Mixing
Find the locus of the foot of perpendicular from the centre of the ellipse.
$begingroup$
Find the locus of the foot of perpendicular from the centre of the ellipse $${x^2over a^2} +{y^2over b^2} = 1$$
on the chord joining the points whose eccentric angles differ by $π/2.$
My approach is:
Consider two points $P$ and $Q$ such that $P(acostheta , bsintheta),; Q(asintheta, -bcostheta).;$
Using these two points I write the equation of the chord and then I use the formula of the foot of a perpendicular from $;(X_1,Y_1);$ to a given line, to find the locus.
But I can't find the locus because this approach is very difficult and takes a lot of time. So please solve this in a simpler way.
conic-sections
edited Jan 11 at 21:10
user376343
3,4033826
asked Jan 7 at 15:53
saket kumarsaket kumar
416
$endgroup$
add a comment |
$begingroup$
Find the locus of the foot of perpendicular from the centre of the ellipse $${x^2over a^2} +{y^2over b^2} = 1$$
on the chord joining the points whose eccentric angles differ by $π/2.$
My approach is:
Consider two points $P$ and $Q$ such that $P(acostheta , bsintheta),; Q(asintheta, -bcostheta).;$
Using these two points I write the equation of the chord and then I use the formula of the foot of a perpendicular from $;(X_1,Y_1);$ to a given line, to find the locus.
But I can't find the locus because this approach is very difficult and takes a lot of time. So please solve this in a simpler way.
conic-sections
edited Jan 11 at 21:10
user376343
3,4033826
asked Jan 7 at 15:53
saket kumarsaket kumar
416
$endgroup$
$begingroup$
What have you tried? Select two points $P_1$ and $P_2$ on the ellipse, what's the relationship between the coordinates if the eccentric angles difference is $pi/2$?
$endgroup$
– Andrei
Jan 7 at 16:41
$begingroup$
My approach is :- Taking two point P and Q, P(acos● , bsin●) and Q (asin●, -bcos●). Using these two point , I write the equation of chord and then I use the Formula of Foot of a perpendicular from (X1,Y1) to a given line, to find the locus. But I can't find the locus because this approach is very difficult and take a lot of time. So plz solve this que in a simple way.
$endgroup$
– saket kumar
Jan 7 at 17:20
1
$begingroup$
OP is correct and you are confusing him. Eccentric angle is exactly the argument in parametric equation.
$endgroup$
– Vasily Mitch
Jan 7 at 18:25
$begingroup$
@saketkumar: Please edit the body of your Question to include the approach given in your Comment. This will greatly improve the Question and put your observations in a clearer context.
$endgroup$
– hardmath
Jan 8 at 21:21
add a comment |
$begingroup$
Find the locus of the foot of perpendicular from the centre of the ellipse $${x^2over a^2} +{y^2over b^2} = 1$$
on the chord joining the points whose eccentric angles differ by $π/2.$
My approach is:
Consider two points $P$ and $Q$ such that $P(acostheta , bsintheta),; Q(asintheta, -bcostheta).;$
Using these two points I write the equation of the chord and then I use the formula of the foot of a perpendicular from $;(X_1,Y_1);$ to a given line, to find the locus.
But I can't find the locus because this approach is very difficult and takes a lot of time. So please solve this in a simpler way.
conic-sections
edited Jan 11 at 21:10
user376343
3,4033826
asked Jan 7 at 15:53
saket kumarsaket kumar
416
$endgroup$
Find the locus of the foot of perpendicular from the centre of the ellipse $${x^2over a^2} +{y^2over b^2} = 1$$
on the chord joining the points whose eccentric angles differ by $π/2.$
My approach is:
Consider two points $P$ and $Q$ such that $P(acostheta , bsintheta),; Q(asintheta, -bcostheta).;$
Using these two points I write the equation of the chord and then I use the formula of the foot of a perpendicular from $;(X_1,Y_1);$ to a given line, to find the locus.
