Mixing
How to solve this ellipse problem?
$begingroup$
Assume that ellipse $C:frac{x^2}{a^2}+frac{y^2}{b^2}=1(a>b>0)$, its left focus is $F$, the line $l$ passing $F$ intersects the ellipse $C$ at $A$ and $B$. The angle of inclination of $l$ is $frac{pi}{3}$, and $overrightarrow{AF}=2overrightarrow{FB}$.
If $|AB|=frac{15}{4}$, how to solve the equation of $C$?
geometry analytic-geometry
edited Feb 1 at 9:42
KReiser
10.1k21435
asked Feb 1 at 8:09
ItoMakotoItoMakoto
135
$endgroup$
add a comment |
$begingroup$
Assume that ellipse $C:frac{x^2}{a^2}+frac{y^2}{b^2}=1(a>b>0)$, its left focus is $F$, the line $l$ passing $F$ intersects the ellipse $C$ at $A$ and $B$. The angle of inclination of $l$ is $frac{pi}{3}$, and $overrightarrow{AF}=2overrightarrow{FB}$.
If $|AB|=frac{15}{4}$, how to solve the equation of $C$?
geometry analytic-geometry
edited Feb 1 at 9:42
KReiser
10.1k21435
asked Feb 1 at 8:09
ItoMakotoItoMakoto
135
$endgroup$
add a comment |
$begingroup$
Assume that ellipse $C:frac{x^2}{a^2}+frac{y^2}{b^2}=1(a>b>0)$, its left focus is $F$, the line $l$ passing $F$ intersects the ellipse $C$ at $A$ and $B$. The angle of inclination of $l$ is $frac{pi}{3}$, and $overrightarrow{AF}=2overrightarrow{FB}$.
If $|AB|=frac{15}{4}$, how to solve the equation of $C$?
geometry analytic-geometry
edited Feb 1 at 9:42
KReiser
10.1k21435
asked Feb 1 at 8:09
ItoMakotoItoMakoto
135
$endgroup$
Assume that ellipse $C:frac{x^2}{a^2}+frac{y^2}{b^2}=1(a>b>0)$, its left focus is $F$, the line $l$ passing $F$ intersects the ellipse $C$ at $A$ and $B$. The angle of inclination of $l$ is $frac{pi}{3}$, and $overrightarrow{AF}=2overrightarrow{FB}$.
If $|AB|=frac{15}{4}$, how to solve the equation of $C$?
geometry analytic-geometry
geometry analytic-geometry
edited Feb 1 at 9:42
KReiser
10.1k21435
asked Feb 1 at 8:09
ItoMakotoItoMakoto
135
edited Feb 1 at 9:42
KReiser
10.1k21435
asked Feb 1 at 8:09
ItoMakotoItoMakoto
135
edited Feb 1 at 9:42
KReiser
10.1k21435
edited Feb 1 at 9:42
KReiser
10.1k21435
edited Feb 1 at 9:42
KReiser
10.1k21435
10.1k21435
asked Feb 1 at 8:09
ItoMakotoItoMakoto
135
asked Feb 1 at 8:09
ItoMakotoItoMakoto
135
asked Feb 1 at 8:09
ItoMakotoItoMakoto
135
135
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
By the assumption we have: $AF=frac{10}{4}$ and $BF=frac{5}{4}$.
We have that left focus is $F=(-c,0)$ where $c^2=a^2-b^2$. Consider the left directrix $d: x=-frac{a}{e}$, where $e= frac{c}{a}$ is the eccentricity, and drop perpendiculars $A',F',B'$ from $A,F,B$ to $d$. The quadrilateral $AA'B'B$ is a trapezoid with the angle $angle BAA'=60^circ$, right angles at vertices $A'$ and $B'$, and basis $AA'=frac{1}{e}AF=frac{10}{4e}$ and $BB'=frac{1}{e}BF= frac{5}{4e}$. Then we easily see (by dropping a perpendicular from $B$ to $AA'$) that $AA'-BB'=ABcos 60^circ$, so $frac{5}{4e}=frac{15}{4}cdot frac{1}{2}$, wherefrom $e=frac{2}{3}$. Therefore, $AA'=frac{15}{4}$ and $BB'=frac{15}{8}$.
Since parallel line $FF'$ to the basis of the trapezoid divides $AB$ in the ratio $2:1$ we have: $FF'=frac{1}{3}(AA'+2BB')$, i.e. $frac{a}{e}-c= frac{1}{3}cdotfrac{30}{4}= frac{10}{4}$, i.e. $frac{a}{e}-ae=a(frac{3}{2}-frac{2}{3})=frac{10}{4}$, wherefrom $a=3$. Now, $c=ae=2$, and $b=sqrt{a^2-c^2}=sqrt{5}$.
answered Feb 1 at 14:07
SMMSMM
3,098512
$endgroup$
$begingroup$
Interesting approach. A little figure could help to follow... I am going to write down a different solution, with the same results !
$endgroup$
– Jean Marie
Feb 1 at 14:24
$begingroup$
Such an interesting solution! Thank you!
$endgroup$
– ItoMakoto
Feb 1 at 14:32
$begingroup$
@SMM Thanks to you, I have been able to use as well the directrix for a (second) solution that is based on different results from yours (for example, I dont use the eccentricity). Your opinion ?
$endgroup$
– Jean Marie
Feb 2 at 11:48
$begingroup$
@JeanMarie Very nice! It seems that qadrics have immense number of very nice properties, and most of them are not commonly known.
$endgroup$
– SMM
Feb 2 at 16:05
$begingroup$
@SMM Yes not known because no longer taught for now half a century. In the first half of the XXth century, in my country, France, they were at the center of the teaching in the final years of high school and at the collegial level. Maybe it was exaggerate ; but now on, the pendulum has gone in a completely reverse sense... although the basic properties of these curves are essential to know, especialy in Physics.
$endgroup$
– Jean Marie
Feb 2 at 16:14
add a comment |
$begingroup$
Fig. 1 :This figure illustrates the 2 answers I give as well as the very interesting answer by @SMM. On the left is figured the directrix line associated with focus $F$. See comments at the end of the first solution.
