Mixing
Proof that the foci of an ellipse are unique
$begingroup$
Given a fixed distance $2a$, and two points $(F_1,F_2)$ in the Euclidean plane, one can define an ellipse as the set of points $E$ such that the sum of the distances $d(E,F_1) + d(E,F_2)$ is equal to $2a$.
How can one prove that those two foci $F_1$ and $F_2$ are unique or not? In other words, prove that for any other set of points $(F_1',F_2') neq (F_1,F_2)$, and some distance $2a'$, there exists at least one point $P$ in the previously defined ellipse $E$, such that $d(P,F_1') + d(P,F_2') neq 2a'$. It would be nice if there was a nice way of finding one or even all points P with that property. That means I wish the solution to be as constructive as possible, even though I will not discard a proof by contradiction.
A weak proof can be made using the Euclidean distance formula $d(A,B)=sqrt{(x_a - x_b)^2 + (y_a - y_b)^2}$. A strong proof would require using a general distance formula using these properties.
conic-sections
asked Jan 18 at 0:43
NighteenNighteen
1192
$endgroup$
|
show 6 more comments
$begingroup$
Given a fixed distance $2a$, and two points $(F_1,F_2)$ in the Euclidean plane, one can define an ellipse as the set of points $E$ such that the sum of the distances $d(E,F_1) + d(E,F_2)$ is equal to $2a$.
How can one prove that those two foci $F_1$ and $F_2$ are unique or not? In other words, prove that for any other set of points $(F_1',F_2') neq (F_1,F_2)$, and some distance $2a'$, there exists at least one point $P$ in the previously defined ellipse $E$, such that $d(P,F_1') + d(P,F_2') neq 2a'$. It would be nice if there was a nice way of finding one or even all points P with that property. That means I wish the solution to be as constructive as possible, even though I will not discard a proof by contradiction.
A weak proof can be made using the Euclidean distance formula $d(A,B)=sqrt{(x_a - x_b)^2 + (y_a - y_b)^2}$. A strong proof would require using a general distance formula using these properties.
conic-sections
asked Jan 18 at 0:43
NighteenNighteen
1192
$endgroup$
$begingroup$
I just realized that my question might be rephrased as "Proof that the definition of an ellipse is unique".
$endgroup$
– Nighteen
Jan 18 at 1:03
$begingroup$
Do you have the distance formula at your disposal? In other words, do you have the Cartesian equation for ellipses? Or must you stick to the definition you've given?
$endgroup$
– Clayton
Jan 18 at 1:12
1
$begingroup$
The foci along with the distance $2a$ uniquely determine the major and minor axes of the ellipse.
$endgroup$
– Théophile
Jan 18 at 1:36
1
$begingroup$
@Théophile You would have to define major and minor axes as the longest and shortest diameter. You would also have to prove that if two ellipses have different axes then they are different (have one point P that is in one and not in the other ellipse).
$endgroup$
– Nighteen
Jan 18 at 1:43
$begingroup$
Isn't it immediate? If the axes are different, you can (very easily) determine a point on one but not the other.
$endgroup$
– Clayton
Jan 18 at 1:50
|
show 6 more comments
$begingroup$
Given a fixed distance $2a$, and two points $(F_1,F_2)$ in the Euclidean plane, one can define an ellipse as the set of points $E$ such that the sum of the distances $d(E,F_1) + d(E,F_2)$ is equal to $2a$.
How can one prove that those two foci $F_1$ and $F_2$ are unique or not? In other words, prove that for any other set of points $(F_1',F_2') neq (F_1,F_2)$, and some distance $2a'$, there exists at least one point $P$ in the previously defined ellipse $E$, such that $d(P,F_1') + d(P,F_2') neq 2a'$. It would be nice if there was a nice way of finding one or even all points P with that property. That means I wish the solution to be as constructive as possible, even though I will not discard a proof by contradiction.
A weak proof can be made using the Euclidean distance formula $d(A,B)=sqrt{(x_a - x_b)^2 + (y_a - y_b)^2}$. A strong proof would require using a general distance formula using these properties.
conic-sections
asked Jan 18 at 0:43
NighteenNighteen
1192
$endgroup$
Given a fixed distance $2a$, and two points $(F_1,F_2)$ in the Euclidean plane, one can define an ellipse as the set of points $E$ such that the sum of the distances $d(E,F_1) + d(E,F_2)$ is equal to $2a$.
How can one prove that those two foci $F_1$ and $F_2$ are unique or not? In other words, prove that for any other set of points $(F_1',F_2') neq (F_1,F_2)$, and some distance $2a'$, there exists at least one point $P$ in the previously defined ellipse $E$, such that $d(P,F_1') + d(P,F_2') neq 2a'$. It would be nice if there was a nice way of finding one or even all points P with that property. That means I wish the solution to be as constructive as possible, even though I will not discard a proof by contradiction.