But I can't find the locus because this approach is very difficult and takes a lot of time. So please solve this in a simpler way.
conic-sections
conic-sections
edited Jan 11 at 21:10
user376343
3,4033826
asked Jan 7 at 15:53
saket kumarsaket kumar
416
edited Jan 11 at 21:10
user376343
3,4033826
asked Jan 7 at 15:53
saket kumarsaket kumar
416
edited Jan 11 at 21:10
user376343
3,4033826
edited Jan 11 at 21:10
user376343
3,4033826
edited Jan 11 at 21:10
user376343
3,4033826
3,4033826
asked Jan 7 at 15:53
saket kumarsaket kumar
416
asked Jan 7 at 15:53
saket kumarsaket kumar
416
asked Jan 7 at 15:53
saket kumarsaket kumar
416
416
$begingroup$
What have you tried? Select two points $P_1$ and $P_2$ on the ellipse, what's the relationship between the coordinates if the eccentric angles difference is $pi/2$?
$endgroup$
– Andrei
Jan 7 at 16:41
$begingroup$
My approach is :- Taking two point P and Q, P(acos● , bsin●) and Q (asin●, -bcos●). Using these two point , I write the equation of chord and then I use the Formula of Foot of a perpendicular from (X1,Y1) to a given line, to find the locus. But I can't find the locus because this approach is very difficult and take a lot of time. So plz solve this que in a simple way.
$endgroup$
– saket kumar
Jan 7 at 17:20
1
$begingroup$
OP is correct and you are confusing him. Eccentric angle is exactly the argument in parametric equation.
$endgroup$
– Vasily Mitch
Jan 7 at 18:25
$begingroup$
@saketkumar: Please edit the body of your Question to include the approach given in your Comment. This will greatly improve the Question and put your observations in a clearer context.
$endgroup$
– hardmath
Jan 8 at 21:21
add a comment |
$begingroup$
What have you tried? Select two points $P_1$ and $P_2$ on the ellipse, what's the relationship between the coordinates if the eccentric angles difference is $pi/2$?
$endgroup$
– Andrei
Jan 7 at 16:41
$begingroup$
My approach is :- Taking two point P and Q, P(acos● , bsin●) and Q (asin●, -bcos●). Using these two point , I write the equation of chord and then I use the Formula of Foot of a perpendicular from (X1,Y1) to a given line, to find the locus. But I can't find the locus because this approach is very difficult and take a lot of time. So plz solve this que in a simple way.
$endgroup$
– saket kumar
Jan 7 at 17:20
1
$begingroup$
OP is correct and you are confusing him. Eccentric angle is exactly the argument in parametric equation.
$endgroup$
– Vasily Mitch
Jan 7 at 18:25
$begingroup$
@saketkumar: Please edit the body of your Question to include the approach given in your Comment. This will greatly improve the Question and put your observations in a clearer context.
$endgroup$
– hardmath
Jan 8 at 21:21
$begingroup$
What have you tried? Select two points $P_1$ and $P_2$ on the ellipse, what's the relationship between the coordinates if the eccentric angles difference is $pi/2$?
$endgroup$
– Andrei
Jan 7 at 16:41
$begingroup$
What have you tried? Select two points $P_1$ and $P_2$ on the ellipse, what's the relationship between the coordinates if the eccentric angles difference is $pi/2$?
$endgroup$
– Andrei
Jan 7 at 16:41
$begingroup$
My approach is :- Taking two point P and Q, P(acos● , bsin●) and Q (asin●, -bcos●). Using these two point , I write the equation of chord and then I use the Formula of Foot of a perpendicular from (X1,Y1) to a given line, to find the locus. But I can't find the locus because this approach is very difficult and take a lot of time. So plz solve this que in a simple way.
$endgroup$
– saket kumar
Jan 7 at 17:20
$begingroup$
My approach is :- Taking two point P and Q, P(acos● , bsin●) and Q (asin●, -bcos●). Using these two point , I write the equation of chord and then I use the Formula of Foot of a perpendicular from (X1,Y1) to a given line, to find the locus. But I can't find the locus because this approach is very difficult and take a lot of time. So plz solve this que in a simple way.