First method
Let us use the following parameterization of the ellipse :
$$begin{cases}x=a cos(t)\y=b sin(t)end{cases}tag{1}$$
Let
$$begin{cases}t_1, c_1=cos(t_1), s_1=sin(t_1) text{correspond to point } A \t_2, c_2=cos(t_2), s_2=sin(t_2) text{correspond to point } Bend{cases}$$
The following system of 6 equations with 6 unknowns takes into account the different constraints of the problem :
$$begin{cases}frac{b(s_1-s_2)}{a(c_1-c_2)}&=&sqrt{3}& (a)\
a^2(c_2-c_1)^2+b^2(s_2-s_1)^2&=&(tfrac{15}{4})^2& (b)\
afrac{c_1+2c_2}{3}&=&-sqrt{a^2-b^2}& (c)\
2s_2+s_1&=&0& (d)\
s_1^2+c_1^2&=&1& (e)\
s_2^2+c_2^2&=&1& (f)end{cases}tag{2}$$
Explanations :
(a) for the slope $tan(tfrac{pi}{3})=sqrt{3},$
(b) for $|overrightarrow{AB}|^2=(tfrac{15}{4})^2,$
(c) and (d) for relationship $tfrac{A+2B}{3}=F$, knowing that the coordinates of this focus are $(c,0)=(-sqrt{a^2-b^2},0).$
No comment for (e) and (f).
I have used a CAS for solving this system, but it could be done by hand (for example, one can at once get rid of $s_1$ using eq. (1)d).
The solutions of system (1) are :
$$a=3, b=sqrt{5}, c_1=-1/4, c_2=-7/8, s_1 =-sqrt{15}/4, s_2=sqrt{15}/8$$
giving the following equation :
$$dfrac{x^2}{9}+dfrac{y^2}{5}=1$$
and the following coordinates for points $A, B, F$:
$$A=(a c_1,b s_1)=(-21/8,-5sqrt{3}/8), B=(a c_2,b s_2)=(-3/4,5sqrt{3}/4)$$
and $F=(-sqrt{a^2-b^2},0)=(-2,0).$
Remarks : First, a point of vocabulary : a chord like $AB$ passing through one of the focii is called a focal chord. One can prove that the pole $P$ of a focal chord (intersection of the tangents in $A$ and $B$) is situated on the directrix. More precisely, the directrix associated with $F$ can be described as the locus of poles of all focal chords passing through $F$. Moreover $PF perp AB$ ; an old reference about all that : https://archive.org/stream/geometricaltreat00drewuoft/geometricaltreat00drewuoft#page/38/mode/1up.
This gives :
2nd method of solution :
Let us first recall that the tangent to an ellipse with equation $dfrac{x^2}{a^2}+dfrac{y^2}{b^2}=1$ at point $(x_k,y_k)$ has the following equation
$$dfrac{xx_k}{a^2}+dfrac{yy_k}{b^2}=1.tag{3}$$
Let us now write the equation of the ellipse under a form that will be more handy for expressing constraints:
$$ux^2+vy^2=1 text{with} u:=1/a^2, v:=1/b^2.tag{4}$$
Let us denote by $(x_0,y_0)$ the coordinates of point $P$ (the so-called "pole").
Then, the unknowns, that are different from the unknowns of system (2), verify :
$$begin{cases}
y_1-y_2&=&sqrt{3}(x_1-x_2)& (a)\
(x_1-x_2)^2+(y_1-y_2)^2&=&225/16& (b)\
(2x_1+x_2)/3&=&f& (c)\
2y_1+y_2&=&0& (d)\
(x_0-f)(x_1-x_2)+y_0(y_1-y_2)&=&0& (e)\
ux_0x_1+vy_0y_1&=&1& (f)\
ux_0x_2+vy_0y_2&=&1& (g)\
ux_1^2+vy_1^2&=&1& (h)\
ux_2^2+vy_2^2&=&1& (i)end{cases}tag{5}$$
Explanations :
(a), (b), (c), (d) : same constraints as in system (2).
(e) : orthogonality between PF and AB.
(f) : $P=(x_0,y_0)$ is on the tangent in $A(x_1,y_1)$ (using (3)).
(g) : In the same way, $P=(x_0,y_0)$ is on the tangent in $B(x_2,y_2)$.
(h) and (i) : $A$ and $B$ belong to the ellipse.
System (5), although bigger than system (2) (9 equations with 9 unknowns) is easily solved by a CAS, or even by hand. It gives in particular
$$P(x_0,y_0)=(-9/2,5sqrt{3}/6).$$
answered Feb 1 at 15:01
Jean MarieJean Marie
31.5k42355
$endgroup$
$begingroup$
I have added a picture. Could you say if my solution is understandable ?
$endgroup$
– Jean Marie
Feb 1 at 15:33
1
$begingroup$
Whoa, I think I've understand it properly, with so many details, thank you for your time! I appreciate it very much!
$endgroup$
– ItoMakoto
Feb 1 at 15:49
1
$begingroup$
Sorry, I calculate it again, the result of a should be 3, and b is square root 5, I'm trying to find something that may cause the problem.
$endgroup$
– ItoMakoto
Feb 1 at 16:12
1
$begingroup$
2:30 am in Tokyo, have a good night in advance! Bon soir :)
$endgroup$
– ItoMakoto
Feb 1 at 17:31
1
$begingroup$
@SMM use the property of the proportional in focus chord. This time I think I would have a deep impression. Interesting property of the focus chord, a lot of things for me to discover, thank you!
$endgroup$
– ItoMakoto
Feb 2 at 9:20
|
show 13 more comments
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%2f3095966%2fhow-to-solve-this-ellipse-problem%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
$begingroup$
By the assumption we have: $AF=frac{10}{4}$ and $BF=frac{5}{4}$.
We have that left focus is $F=(-c,0)$ where $c^2=a^2-b^2$. Consider the left directrix $d: x=-frac{a}{e}$, where $e= frac{c}{a}$ is the eccentricity, and drop perpendiculars $A',F',B'$ from $A,F,B$ to $d$. The quadrilateral $AA'B'B$ is a trapezoid with the angle $angle BAA'=60^circ$, right angles at vertices $A'$ and $B'$, and basis $AA'=frac{1}{e}AF=frac{10}{4e}$ and $BB'=frac{1}{e}BF= frac{5}{4e}$. Then we easily see (by dropping a perpendicular from $B$ to $AA'$) that $AA'-BB'=ABcos 60^circ$, so $frac{5}{4e}=frac{15}{4}cdot frac{1}{2}$, wherefrom $e=frac{2}{3}$. Therefore, $AA'=frac{15}{4}$ and $BB'=frac{15}{8}$.
