5sqrt2,5.4+1.5sqrt2)$ and $(5.2-4.5sqrt2,5.4-1.5sqrt2)$ also lie on the ellipse, and they also lie on the major axis. The minor axis is $y=-3x+10.2$ and the centre of the ellipse (where the major axis and minor axis meet) is at $(5.2,5.4)$. Can you continue from here?
$endgroup$
– Mohammad Zuhair Khan
Dec 28 '18 at 15:57
$begingroup$
I don't even know how to begin writing the transformation...
$endgroup$
– nene123
Dec 28 '18 at 17:18
add a comment |
$begingroup$
Let $K$ be a circle $(x-1)^2+(y-4)^2=5$ and $P$ a line with the equation $y=-3x$.
We transform the circle $K$ with a transformation, defined with the next conditions:
points are tranformed, so that their distance from the line $P$ gets
multiplied by the factor $3$vector from the original point to the image is perpendicular to $P$
- the original points and their images are on the same side of $P$
Let $E$ be the new curve we get by tranformimg $K$ like this.
What is the equation of $E$?
circle conic-sections transformation
asked Dec 28 '18 at 14:12
nene123nene123
284
$endgroup$
Let $K$ be a circle $(x-1)^2+(y-4)^2=5$ and $P$ a line with the equation $y=-3x$.
We transform the circle $K$ with a transformation, defined with the next conditions:
points are tranformed, so that their distance from the line $P$ gets
multiplied by the factor $3$vector from the original point to the image is perpendicular to $P$
- the original points and their images are on the same side of $P$
Let $E$ be the new curve we get by tranformimg $K$ like this.
What is the equation of $E$?
circle conic-sections transformation
circle conic-sections transformation
asked Dec 28 '18 at 14:12
nene123nene123
284
asked Dec 28 '18 at 14:12
nene123nene123
284
asked Dec 28 '18 at 14:12
nene123nene123
284
asked Dec 28 '18 at 14:12
nene123nene123
284
asked Dec 28 '18 at 14:12
nene123nene123
284
284
$begingroup$
The only idea I have is that an ellipse with major axis $x=3y-11$ forms and that $(-1.2,3.6)$ and $(-1,3)$ lie on that ellipse. Also, the minor axis is parallel to $y=-3x$. I am currently working on the solution.
$endgroup$
– Mohammad Zuhair Khan
Dec 28 '18 at 15:03
$begingroup$
Interesting question, but what work have you done on this yourself? Where are you getting stuck?
$endgroup$
– Théophile
Dec 28 '18 at 15:52
$begingroup$
A lot of effort later, I have found that $(5.2+4.5sqrt2,5.4+1.5sqrt2)$ and $(5.2-4.5sqrt2,5.4-1.5sqrt2)$ also lie on the ellipse, and they also lie on the major axis. The minor axis is $y=-3x+10.2$ and the centre of the ellipse (where the major axis and minor axis meet) is at $(5.2,5.4)$. Can you continue from here?
$endgroup$
– Mohammad Zuhair Khan
Dec 28 '18 at 15:57
$begingroup$
I don't even know how to begin writing the transformation...
$endgroup$
– nene123
Dec 28 '18 at 17:18
add a comment |
$begingroup$
The only idea I have is that an ellipse with major axis $x=3y-11$ forms and that $(-1.2,3.6)$ and $(-1,3)$ lie on that ellipse. Also, the minor axis is parallel to $y=-3x$. I am currently working on the solution.
$endgroup$
– Mohammad Zuhair Khan
Dec 28 '18 at 15:03
$begingroup$
Interesting question, but what work have you done on this yourself? Where are you getting stuck?
$endgroup$
– Théophile
Dec 28 '18 at 15:52
$begingroup$
A lot of effort later, I have found that $(5.2+4.5sqrt2,5.4+1.5sqrt2)$ and $(5.2-4.5sqrt2,5.4-1.5sqrt2)$ also lie on the ellipse, and they also lie on the major axis. The minor axis is $y=-3x+10.2$ and the centre of the ellipse (where the major axis and minor axis meet) is at $(5.2,5.4)$. Can you continue from here?
$endgroup$
– Mohammad Zuhair Khan
Dec 28 '18 at 15:57
$begingroup$
I don't even know how to begin writing the transformation...
$endgroup$
– nene123
Dec 28 '18 at 17:18
$begingroup$
The only idea I have is that an ellipse with major axis $x=3y-11$ forms and that $(-1.2,3.6)$ and $(-1,3)$ lie on that ellipse. Also, the minor axis is parallel to $y=-3x$. I am currently working on the solution.
$endgroup$
– Mohammad Zuhair Khan
Dec 28 '18 at 15:03
$begingroup$
The only idea I have is that an ellipse with major axis $x=3y-11$ forms and that $(-1.2,3.6)$ and $(-1,3)$ lie on that ellipse. Also, the minor axis is parallel to $y=-3x$. I am currently working on the solution.