$endgroup$
– saket kumar
Jan 7 at 17:20
1
1
$begingroup$
OP is correct and you are confusing him. Eccentric angle is exactly the argument in parametric equation.
$endgroup$
– Vasily Mitch
Jan 7 at 18:25
$begingroup$
OP is correct and you are confusing him. Eccentric angle is exactly the argument in parametric equation.
$endgroup$
– Vasily Mitch
Jan 7 at 18:25
$begingroup$
@saketkumar: Please edit the body of your Question to include the approach given in your Comment. This will greatly improve the Question and put your observations in a clearer context.
$endgroup$
– hardmath
Jan 8 at 21:21
$begingroup$
@saketkumar: Please edit the body of your Question to include the approach given in your Comment. This will greatly improve the Question and put your observations in a clearer context.
$endgroup$
– hardmath
Jan 8 at 21:21
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let the centre of the ellipse be O and the foot be R. Also let point P is $(acostheta, bsintheta)$ then Q is $(acos(theta+frac{pi}{2}), bsin(theta+frac{pi}{2})$ or $(-asintheta, bcostheta)$ then PQ is
$$frac{y - bcostheta}{x + asintheta} = frac{b(sintheta - costheta)}{a(sintheta + costheta)} or frac{frac{y}{b} - costheta}{frac{x}{a} + sintheta} = frac{sintheta - costheta}{sintheta + costheta}$$
Also OR is
$$frac{y}{x} = - frac{a(sintheta + costheta)}{b(sintheta - costheta)} or frac{by}{ax} = -frac{sintheta + costheta}{sintheta - costheta}$$
From these two equations we have
$$x^2 + y^2 = - axsintheta + bycostheta$$
Divide both sides by $sqrt{a^2x^2 + b^2y^2}$ and let $tanphi = frac{by}{ax}$ then we have
$sin(theta - phi) = frac{x^2 + y^2}{sqrt{a^2x^2 + b^2y^2}}$ or $theta = sin^{-1}(frac{x^2 + y^2}{sqrt{a^2x^2 + b^2y^2}}) + phi$
Also from the second equation
$$frac{y}{x} = - frac{a}{b}frac{tantheta + 1}{-1 + tantheta}$$
Substitute $theta$ in the last equation using tangent compound angles and simplify, we have the following equation of the locus provided x and y not equal to zero.
$$a^2x^2 + b^2y^2 = 2(x^2 + y^2)^2$$
answered Jan 8 at 2:30
KY TangKY Tang
1495
$endgroup$
$begingroup$
Why you Divide both sides by √a^2x^2+b^2y^2 and also why you let tan●= by/ax
$endgroup$
– saket kumar
Jan 8 at 3:44
1
$begingroup$
In order to express $theta$ in terms of x and y I have to define $phi$ so that sin the sum of angles can be used. The same for tan. = by/ax. This is the only way to solve x and y from the 2 equations.
$endgroup$
– KY Tang
Jan 8 at 23:07
1
$begingroup$
The expression is obtained by the sqrt of the sum of squares of the coefficient of sin theta and cos theta in order to make the whole expression equal to sin theta + phi.
$endgroup$
– KY Tang
Jan 8 at 23:49
add a comment |
$begingroup$
The envelop of all your chords is the ellipse itself scaled down by $sqrt2$. It's easy to see by affine transforming ellipse into circle, drawing all the chords, and transforming it back.
All the chords are now tangents to smaller ellipse, so the locus of perpendicular foot has a special name pedal curve. Ellipse pedal curves have well-known formulae.
If you want to get to the formula yourself, you should really follow your approach. It's not that hard, just a little bit of algebra involved.
answered Jan 7 at 18:34
Vasily MitchVasily Mitch
1,99039
$endgroup$
$begingroup$
Can you solve this question according to my approach because I can't find my mistake. So plz solve for me, I know it is not a correct way to handle the mathematics question but I am very frustrated after solving same question 5 times and don't getting the correct answer.