Since parallel line $FF'$ to the basis of the trapezoid divides $AB$ in the ratio $2:1$ we have: $FF'=frac{1}{3}(AA'+2BB')$, i.e. $frac{a}{e}-c= frac{1}{3}cdotfrac{30}{4}= frac{10}{4}$, i.e. $frac{a}{e}-ae=a(frac{3}{2}-frac{2}{3})=frac{10}{4}$, wherefrom $a=3$. Now, $c=ae=2$, and $b=sqrt{a^2-c^2}=sqrt{5}$.
answered Feb 1 at 14:07
SMMSMM
3,098512
$endgroup$
$begingroup$
Interesting approach. A little figure could help to follow... I am going to write down a different solution, with the same results !
$endgroup$
– Jean Marie
Feb 1 at 14:24
$begingroup$
Such an interesting solution! Thank you!
$endgroup$
– ItoMakoto
Feb 1 at 14:32
$begingroup$
@SMM Thanks to you, I have been able to use as well the directrix for a (second) solution that is based on different results from yours (for example, I dont use the eccentricity). Your opinion ?
$endgroup$
– Jean Marie
Feb 2 at 11:48
$begingroup$
@JeanMarie Very nice! It seems that qadrics have immense number of very nice properties, and most of them are not commonly known.
$endgroup$
– SMM
Feb 2 at 16:05
$begingroup$
@SMM Yes not known because no longer taught for now half a century. In the first half of the XXth century, in my country, France, they were at the center of the teaching in the final years of high school and at the collegial level. Maybe it was exaggerate ; but now on, the pendulum has gone in a completely reverse sense... although the basic properties of these curves are essential to know, especialy in Physics.
$endgroup$
– Jean Marie
Feb 2 at 16:14
add a comment |
$begingroup$
By the assumption we have: $AF=frac{10}{4}$ and $BF=frac{5}{4}$.
We have that left focus is $F=(-c,0)$ where $c^2=a^2-b^2$. Consider the left directrix $d: x=-frac{a}{e}$, where $e= frac{c}{a}$ is the eccentricity, and drop perpendiculars $A',F',B'$ from $A,F,B$ to $d$. The quadrilateral $AA'B'B$ is a trapezoid with the angle $angle BAA'=60^circ$, right angles at vertices $A'$ and $B'$, and basis $AA'=frac{1}{e}AF=frac{10}{4e}$ and $BB'=frac{1}{e}BF= frac{5}{4e}$. Then we easily see (by dropping a perpendicular from $B$ to $AA'$) that $AA'-BB'=ABcos 60^circ$, so $frac{5}{4e}=frac{15}{4}cdot frac{1}{2}$, wherefrom $e=frac{2}{3}$. Therefore, $AA'=frac{15}{4}$ and $BB'=frac{15}{8}$.
Since parallel line $FF'$ to the basis of the trapezoid divides $AB$ in the ratio $2:1$ we have: $FF'=frac{1}{3}(AA'+2BB')$, i.e. $frac{a}{e}-c= frac{1}{3}cdotfrac{30}{4}= frac{10}{4}$, i.e. $frac{a}{e}-ae=a(frac{3}{2}-frac{2}{3})=frac{10}{4}$, wherefrom $a=3$. Now, $c=ae=2$, and $b=sqrt{a^2-c^2}=sqrt{5}$.
answered Feb 1 at 14:07
SMMSMM
3,098512
$endgroup$
$begingroup$
Interesting approach. A little figure could help to follow... I am going to write down a different solution, with the same results !
$endgroup$
– Jean Marie
Feb 1 at 14:24
$begingroup$
Such an interesting solution! Thank you!
$endgroup$
– ItoMakoto
Feb 1 at 14:32
$begingroup$
@SMM Thanks to you, I have been able to use as well the directrix for a (second) solution that is based on different results from yours (for example, I dont use the eccentricity). Your opinion ?
$endgroup$
– Jean Marie
Feb 2 at 11:48
$begingroup$
@JeanMarie Very nice! It seems that qadrics have immense number of very nice properties, and most of them are not commonly known.
$endgroup$
– SMM
Feb 2 at 16:05
$begingroup$
@SMM Yes not known because no longer taught for now half a century. In the first half of the XXth century, in my country, France, they were at the center of the teaching in the final years of high school and at the collegial level. Maybe it was exaggerate ; but now on, the pendulum has gone in a completely reverse sense... although the basic properties of these curves are essential to know, especialy in Physics.
$endgroup$
– Jean Marie
Feb 2 at 16:14
add a comment |
$begingroup$
By the assumption we have: $AF=frac{10}{4}$ and $BF=frac{5}{4}$.
We have that left focus is $F=(-c,0)$ where $c^2=a^2-b^2$. Consider the left directrix $d: x=-frac{a}{e}$, where $e= frac{c}{a}$ is the eccentricity, and drop perpendiculars $A',F',B'$ from $A,F,B$ to $d$. The quadrilateral $AA'B'B$ is a trapezoid with the angle $angle BAA'=60^circ$, right angles at vertices $A'$ and $B'$, and basis $AA'=frac{1}{e}AF=frac{10}{4e}$ and $BB'=frac{1}{e}BF= frac{5}{4e}$. Then we easily see (by dropping a perpendicular from $B$ to $AA'$) that $AA'-BB'=ABcos 60^circ$, so $frac{5}{4e}=frac{15}{4}cdot frac{1}{2}$, wherefrom $e=frac{2}{3}$. Therefore, $AA'=frac{15}{4}$ and $BB'=frac{15}{8}$.
Since parallel line $FF'$ to the basis of the trapezoid divides $AB$ in the ratio $2:1$ we have: $FF'=frac{1}{3}(AA'+2BB')$, i.e. $frac{a}{e}-c= frac{1}{3}cdotfrac{30}{4}= frac{10}{4}$, i.e. $frac{a}{e}-ae=a(frac{3}{2}-frac{2}{3})=frac{10}{4}$, wherefrom $a=3$. Now, $c=ae=2$, and $b=sqrt{a^2-c^2}=sqrt{5}$.
answered Feb 1 at 14:07
SMMSMM
3,098512
$endgroup$
By the assumption we have: $AF=frac{10}{4}$ and $BF=frac{5}{4}$.