$endgroup$
– Mohammad Zuhair Khan
Dec 28 '18 at 15:03
$begingroup$
Interesting question, but what work have you done on this yourself? Where are you getting stuck?
$endgroup$
– Théophile
Dec 28 '18 at 15:52
$begingroup$
Interesting question, but what work have you done on this yourself? Where are you getting stuck?
$endgroup$
– Théophile
Dec 28 '18 at 15:52
$begingroup$
A lot of effort later, I have found that $(5.2+4.5sqrt2,5.4+1.5sqrt2)$ and $(5.2-4.5sqrt2,5.4-1.5sqrt2)$ also lie on the ellipse, and they also lie on the major axis. The minor axis is $y=-3x+10.2$ and the centre of the ellipse (where the major axis and minor axis meet) is at $(5.2,5.4)$. Can you continue from here?
$endgroup$
– Mohammad Zuhair Khan
Dec 28 '18 at 15:57
$begingroup$
A lot of effort later, I have found that $(5.2+4.5sqrt2,5.4+1.5sqrt2)$ and $(5.2-4.5sqrt2,5.4-1.5sqrt2)$ also lie on the ellipse, and they also lie on the major axis. The minor axis is $y=-3x+10.2$ and the centre of the ellipse (where the major axis and minor axis meet) is at $(5.2,5.4)$. Can you continue from here?
$endgroup$
– Mohammad Zuhair Khan
Dec 28 '18 at 15:57
$begingroup$
I don't even know how to begin writing the transformation...
$endgroup$
– nene123
Dec 28 '18 at 17:18
$begingroup$
I don't even know how to begin writing the transformation...
$endgroup$
– nene123
Dec 28 '18 at 17:18
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The transformation is a dilation perpendicular to the given line. There are several ways to come up with a formula for it. For example, you could try rotating the coordinate system so that the line becomes the $x'$-axis. The dilation is then just a matter of multiplying the $y'$-coordinate by $3$, after which you rotate back the other way. Once you have the transformation formula, invert it and substitute into the circle equation.
You will find that the result is an ellipse. Indeed, the image of a circle under any affine transformation is some sort of ellipse. Knowing this, you can take a somewhat simpler approach. The circle is being stretched in a direction perpendicular to the given line. From this you can deduce both the principal axis directions and semiaxis lengths of the resulting ellipse. Since it’s not centered at the origin, its center will also move, so work out where the transformation moves this point and you should be able to derive an equation for the resulting ellipse without too much more work.
answered Dec 28 '18 at 19:49
amdamd
30.4k21050
$endgroup$
$begingroup$
is it the $y'-$coordinate that is multiplied by $3$?
$endgroup$
– user376343
Dec 31 '18 at 20:58
1
$begingroup$
@user376343 Yes. Good catch.
$endgroup$
– amd
Jan 5 at 19:41
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%2f3054925%2fcircle-transformation%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
$begingroup$
The transformation is a dilation perpendicular to the given line. There are several ways to come up with a formula for it. For example, you could try rotating the coordinate system so that the line becomes the $x'$-axis. The dilation is then just a matter of multiplying the $y'$-coordinate by $3$, after which you rotate back the other way. Once you have the transformation formula, invert it and substitute into the circle equation.
You will find that the result is an ellipse. Indeed, the image of a circle under any affine transformation is some sort of ellipse. Knowing this, you can take a somewhat simpler approach. The circle is being stretched in a direction perpendicular to the given line. From this you can deduce both the principal axis directions and semiaxis lengths of the resulting ellipse. Since it’s not centered at the origin, its center will also move, so work out where the transformation moves this point and you should be able to derive an equation for the resulting ellipse without too much more work.
answered Dec 28 '18 at 19:49
amdamd
30.4k21050
$endgroup$
$begingroup$
is it the $y'-$coordinate that is multiplied by $3$?
$endgroup$
– user376343
Dec 31 '18 at 20:58
1
$begingroup$
@user376343 Yes. Good catch.
$endgroup$
– amd
Jan 5 at 19:41
add a comment |
$begingroup$
The transformation is a dilation perpendicular to the given line. There are several ways to come up with a formula for it. For example, you could try rotating the coordinate system so that the line becomes the $x'$-axis. The dilation is then just a matter of multiplying the $y'$-coordinate by $3$, after which you rotate back the other way. Once you have the transformation formula, invert it and substitute into the circle equation.