$endgroup$
– saket kumar
Jan 7 at 18:48
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%2f3065152%2ffind-the-locus-of-the-foot-of-perpendicular-from-the-centre-of-the-ellipse%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let the centre of the ellipse be O and the foot be R. Also let point P is $(acostheta, bsintheta)$ then Q is $(acos(theta+frac{pi}{2}), bsin(theta+frac{pi}{2})$ or $(-asintheta, bcostheta)$ then PQ is
$$frac{y - bcostheta}{x + asintheta} = frac{b(sintheta - costheta)}{a(sintheta + costheta)} or frac{frac{y}{b} - costheta}{frac{x}{a} + sintheta} = frac{sintheta - costheta}{sintheta + costheta}$$
Also OR is
$$frac{y}{x} = - frac{a(sintheta + costheta)}{b(sintheta - costheta)} or frac{by}{ax} = -frac{sintheta + costheta}{sintheta - costheta}$$
From these two equations we have
$$x^2 + y^2 = - axsintheta + bycostheta$$
Divide both sides by $sqrt{a^2x^2 + b^2y^2}$ and let $tanphi = frac{by}{ax}$ then we have
$sin(theta - phi) = frac{x^2 + y^2}{sqrt{a^2x^2 + b^2y^2}}$ or $theta = sin^{-1}(frac{x^2 + y^2}{sqrt{a^2x^2 + b^2y^2}}) + phi$
Also from the second equation
$$frac{y}{x} = - frac{a}{b}frac{tantheta + 1}{-1 + tantheta}$$
Substitute $theta$ in the last equation using tangent compound angles and simplify, we have the following equation of the locus provided x and y not equal to zero.
$$a^2x^2 + b^2y^2 = 2(x^2 + y^2)^2$$
answered Jan 8 at 2:30
KY TangKY Tang
1495
$endgroup$
$begingroup$
Why you Divide both sides by √a^2x^2+b^2y^2 and also why you let tan●= by/ax
$endgroup$
– saket kumar
Jan 8 at 3:44
1
$begingroup$
In order to express $theta$ in terms of x and y I have to define $phi$ so that sin the sum of angles can be used. The same for tan. = by/ax. This is the only way to solve x and y from the 2 equations.
$endgroup$
– KY Tang
Jan 8 at 23:07
1
$begingroup$
The expression is obtained by the sqrt of the sum of squares of the coefficient of sin theta and cos theta in order to make the whole expression equal to sin theta + phi.
$endgroup$
– KY Tang
Jan 8 at 23:49
add a comment |
$begingroup$
Let the centre of the ellipse be O and the foot be R. Also let point P is $(acostheta, bsintheta)$ then Q is $(acos(theta+frac{pi}{2}), bsin(theta+frac{pi}{2})$ or $(-asintheta, bcostheta)$ then PQ is
$$frac{y - bcostheta}{x + asintheta} = frac{b(sintheta - costheta)}{a(sintheta + costheta)} or frac{frac{y}{b} - costheta}{frac{x}{a} + sintheta} = frac{sintheta - costheta}{sintheta + costheta}$$
Also OR is
$$frac{y}{x} = - frac{a(sintheta + costheta)}{b(sintheta - costheta)} or frac{by}{ax} = -frac{sintheta + costheta}{sintheta - costheta}$$
From these two equations we have
$$x^2 + y^2 = - axsintheta + bycostheta$$
Divide both sides by $sqrt{a^2x^2 + b^2y^2}$ and let $tanphi = frac{by}{ax}$ then we have
$sin(theta - phi) = frac{x^2 + y^2}{sqrt{a^2x^2 + b^2y^2}}$ or $theta = sin^{-1}(frac{x^2 + y^2}{sqrt{a^2x^2 + b^2y^2}}) + phi$
Also from the second equation
$$frac{y}{x} = - frac{a}{b}frac{tantheta + 1}{-1 + tantheta}$$
Substitute $theta$ in the last equation using tangent compound angles and simplify, we have the following equation of the locus provided x and y not equal to zero.
$$a^2x^2 + b^2y^2 = 2(x^2 + y^2)^2$$
answered Jan 8 at 2:30
KY TangKY Tang
1495
$endgroup$
$begingroup$
Why you Divide both sides by √a^2x^2+b^2y^2 and also why you let tan●= by/ax
$endgroup$
– saket kumar
Jan 8 at 3:44
1
$begingroup$
In order to express $theta$ in terms of x and y I have to define $phi$ so that sin the sum of angles can be used. The same for tan. = by/ax. This is the only way to solve x and y from the 2 equations.