We have that left focus is $F=(-c,0)$ where $c^2=a^2-b^2$. Consider the left directrix $d: x=-frac{a}{e}$, where $e= frac{c}{a}$ is the eccentricity, and drop perpendiculars $A',F',B'$ from $A,F,B$ to $d$. The quadrilateral $AA'B'B$ is a trapezoid with the angle $angle BAA'=60^circ$, right angles at vertices $A'$ and $B'$, and basis $AA'=frac{1}{e}AF=frac{10}{4e}$ and $BB'=frac{1}{e}BF= frac{5}{4e}$. Then we easily see (by dropping a perpendicular from $B$ to $AA'$) that $AA'-BB'=ABcos 60^circ$, so $frac{5}{4e}=frac{15}{4}cdot frac{1}{2}$, wherefrom $e=frac{2}{3}$. Therefore, $AA'=frac{15}{4}$ and $BB'=frac{15}{8}$.
Since parallel line $FF'$ to the basis of the trapezoid divides $AB$ in the ratio $2:1$ we have: $FF'=frac{1}{3}(AA'+2BB')$, i.e. $frac{a}{e}-c= frac{1}{3}cdotfrac{30}{4}= frac{10}{4}$, i.e. $frac{a}{e}-ae=a(frac{3}{2}-frac{2}{3})=frac{10}{4}$, wherefrom $a=3$. Now, $c=ae=2$, and $b=sqrt{a^2-c^2}=sqrt{5}$.
answered Feb 1 at 14:07
SMMSMM
3,098512
answered Feb 1 at 14:07
SMMSMM
3,098512
answered Feb 1 at 14:07
SMMSMM
3,098512
answered Feb 1 at 14:07
SMMSMM
3,098512
3,098512
$begingroup$
Interesting approach. A little figure could help to follow... I am going to write down a different solution, with the same results !
$endgroup$
– Jean Marie
Feb 1 at 14:24
$begingroup$
Such an interesting solution! Thank you!
$endgroup$
– ItoMakoto
Feb 1 at 14:32
$begingroup$
@SMM Thanks to you, I have been able to use as well the directrix for a (second) solution that is based on different results from yours (for example, I dont use the eccentricity). Your opinion ?
$endgroup$
– Jean Marie
Feb 2 at 11:48
$begingroup$
@JeanMarie Very nice! It seems that qadrics have immense number of very nice properties, and most of them are not commonly known.
$endgroup$
– SMM
Feb 2 at 16:05
$begingroup$
@SMM Yes not known because no longer taught for now half a century. In the first half of the XXth century, in my country, France, they were at the center of the teaching in the final years of high school and at the collegial level. Maybe it was exaggerate ; but now on, the pendulum has gone in a completely reverse sense... although the basic properties of these curves are essential to know, especialy in Physics.
$endgroup$
– Jean Marie
Feb 2 at 16:14
add a comment |
$begingroup$
Interesting approach. A little figure could help to follow... I am going to write down a different solution, with the same results !
$endgroup$
– Jean Marie
Feb 1 at 14:24
$begingroup$
Such an interesting solution! Thank you!
$endgroup$
– ItoMakoto
Feb 1 at 14:32
$begingroup$
@SMM Thanks to you, I have been able to use as well the directrix for a (second) solution that is based on different results from yours (for example, I dont use the eccentricity). Your opinion ?
$endgroup$
– Jean Marie
Feb 2 at 11:48
$begingroup$
@JeanMarie Very nice! It seems that qadrics have immense number of very nice properties, and most of them are not commonly known.
$endgroup$
– SMM
Feb 2 at 16:05
$begingroup$
@SMM Yes not known because no longer taught for now half a century. In the first half of the XXth century, in my country, France, they were at the center of the teaching in the final years of high school and at the collegial level. Maybe it was exaggerate ; but now on, the pendulum has gone in a completely reverse sense... although the basic properties of these curves are essential to know, especialy in Physics.
$endgroup$
– Jean Marie
Feb 2 at 16:14
$begingroup$
Interesting approach. A little figure could help to follow... I am going to write down a different solution, with the same results !
$endgroup$
– Jean Marie
Feb 1 at 14:24
$begingroup$
Interesting approach. A little figure could help to follow... I am going to write down a different solution, with the same results !
$endgroup$
– Jean Marie
Feb 1 at 14:24
$begingroup$
Such an interesting solution! Thank you!
$endgroup$
– ItoMakoto
Feb 1 at 14:32
$begingroup$
Such an interesting solution! Thank you!
$endgroup$
– ItoMakoto
Feb 1 at 14:32
$begingroup$
@SMM Thanks to you, I have been able to use as well the directrix for a (second) solution that is based on different results from yours (for example, I dont use the eccentricity). Your opinion ?
$endgroup$
– Jean Marie
Feb 2 at 11:48
$begingroup$
@SMM Thanks to you, I have been able to use as well the directrix for a (second) solution that is based on different results from yours (for example, I dont use the eccentricity). Your opinion ?
$endgroup$
– Jean Marie
Feb 2 at 11:48
$begingroup$
@JeanMarie Very nice! It seems that qadrics have immense number of very nice properties, and most of them are not commonly known.
$endgroup$
– SMM
Feb 2 at 16:05
$begingroup$
@JeanMarie Very nice! It seems that qadrics have immense number of very nice properties, and most of them are not commonly known.
$endgroup$
– SMM
Feb 2 at 16:05
$begingroup$
@SMM Yes not known because no longer taught for now half a century. In the first half of the XXth century, in my country, France, they were at the center of the teaching in the final years of high school and at the collegial level. Maybe it was exaggerate ; but now on, the pendulum has gone in a completely reverse sense... although the basic properties of these curves are essential to know, especialy in Physics.
$endgroup$
– Jean Marie
Feb 2 at 16:14
$begingroup$
@SMM Yes not known because no longer taught for now half a century. In the first half of the XXth century, in my country, France, they were at the center of the teaching in the final years of high school and at the collegial level. Maybe it was exaggerate ; but now on, the pendulum has gone in a completely reverse sense... although the basic properties of these curves are essential to know, especialy in Physics.