You will find that the result is an ellipse. Indeed, the image of a circle under any affine transformation is some sort of ellipse. Knowing this, you can take a somewhat simpler approach. The circle is being stretched in a direction perpendicular to the given line. From this you can deduce both the principal axis directions and semiaxis lengths of the resulting ellipse. Since it’s not centered at the origin, its center will also move, so work out where the transformation moves this point and you should be able to derive an equation for the resulting ellipse without too much more work.
answered Dec 28 '18 at 19:49
amdamd
30.4k21050
$endgroup$
$begingroup$
is it the $y'-$coordinate that is multiplied by $3$?
$endgroup$
– user376343
Dec 31 '18 at 20:58
1
$begingroup$
@user376343 Yes. Good catch.
$endgroup$
– amd
Jan 5 at 19:41
add a comment |
$begingroup$
The transformation is a dilation perpendicular to the given line. There are several ways to come up with a formula for it. For example, you could try rotating the coordinate system so that the line becomes the $x'$-axis. The dilation is then just a matter of multiplying the $y'$-coordinate by $3$, after which you rotate back the other way. Once you have the transformation formula, invert it and substitute into the circle equation.
You will find that the result is an ellipse. Indeed, the image of a circle under any affine transformation is some sort of ellipse. Knowing this, you can take a somewhat simpler approach. The circle is being stretched in a direction perpendicular to the given line. From this you can deduce both the principal axis directions and semiaxis lengths of the resulting ellipse. Since it’s not centered at the origin, its center will also move, so work out where the transformation moves this point and you should be able to derive an equation for the resulting ellipse without too much more work.
answered Dec 28 '18 at 19:49
amdamd
30.4k21050
$endgroup$
The transformation is a dilation perpendicular to the given line. There are several ways to come up with a formula for it. For example, you could try rotating the coordinate system so that the line becomes the $x'$-axis. The dilation is then just a matter of multiplying the $y'$-coordinate by $3$, after which you rotate back the other way. Once you have the transformation formula, invert it and substitute into the circle equation.
You will find that the result is an ellipse. Indeed, the image of a circle under any affine transformation is some sort of ellipse. Knowing this, you can take a somewhat simpler approach. The circle is being stretched in a direction perpendicular to the given line. From this you can deduce both the principal axis directions and semiaxis lengths of the resulting ellipse. Since it’s not centered at the origin, its center will also move, so work out where the transformation moves this point and you should be able to derive an equation for the resulting ellipse without too much more work.
answered Dec 28 '18 at 19:49
amdamd
30.4k21050
edited Jan 5 at 19:41
answered Dec 28 '18 at 19:49
amdamd
30.4k21050
answered Dec 28 '18 at 19:49
amdamd
30.4k21050
answered Dec 28 '18 at 19:49
amdamd
30.4k21050
30.4k21050
$begingroup$
is it the $y'-$coordinate that is multiplied by $3$?
$endgroup$
– user376343
Dec 31 '18 at 20:58
1
$begingroup$
@user376343 Yes. Good catch.
$endgroup$
– amd
Jan 5 at 19:41
add a comment |
$begingroup$
is it the $y'-$coordinate that is multiplied by $3$?
$endgroup$
– user376343
Dec 31 '18 at 20:58
1
$begingroup$
@user376343 Yes. Good catch.
$endgroup$
– amd
Jan 5 at 19:41
$begingroup$
is it the $y'-$coordinate that is multiplied by $3$?
$endgroup$
– user376343
Dec 31 '18 at 20:58
$begingroup$
is it the $y'-$coordinate that is multiplied by $3$?
$endgroup$
– user376343
Dec 31 '18 at 20:58
1
1
$begingroup$
@user376343 Yes. Good catch.
$endgroup$
– amd
Jan 5 at 19:41
$begingroup$
@user376343 Yes. Good catch.
$endgroup$
– amd
Jan 5 at 19:41
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%2f3054925%2fcircle-transformation%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
$begingroup$
The only idea I have is that an ellipse with major axis $x=3y-11$ forms and that $(-1.2,3.6)$ and $(-1,3)$ lie on that ellipse. Also, the minor axis is parallel to $y=-3x$. I am currently working on the solution.
$endgroup$
– Mohammad Zuhair Khan
Dec 28 '18 at 15:03
$begingroup$
Interesting question, but what work have you done on this yourself? Where are you getting stuck?
$endgroup$
– Théophile
Dec 28 '18 at 15:52
$begingroup$
A lot of effort later, I have found that $(5.2+4.5sqrt2,5.4+1.5sqrt2)$ and $(5.2-4.5sqrt2,5.4-1.5sqrt2)$ also lie on the ellipse, and they also lie on the major axis. The minor axis is $y=-3x+10.2$ and the centre of the ellipse (where the major axis and minor axis meet) is at $(5.2,5.4)$. Can you continue from here?
$endgroup$
– Mohammad Zuhair Khan
Dec 28 '18 at 15:57
$begingroup$
I don't even know how to begin writing the transformation...
$endgroup$
– nene123
Dec 28 '18 at 17:18