$endgroup$
– KY Tang
Jan 8 at 23:07
1
$begingroup$
The expression is obtained by the sqrt of the sum of squares of the coefficient of sin theta and cos theta in order to make the whole expression equal to sin theta + phi.
$endgroup$
– KY Tang
Jan 8 at 23:49
add a comment |
$begingroup$
Let the centre of the ellipse be O and the foot be R. Also let point P is $(acostheta, bsintheta)$ then Q is $(acos(theta+frac{pi}{2}), bsin(theta+frac{pi}{2})$ or $(-asintheta, bcostheta)$ then PQ is
$$frac{y - bcostheta}{x + asintheta} = frac{b(sintheta - costheta)}{a(sintheta + costheta)} or frac{frac{y}{b} - costheta}{frac{x}{a} + sintheta} = frac{sintheta - costheta}{sintheta + costheta}$$
Also OR is
$$frac{y}{x} = - frac{a(sintheta + costheta)}{b(sintheta - costheta)} or frac{by}{ax} = -frac{sintheta + costheta}{sintheta - costheta}$$
From these two equations we have
$$x^2 + y^2 = - axsintheta + bycostheta$$
Divide both sides by $sqrt{a^2x^2 + b^2y^2}$ and let $tanphi = frac{by}{ax}$ then we have
$sin(theta - phi) = frac{x^2 + y^2}{sqrt{a^2x^2 + b^2y^2}}$ or $theta = sin^{-1}(frac{x^2 + y^2}{sqrt{a^2x^2 + b^2y^2}}) + phi$
Also from the second equation
$$frac{y}{x} = - frac{a}{b}frac{tantheta + 1}{-1 + tantheta}$$
Substitute $theta$ in the last equation using tangent compound angles and simplify, we have the following equation of the locus provided x and y not equal to zero.
$$a^2x^2 + b^2y^2 = 2(x^2 + y^2)^2$$
answered Jan 8 at 2:30
KY TangKY Tang
1495
$endgroup$
Let the centre of the ellipse be O and the foot be R. Also let point P is $(acostheta, bsintheta)$ then Q is $(acos(theta+frac{pi}{2}), bsin(theta+frac{pi}{2})$ or $(-asintheta, bcostheta)$ then PQ is
$$frac{y - bcostheta}{x + asintheta} = frac{b(sintheta - costheta)}{a(sintheta + costheta)} or frac{frac{y}{b} - costheta}{frac{x}{a} + sintheta} = frac{sintheta - costheta}{sintheta + costheta}$$
Also OR is
$$frac{y}{x} = - frac{a(sintheta + costheta)}{b(sintheta - costheta)} or frac{by}{ax} = -frac{sintheta + costheta}{sintheta - costheta}$$
From these two equations we have
$$x^2 + y^2 = - axsintheta + bycostheta$$
Divide both sides by $sqrt{a^2x^2 + b^2y^2}$ and let $tanphi = frac{by}{ax}$ then we have
$sin(theta - phi) = frac{x^2 + y^2}{sqrt{a^2x^2 + b^2y^2}}$ or $theta = sin^{-1}(frac{x^2 + y^2}{sqrt{a^2x^2 + b^2y^2}}) + phi$
Also from the second equation
$$frac{y}{x} = - frac{a}{b}frac{tantheta + 1}{-1 + tantheta}$$
Substitute $theta$ in the last equation using tangent compound angles and simplify, we have the following equation of the locus provided x and y not equal to zero.
$$a^2x^2 + b^2y^2 = 2(x^2 + y^2)^2$$
answered Jan 8 at 2:30
KY TangKY Tang
1495
edited Jan 8 at 3:55
answered Jan 8 at 2:30
KY TangKY Tang
1495
answered Jan 8 at 2:30
KY TangKY Tang
1495
answered Jan 8 at 2:30
KY TangKY Tang
1495
1495
$begingroup$
Why you Divide both sides by √a^2x^2+b^2y^2 and also why you let tan●= by/ax
$endgroup$
– saket kumar
Jan 8 at 3:44
1
$begingroup$
In order to express $theta$ in terms of x and y I have to define $phi$ so that sin the sum of angles can be used. The same for tan. = by/ax. This is the only way to solve x and y from the 2 equations.