$endgroup$
– Jean Marie
Feb 2 at 16:14
add a comment |
$begingroup$
Fig. 1 :This figure illustrates the 2 answers I give as well as the very interesting answer by @SMM. On the left is figured the directrix line associated with focus $F$. See comments at the end of the first solution.
First method
Let us use the following parameterization of the ellipse :
$$begin{cases}x=a cos(t)\y=b sin(t)end{cases}tag{1}$$
Let
$$begin{cases}t_1, c_1=cos(t_1), s_1=sin(t_1) text{correspond to point } A \t_2, c_2=cos(t_2), s_2=sin(t_2) text{correspond to point } Bend{cases}$$
The following system of 6 equations with 6 unknowns takes into account the different constraints of the problem :
$$begin{cases}frac{b(s_1-s_2)}{a(c_1-c_2)}&=&sqrt{3}& (a)\
a^2(c_2-c_1)^2+b^2(s_2-s_1)^2&=&(tfrac{15}{4})^2& (b)\
afrac{c_1+2c_2}{3}&=&-sqrt{a^2-b^2}& (c)\
2s_2+s_1&=&0& (d)\
s_1^2+c_1^2&=&1& (e)\
s_2^2+c_2^2&=&1& (f)end{cases}tag{2}$$
Explanations :
(a) for the slope $tan(tfrac{pi}{3})=sqrt{3},$
(b) for $|overrightarrow{AB}|^2=(tfrac{15}{4})^2,$
(c) and (d) for relationship $tfrac{A+2B}{3}=F$, knowing that the coordinates of this focus are $(c,0)=(-sqrt{a^2-b^2},0).$
No comment for (e) and (f).
I have used a CAS for solving this system, but it could be done by hand (for example, one can at once get rid of $s_1$ using eq. (1)d).
The solutions of system (1) are :
$$a=3, b=sqrt{5}, c_1=-1/4, c_2=-7/8, s_1 =-sqrt{15}/4, s_2=sqrt{15}/8$$
giving the following equation :
$$dfrac{x^2}{9}+dfrac{y^2}{5}=1$$
and the following coordinates for points $A, B, F$:
$$A=(a c_1,b s_1)=(-21/8,-5sqrt{3}/8), B=(a c_2,b s_2)=(-3/4,5sqrt{3}/4)$$
and $F=(-sqrt{a^2-b^2},0)=(-2,0).$
Remarks : First, a point of vocabulary : a chord like $AB$ passing through one of the focii is called a focal chord. One can prove that the pole $P$ of a focal chord (intersection of the tangents in $A$ and $B$) is situated on the directrix. More precisely, the directrix associated with $F$ can be described as the locus of poles of all focal chords passing through $F$. Moreover $PF perp AB$ ; an old reference about all that : https://archive.org/stream/geometricaltreat00drewuoft/geometricaltreat00drewuoft#page/38/mode/1up.
This gives :
2nd method of solution :
Let us first recall that the tangent to an ellipse with equation $dfrac{x^2}{a^2}+dfrac{y^2}{b^2}=1$ at point $(x_k,y_k)$ has the following equation
$$dfrac{xx_k}{a^2}+dfrac{yy_k}{b^2}=1.tag{3}$$
Let us now write the equation of the ellipse under a form that will be more handy for expressing constraints:
$$ux^2+vy^2=1 text{with} u:=1/a^2, v:=1/b^2.tag{4}$$
Let us denote by $(x_0,y_0)$ the coordinates of point $P$ (the so-called "pole").
Then, the unknowns, that are different from the unknowns of system (2), verify :
$$begin{cases}
y_1-y_2&=&sqrt{3}(x_1-x_2)& (a)\
(x_1-x_2)^2+(y_1-y_2)^2&=&225/16& (b)\
(2x_1+x_2)/3&=&f& (c)\
2y_1+y_2&=&0& (d)\
(x_0-f)(x_1-x_2)+y_0(y_1-y_2)&=&0& (e)\
ux_0x_1+vy_0y_1&=&1& (f)\
ux_0x_2+vy_0y_2&=&1& (g)\
ux_1^2+vy_1^2&=&1& (h)\
ux_2^2+vy_2^2&=&1& (i)end{cases}tag{5}$$
Explanations :
(a), (b), (c), (d) : same constraints as in system (2).
(e) : orthogonality between PF and AB.
(f) : $P=(x_0,y_0)$ is on the tangent in $A(x_1,y_1)$ (using (3)).
(g) : In the same way, $P=(x_0,y_0)$ is on the tangent in $B(x_2,y_2)$.
(h) and (i) : $A$ and $B$ belong to the ellipse.
System (5), although bigger than system (2) (9 equations with 9 unknowns) is easily solved by a CAS, or even by hand. It gives in particular
$$P(x_0,y_0)=(-9/2,5sqrt{3}/6).$$
answered Feb 1 at 15:01
Jean MarieJean Marie
31.5k42355
$endgroup$
$begingroup$
I have added a picture. Could you say if my solution is understandable ?
$endgroup$
– Jean Marie
Feb 1 at 15:33
1
$begingroup$
Whoa, I think I've understand it properly, with so many details, thank you for your time! I appreciate it very much!
$endgroup$
– ItoMakoto
Feb 1 at 15:49
1
$begingroup$
Sorry, I calculate it again, the result of a should be 3, and b is square root 5, I'm trying to find something that may cause the problem.
$endgroup$
– ItoMakoto
Feb 1 at 16:12
1
$begingroup$
2:30 am in Tokyo, have a good night in advance! Bon soir :)
$endgroup$
– ItoMakoto
Feb 1 at 17:31
1
$begingroup$
@SMM use the property of the proportional in focus chord. This time I think I would have a deep impression. Interesting property of the focus chord, a lot of things for me to discover, thank you!
$endgroup$
– ItoMakoto
Feb 2 at 9:20
|
show 13 more comments
$begingroup$
Fig. 1 :This figure illustrates the 2 answers I give as well as the very interesting answer by @SMM. On the left is figured the directrix line associated with focus $F$. See comments at the end of the first solution.
First method
Let us use the following parameterization of the ellipse :
$$begin{cases}x=a cos(t)\y=b sin(t)end{cases}tag{1}$$
Let
$$begin{cases}t_1, c_1=cos(t_1), s_1=sin(t_1) text{correspond to point } A \t_2, c_2=cos(t_2), s_2=sin(t_2) text{correspond to point } Bend{cases}$$
The following system of 6 equations with 6 unknowns takes into account the different constraints of the problem :
$$begin{cases}frac{b(s_1-s_2)}{a(c_1-c_2)}&=&sqrt{3}& (a)\
a^2(c_2-c_1)^2+b^2(s_2-s_1)^2&=&(tfrac{15}{4})^2& (b)\
afrac{c_1+2c_2}{3}&=&-sqrt{a^2-b^2}& (c)\
2s_2+s_1&=&0& (d)\
s_1^2+c_1^2&=&1& (e)\
s_2^2+c_2^2&=&1& (f)end{cases}tag{2}$$
Explanations :
(a) for the slope $tan(tfrac{pi}{3})=sqrt{3},$
(b) for $|overrightarrow{AB}|^2=(tfrac{15}{4})^2,$
(c) and (d) for relationship $tfrac{A+2B}{3}=F$, knowing that the coordinates of this focus are $(c,0)=(-sqrt{a^2-b^2},0).$
No comment for (e) and (f).
I have used a CAS for solving this system, but it could be done by hand (for example, one can at once get rid of $s_1$ using eq. (1)d).
The solutions of system (1) are :
$$a=3, b=sqrt{5}, c_1=-1/4, c_2=-7/8, s_1 =-sqrt{15}/4, s_2=sqrt{15}/8$$
giving the following equation :
$$dfrac{x^2}{9}+dfrac{y^2}{5}=1$$
and the following coordinates for points $A, B, F$:
$$A=(a c_1,b s_1)=(-21/8,-5sqrt{3}/8), B=(a c_2,b s_2)=(-3/4,5sqrt{3}/4)$$
and $F=(-sqrt{a^2-b^2},0)=(-2,0).$
Remarks : First, a point of vocabulary : a chord like $AB$ passing through one of the focii is called a focal chord. One can prove that the pole $P$ of a focal chord (intersection of the tangents in $A$ and $B$) is situated on the directrix. More precisely, the directrix associated with $F$ can be described as the locus of poles of all focal chords passing through $F$. Moreover $PF perp AB$ ; an old reference about all that : https://archive.org/stream/geometricaltreat00drewuoft/geometricaltreat00drewuoft#page/38/mode/1up.
This gives :
2nd method of solution :
Let us first recall that the tangent to an ellipse with equation $dfrac{x^2}{a^2}+dfrac{y^2}{b^2}=1$ at point $(x_k,y_k)$ has the following equation
$$dfrac{xx_k}{a^2}+dfrac{yy_k}{b^2}=1.tag{3}$$
Let us now write the equation of the ellipse under a form that will be more handy for expressing constraints:
$$ux^2+vy^2=1 text{with} u:=1/a^2, v:=1/b^2.tag{4}$$
Let us denote by $(x_0,y_0)$ the coordinates of point $P$ (the so-called "pole").
Then, the unknowns, that are different from the unknowns of system (2), verify :
$$begin{cases}
y_1-y_2&=&sqrt{3}(x_1-x_2)& (a)\
(x_1-x_2)^2+(y_1-y_2)^2&=&225/16& (b)\
(2x_1+x_2)/3&=&f& (c)\
2y_1+y_2&=&0& (d)\
(x_0-f)(x_1-x_2)+y_0(y_1-y_2)&=&0& (e)\
ux_0x_1+vy_0y_1&=&1& (f)\
ux_0x_2+vy_0y_2&=&1& (g)\
ux_1^2+vy_1^2&=&1& (h)\
ux_2^2+vy_2^2&=&1& (i)end{cases}tag{5}$$
Explanations :
(a), (b), (c), (d) : same constraints as in system (2).
(e) : orthogonality between PF and AB.
(f) : $P=(x_0,y_0)$ is on the tangent in $A(x_1,y_1)$ (using (3)).
(g) : In the same way, $P=(x_0,y_0)$ is on the tangent in $B(x_2,y_2)$.
(h) and (i) : $A$ and $B$ belong to the ellipse.
System (5), although bigger than system (2) (9 equations with 9 unknowns) is easily solved by a CAS, or even by hand. It gives in particular
$$P(x_0,y_0)=(-9/2,5sqrt{3}/6).$$
answered Feb 1 at 15:01
Jean MarieJean Marie
31.5k42355
$endgroup$
$begingroup$
I have added a picture. Could you say if my solution is understandable ?
$endgroup$
– Jean Marie
Feb 1 at 15:33
1
$begingroup$
Whoa, I think I've understand it properly, with so many details, thank you for your time! I appreciate it very much!
$endgroup$
– ItoMakoto
Feb 1 at 15:49
1
$begingroup$
Sorry, I calculate it again, the result of a should be 3, and b is square root 5, I'm trying to find something that may cause the problem.
$endgroup$
– ItoMakoto
Feb 1 at 16:12
1
$begingroup$
2:30 am in Tokyo, have a good night in advance! Bon soir :)
$endgroup$
– ItoMakoto
Feb 1 at 17:31
1
$begingroup$
@SMM use the property of the proportional in focus chord. This time I think I would have a deep impression. Interesting property of the focus chord, a lot of things for me to discover, thank you!
$endgroup$
– ItoMakoto
Feb 2 at 9:20
|
show 13 more comments
$begingroup$
Fig. 1 :This figure illustrates the 2 answers I give as well as the very interesting answer by @SMM. On the left is figured the directrix line associated with focus $F$. See comments at the end of the first solution.
First method
Let us use the following parameterization of the ellipse :
$$begin{cases}x=a cos(t)\y=b sin(t)end{cases}tag{1}$$
Let
$$begin{cases}t_1, c_1=cos(t_1), s_1=sin(t_1) text{correspond to point } A \t_2, c_2=cos(t_2), s_2=sin(t_2) text{correspond to point } Bend{cases}$$
The following system of 6 equations with 6 unknowns takes into account the different constraints of the problem :
$$begin{cases}frac{b(s_1-s_2)}{a(c_1-c_2)}&=&sqrt{3}& (a)\
a^2(c_2-c_1)^2+b^2(s_2-s_1)^2&=&(tfrac{15}{4})^2& (b)\
afrac{c_1+2c_2}{3}&=&-sqrt{a^2-b^2}& (c)\
2s_2+s_1&=&0& (d)\
s_1^2+c_1^2&=&1& (e)\
s_2^2+c_2^2&=&1& (f)end{cases}tag{2}$$
Explanations :
(a) for the slope $tan(tfrac{pi}{3})=sqrt{3},$
(b) for $|overrightarrow{AB}|^2=(tfrac{15}{4})^2,$
(c) and (d) for relationship $tfrac{A+2B}{3}=F$, knowing that the coordinates of this focus are $(c,0)=(-sqrt{a^2-b^2},0).$
No comment for (e) and (f).