$endgroup$
– KY Tang
Jan 8 at 23:07
1
$begingroup$
The expression is obtained by the sqrt of the sum of squares of the coefficient of sin theta and cos theta in order to make the whole expression equal to sin theta + phi.
$endgroup$
– KY Tang
Jan 8 at 23:49
add a comment |
$begingroup$
Why you Divide both sides by √a^2x^2+b^2y^2 and also why you let tan●= by/ax
$endgroup$
– saket kumar
Jan 8 at 3:44
1
$begingroup$
In order to express $theta$ in terms of x and y I have to define $phi$ so that sin the sum of angles can be used. The same for tan. = by/ax. This is the only way to solve x and y from the 2 equations.
$endgroup$
– KY Tang
Jan 8 at 23:07
1
$begingroup$
The expression is obtained by the sqrt of the sum of squares of the coefficient of sin theta and cos theta in order to make the whole expression equal to sin theta + phi.
$endgroup$
– KY Tang
Jan 8 at 23:49
$begingroup$
Why you Divide both sides by √a^2x^2+b^2y^2 and also why you let tan●= by/ax
$endgroup$
– saket kumar
Jan 8 at 3:44
$begingroup$
Why you Divide both sides by √a^2x^2+b^2y^2 and also why you let tan●= by/ax
$endgroup$
– saket kumar
Jan 8 at 3:44
1
1
$begingroup$
In order to express $theta$ in terms of x and y I have to define $phi$ so that sin the sum of angles can be used. The same for tan. = by/ax. This is the only way to solve x and y from the 2 equations.
$endgroup$
– KY Tang
Jan 8 at 23:07
$begingroup$
In order to express $theta$ in terms of x and y I have to define $phi$ so that sin the sum of angles can be used. The same for tan. = by/ax. This is the only way to solve x and y from the 2 equations.
$endgroup$
– KY Tang
Jan 8 at 23:07
1
1
$begingroup$
The expression is obtained by the sqrt of the sum of squares of the coefficient of sin theta and cos theta in order to make the whole expression equal to sin theta + phi.
$endgroup$
– KY Tang
Jan 8 at 23:49
$begingroup$
The expression is obtained by the sqrt of the sum of squares of the coefficient of sin theta and cos theta in order to make the whole expression equal to sin theta + phi.
$endgroup$
– KY Tang
Jan 8 at 23:49
add a comment |
$begingroup$
The envelop of all your chords is the ellipse itself scaled down by $sqrt2$. It's easy to see by affine transforming ellipse into circle, drawing all the chords, and transforming it back.
All the chords are now tangents to smaller ellipse, so the locus of perpendicular foot has a special name pedal curve. Ellipse pedal curves have well-known formulae.
If you want to get to the formula yourself, you should really follow your approach. It's not that hard, just a little bit of algebra involved.
answered Jan 7 at 18:34
Vasily MitchVasily Mitch
1,99039
$endgroup$
$begingroup$
Can you solve this question according to my approach because I can't find my mistake. So plz solve for me, I know it is not a correct way to handle the mathematics question but I am very frustrated after solving same question 5 times and don't getting the correct answer.
$endgroup$
– saket kumar
Jan 7 at 18:48
add a comment |
$begingroup$
The envelop of all your chords is the ellipse itself scaled down by $sqrt2$. It's easy to see by affine transforming ellipse into circle, drawing all the chords, and transforming it back.
All the chords are now tangents to smaller ellipse, so the locus of perpendicular foot has a special name pedal curve. Ellipse pedal curves have well-known formulae.