I have used a CAS for solving this system, but it could be done by hand (for example, one can at once get rid of $s_1$ using eq. (1)d).
The solutions of system (1) are :
$$a=3, b=sqrt{5}, c_1=-1/4, c_2=-7/8, s_1 =-sqrt{15}/4, s_2=sqrt{15}/8$$
giving the following equation :
$$dfrac{x^2}{9}+dfrac{y^2}{5}=1$$
and the following coordinates for points $A, B, F$:
$$A=(a c_1,b s_1)=(-21/8,-5sqrt{3}/8), B=(a c_2,b s_2)=(-3/4,5sqrt{3}/4)$$
and $F=(-sqrt{a^2-b^2},0)=(-2,0).$
Remarks : First, a point of vocabulary : a chord like $AB$ passing through one of the focii is called a focal chord. One can prove that the pole $P$ of a focal chord (intersection of the tangents in $A$ and $B$) is situated on the directrix. More precisely, the directrix associated with $F$ can be described as the locus of poles of all focal chords passing through $F$. Moreover $PF perp AB$ ; an old reference about all that : https://archive.org/stream/geometricaltreat00drewuoft/geometricaltreat00drewuoft#page/38/mode/1up.
This gives :
2nd method of solution :
Let us first recall that the tangent to an ellipse with equation $dfrac{x^2}{a^2}+dfrac{y^2}{b^2}=1$ at point $(x_k,y_k)$ has the following equation
$$dfrac{xx_k}{a^2}+dfrac{yy_k}{b^2}=1.tag{3}$$
Let us now write the equation of the ellipse under a form that will be more handy for expressing constraints:
$$ux^2+vy^2=1 text{with} u:=1/a^2, v:=1/b^2.tag{4}$$
Let us denote by $(x_0,y_0)$ the coordinates of point $P$ (the so-called "pole").
Then, the unknowns, that are different from the unknowns of system (2), verify :
$$begin{cases}
y_1-y_2&=&sqrt{3}(x_1-x_2)& (a)\
(x_1-x_2)^2+(y_1-y_2)^2&=&225/16& (b)\
(2x_1+x_2)/3&=&f& (c)\
2y_1+y_2&=&0& (d)\
(x_0-f)(x_1-x_2)+y_0(y_1-y_2)&=&0& (e)\
ux_0x_1+vy_0y_1&=&1& (f)\
ux_0x_2+vy_0y_2&=&1& (g)\
ux_1^2+vy_1^2&=&1& (h)\
ux_2^2+vy_2^2&=&1& (i)end{cases}tag{5}$$
Explanations :
(a), (b), (c), (d) : same constraints as in system (2).
(e) : orthogonality between PF and AB.
(f) : $P=(x_0,y_0)$ is on the tangent in $A(x_1,y_1)$ (using (3)).
(g) : In the same way, $P=(x_0,y_0)$ is on the tangent in $B(x_2,y_2)$.
(h) and (i) : $A$ and $B$ belong to the ellipse.
System (5), although bigger than system (2) (9 equations with 9 unknowns) is easily solved by a CAS, or even by hand. It gives in particular
$$P(x_0,y_0)=(-9/2,5sqrt{3}/6).$$
answered Feb 1 at 15:01
Jean MarieJean Marie
31.5k42355
$endgroup$
Fig. 1 :This figure illustrates the 2 answers I give as well as the very interesting answer by @SMM. On the left is figured the directrix line associated with focus $F$. See comments at the end of the first solution.
First method
Let us use the following parameterization of the ellipse :
$$begin{cases}x=a cos(t)\y=b sin(t)end{cases}tag{1}$$
Let
$$begin{cases}t_1, c_1=cos(t_1), s_1=sin(t_1) text{correspond to point } A \t_2, c_2=cos(t_2), s_2=sin(t_2) text{correspond to point } Bend{cases}$$
The following system of 6 equations with 6 unknowns takes into account the different constraints of the problem :
$$begin{cases}frac{b(s_1-s_2)}{a(c_1-c_2)}&=&sqrt{3}& (a)\
a^2(c_2-c_1)^2+b^2(s_2-s_1)^2&=&(tfrac{15}{4})^2& (b)\
afrac{c_1+2c_2}{3}&=&-sqrt{a^2-b^2}& (c)\
2s_2+s_1&=&0& (d)\
s_1^2+c_1^2&=&1& (e)\
s_2^2+c_2^2&=&1& (f)end{cases}tag{2}$$
Explanations :
(a) for the slope $tan(tfrac{pi}{3})=sqrt{3},$
(b) for $|overrightarrow{AB}|^2=(tfrac{15}{4})^2,$
(c) and (d) for relationship $tfrac{A+2B}{3}=F$, knowing that the coordinates of this focus are $(c,0)=(-sqrt{a^2-b^2},0).$
No comment for (e) and (f).
I have used a CAS for solving this system, but it could be done by hand (for example, one can at once get rid of $s_1$ using eq. (1)d).
The solutions of system (1) are :
$$a=3, b=sqrt{5}, c_1=-1/4, c_2=-7/8, s_1 =-sqrt{15}/4, s_2=sqrt{15}/8$$
giving the following equation :
$$dfrac{x^2}{9}+dfrac{y^2}{5}=1$$
and the following coordinates for points $A, B, F$:
$$A=(a c_1,b s_1)=(-21/8,-5sqrt{3}/8), B=(a c_2,b s_2)=(-3/4,5sqrt{3}/4)$$
and $F=(-sqrt{a^2-b^2},0)=(-2,0).$
Remarks : First, a point of vocabulary : a chord like $AB$ passing through one of the focii is called a focal chord. One can prove that the pole $P$ of a focal chord (intersection of the tangents in $A$ and $B$) is situated on the directrix. More precisely, the directrix associated with $F$ can be described as the locus of poles of all focal chords passing through $F$. Moreover $PF perp AB$ ; an old reference about all that : https://archive.org/stream/geometricaltreat00drewuoft/geometricaltreat00drewuoft#page/38/mode/1up.
This gives :
2nd method of solution :
Let us first recall that the tangent to an ellipse with equation $dfrac{x^2}{a^2}+dfrac{y^2}{b^2}=1$ at point $(x_k,y_k)$ has the following equation
$$dfrac{xx_k}{a^2}+dfrac{yy_k}{b^2}=1.tag{3}$$
Let us now write the equation of the ellipse under a form that will be more handy for expressing constraints:
$$ux^2+vy^2=1 text{with} u:=1/a^2, v:=1/b^2.tag{4}$$
Let us denote by $(x_0,y_0)$ the coordinates of point $P$ (the so-called "pole").
Then, the unknowns, that are different from the unknowns of system (2), verify :
$$begin{cases}
y_1-y_2&=&sqrt{3}(x_1-x_2)& (a)\
(x_1-x_2)^2+(y_1-y_2)^2&=&225/16& (b)\
(2x_1+x_2)/3&=&f& (c)\
2y_1+y_2&=&0& (d)\
(x_0-f)(x_1-x_2)+y_0(y_1-y_2)&=&0& (e)\
ux_0x_1+vy_0y_1&=&1& (f)\
ux_0x_2+vy_0y_2&=&1& (g)\
ux_1^2+vy_1^2&=&1& (h)\
ux_2^2+vy_2^2&=&1& (i)end{cases}tag{5}$$
Explanations :
(a), (b), (c), (d) : same constraints as in system (2).
(e) : orthogonality between PF and AB.
(f) : $P=(x_0,y_0)$ is on the tangent in $A(x_1,y_1)$ (using (3)).
(g) : In the same way, $P=(x_0,y_0)$ is on the tangent in $B(x_2,y_2)$.
(h) and (i) : $A$ and $B$ belong to the ellipse.
System (5), although bigger than system (2) (9 equations with 9 unknowns) is easily solved by a CAS, or even by hand. It gives in particular
$$P(x_0,y_0)=(-9/2,5sqrt{3}/6).$$
answered Feb 1 at 15:01
Jean MarieJean Marie
31.5k42355
edited Feb 2 at 16:06
answered Feb 1 at 15:01
Jean MarieJean Marie
31.5k42355
answered Feb 1 at 15:01
Jean MarieJean Marie
31.5k42355
answered Feb 1 at 15:01
Jean MarieJean Marie
31.5k42355
31.5k42355
$begingroup$
I have added a picture. Could you say if my solution is understandable ?
$endgroup$
– Jean Marie
Feb 1 at 15:33
1
$begingroup$
Whoa, I think I've understand it properly, with so many details, thank you for your time! I appreciate it very much!
$endgroup$
– ItoMakoto
Feb 1 at 15:49
1
$begingroup$
Sorry, I calculate it again, the result of a should be 3, and b is square root 5, I'm trying to find something that may cause the problem.
$endgroup$
– ItoMakoto
Feb 1 at 16:12
1
$begingroup$
2:30 am in Tokyo, have a good night in advance! Bon soir :)
$endgroup$
– ItoMakoto
Feb 1 at 17:31
1
$begingroup$
@SMM use the property of the proportional in focus chord. This time I think I would have a deep impression. Interesting property of the focus chord, a lot of things for me to discover, thank you!
$endgroup$
– ItoMakoto
Feb 2 at 9:20
|
show 13 more comments
$begingroup$
I have added a picture. Could you say if my solution is understandable ?
$endgroup$
– Jean Marie
Feb 1 at 15:33
1
$begingroup$
Whoa, I think I've understand it properly, with so many details, thank you for your time! I appreciate it very much!
$endgroup$
– ItoMakoto
Feb 1 at 15:49
1
$begingroup$
Sorry, I calculate it again, the result of a should be 3, and b is square root 5, I'm trying to find something that may cause the problem.
$endgroup$
– ItoMakoto
Feb 1 at 16:12
1
$begingroup$
2:30 am in Tokyo, have a good night in advance! Bon soir :)
$endgroup$
– ItoMakoto
Feb 1 at 17:31
1
$begingroup$
@SMM use the property of the proportional in focus chord. This time I think I would have a deep impression. Interesting property of the focus chord, a lot of things for me to discover, thank you!
$endgroup$
– ItoMakoto
Feb 2 at 9:20
$begingroup$
I have added a picture. Could you say if my solution is understandable ?
$endgroup$
– Jean Marie
Feb 1 at 15:33
$begingroup$
I have added a picture. Could you say if my solution is understandable ?
$endgroup$
– Jean Marie
Feb 1 at 15:33
1
1
$begingroup$
Whoa, I think I've understand it properly, with so many details, thank you for your time! I appreciate it very much!
$endgroup$
– ItoMakoto
Feb 1 at 15:49
$begingroup$
Whoa, I think I've understand it properly, with so many details, thank you for your time! I appreciate it very much!
$endgroup$
– ItoMakoto
Feb 1 at 15:49
1
1
$begingroup$
Sorry, I calculate it again, the result of a should be 3, and b is square root 5, I'm trying to find something that may cause the problem.
$endgroup$
– ItoMakoto
Feb 1 at 16:12
$begingroup$
Sorry, I calculate it again, the result of a should be 3, and b is square root 5, I'm trying to find something that may cause the problem.
$endgroup$
– ItoMakoto
Feb 1 at 16:12
1
1
$begingroup$
2:30 am in Tokyo, have a good night in advance! Bon soir :)
$endgroup$
– ItoMakoto
Feb 1 at 17:31
$begingroup$
2:30 am in Tokyo, have a good night in advance! Bon soir :)
$endgroup$
– ItoMakoto
Feb 1 at 17:31
1
1
$begingroup$
@SMM use the property of the proportional in focus chord. This time I think I would have a deep impression. Interesting property of the focus chord, a lot of things for me to discover, thank you!
$endgroup$
– ItoMakoto
Feb 2 at 9:20
$begingroup$
@SMM use the property of the proportional in focus chord. This time I think I would have a deep impression. Interesting property of the focus chord, a lot of things for me to discover, thank you!
$endgroup$
– ItoMakoto
Feb 2 at 9:20
|
show 13 more comments
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%2f3095966%2fhow-to-solve-this-ellipse-problem%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