If you want to get to the formula yourself, you should really follow your approach. It's not that hard, just a little bit of algebra involved.
answered Jan 7 at 18:34
Vasily MitchVasily Mitch
1,99039
$endgroup$
$begingroup$
Can you solve this question according to my approach because I can't find my mistake. So plz solve for me, I know it is not a correct way to handle the mathematics question but I am very frustrated after solving same question 5 times and don't getting the correct answer.
$endgroup$
– saket kumar
Jan 7 at 18:48
add a comment |
$begingroup$
The envelop of all your chords is the ellipse itself scaled down by $sqrt2$. It's easy to see by affine transforming ellipse into circle, drawing all the chords, and transforming it back.
All the chords are now tangents to smaller ellipse, so the locus of perpendicular foot has a special name pedal curve. Ellipse pedal curves have well-known formulae.
If you want to get to the formula yourself, you should really follow your approach. It's not that hard, just a little bit of algebra involved.
answered Jan 7 at 18:34
Vasily MitchVasily Mitch
1,99039
$endgroup$
The envelop of all your chords is the ellipse itself scaled down by $sqrt2$. It's easy to see by affine transforming ellipse into circle, drawing all the chords, and transforming it back.
All the chords are now tangents to smaller ellipse, so the locus of perpendicular foot has a special name pedal curve. Ellipse pedal curves have well-known formulae.
If you want to get to the formula yourself, you should really follow your approach. It's not that hard, just a little bit of algebra involved.
answered Jan 7 at 18:34
Vasily MitchVasily Mitch
1,99039
answered Jan 7 at 18:34
Vasily MitchVasily Mitch
1,99039
answered Jan 7 at 18:34
Vasily MitchVasily Mitch
1,99039
answered Jan 7 at 18:34
Vasily MitchVasily Mitch
1,99039
1,99039
$begingroup$
Can you solve this question according to my approach because I can't find my mistake. So plz solve for me, I know it is not a correct way to handle the mathematics question but I am very frustrated after solving same question 5 times and don't getting the correct answer.
$endgroup$
– saket kumar
Jan 7 at 18:48
add a comment |
$begingroup$
Can you solve this question according to my approach because I can't find my mistake. So plz solve for me, I know it is not a correct way to handle the mathematics question but I am very frustrated after solving same question 5 times and don't getting the correct answer.
$endgroup$
– saket kumar
Jan 7 at 18:48
$begingroup$
Can you solve this question according to my approach because I can't find my mistake. So plz solve for me, I know it is not a correct way to handle the mathematics question but I am very frustrated after solving same question 5 times and don't getting the correct answer.
$endgroup$
– saket kumar
Jan 7 at 18:48
$begingroup$
Can you solve this question according to my approach because I can't find my mistake. So plz solve for me, I know it is not a correct way to handle the mathematics question but I am very frustrated after solving same question 5 times and don't getting the correct answer.
$endgroup$
– saket kumar
Jan 7 at 18:48
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%2f3065152%2ffind-the-locus-of-the-foot-of-perpendicular-from-the-centre-of-the-ellipse%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
$begingroup$
What have you tried? Select two points $P_1$ and $P_2$ on the ellipse, what's the relationship between the coordinates if the eccentric angles difference is $pi/2$?
$endgroup$
– Andrei
Jan 7 at 16:41
$begingroup$
My approach is :- Taking two point P and Q, P(acos● , bsin●) and Q (asin●, -bcos●). Using these two point , I write the equation of chord and then I use the Formula of Foot of a perpendicular from (X1,Y1) to a given line, to find the locus. But I can't find the locus because this approach is very difficult and take a lot of time. So plz solve this que in a simple way.
$endgroup$
– saket kumar
Jan 7 at 17:20
1
$begingroup$
OP is correct and you are confusing him. Eccentric angle is exactly the argument in parametric equation.
$endgroup$
– Vasily Mitch
Jan 7 at 18:25
$begingroup$
@saketkumar: Please edit the body of your Question to include the approach given in your Comment. This will greatly improve the Question and put your observations in a clearer context.
$endgroup$
– hardmath
Jan 8 at 21:21