A weak proof can be made using the Euclidean distance formula $d(A,B)=sqrt{(x_a - x_b)^2 + (y_a - y_b)^2}$. A strong proof would require using a general distance formula using these properties.
conic-sections
conic-sections
asked Jan 18 at 0:43
NighteenNighteen
1192
asked Jan 18 at 0:43
NighteenNighteen
1192
edited Jan 18 at 1:48
Nighteen
asked Jan 18 at 0:43
NighteenNighteen
1192
asked Jan 18 at 0:43
NighteenNighteen
1192
asked Jan 18 at 0:43
NighteenNighteen
1192
1192
$begingroup$
I just realized that my question might be rephrased as "Proof that the definition of an ellipse is unique".
$endgroup$
– Nighteen
Jan 18 at 1:03
$begingroup$
Do you have the distance formula at your disposal? In other words, do you have the Cartesian equation for ellipses? Or must you stick to the definition you've given?
$endgroup$
– Clayton
Jan 18 at 1:12
1
$begingroup$
The foci along with the distance $2a$ uniquely determine the major and minor axes of the ellipse.
$endgroup$
– Théophile
Jan 18 at 1:36
1
$begingroup$
@Théophile You would have to define major and minor axes as the longest and shortest diameter. You would also have to prove that if two ellipses have different axes then they are different (have one point P that is in one and not in the other ellipse).
$endgroup$
– Nighteen
Jan 18 at 1:43
$begingroup$
Isn't it immediate? If the axes are different, you can (very easily) determine a point on one but not the other.
$endgroup$
– Clayton
Jan 18 at 1:50
|
show 6 more comments
$begingroup$
I just realized that my question might be rephrased as "Proof that the definition of an ellipse is unique".
$endgroup$
– Nighteen
Jan 18 at 1:03
$begingroup$
Do you have the distance formula at your disposal? In other words, do you have the Cartesian equation for ellipses? Or must you stick to the definition you've given?
$endgroup$
– Clayton
Jan 18 at 1:12
1
$begingroup$
The foci along with the distance $2a$ uniquely determine the major and minor axes of the ellipse.
$endgroup$
– Théophile
Jan 18 at 1:36
1
$begingroup$
@Théophile You would have to define major and minor axes as the longest and shortest diameter. You would also have to prove that if two ellipses have different axes then they are different (have one point P that is in one and not in the other ellipse).
$endgroup$
– Nighteen
Jan 18 at 1:43
$begingroup$
Isn't it immediate? If the axes are different, you can (very easily) determine a point on one but not the other.
$endgroup$
– Clayton
Jan 18 at 1:50
$begingroup$
I just realized that my question might be rephrased as "Proof that the definition of an ellipse is unique".
$endgroup$
– Nighteen
Jan 18 at 1:03
$begingroup$
I just realized that my question might be rephrased as "Proof that the definition of an ellipse is unique".
$endgroup$
– Nighteen
Jan 18 at 1:03
$begingroup$
Do you have the distance formula at your disposal? In other words, do you have the Cartesian equation for ellipses? Or must you stick to the definition you've given?
$endgroup$
– Clayton
Jan 18 at 1:12
$begingroup$
Do you have the distance formula at your disposal? In other words, do you have the Cartesian equation for ellipses? Or must you stick to the definition you've given?
$endgroup$
– Clayton
Jan 18 at 1:12
1
1
$begingroup$
The foci along with the distance $2a$ uniquely determine the major and minor axes of the ellipse.
$endgroup$
– Théophile
Jan 18 at 1:36
$begingroup$
The foci along with the distance $2a$ uniquely determine the major and minor axes of the ellipse.
$endgroup$
– Théophile
Jan 18 at 1:36
1
1
$begingroup$
@Théophile You would have to define major and minor axes as the longest and shortest diameter. You would also have to prove that if two ellipses have different axes then they are different (have one point P that is in one and not in the other ellipse).
$endgroup$
– Nighteen
Jan 18 at 1:43
$begingroup$
@Théophile You would have to define major and minor axes as the longest and shortest diameter. You would also have to prove that if two ellipses have different axes then they are different (have one point P that is in one and not in the other ellipse).
$endgroup$
– Nighteen
Jan 18 at 1:43
$begingroup$
Isn't it immediate? If the axes are different, you can (very easily) determine a point on one but not the other.
$endgroup$
– Clayton
Jan 18 at 1:50
$begingroup$
Isn't it immediate? If the axes are different, you can (very easily) determine a point on one but not the other.
$endgroup$
– Clayton
Jan 18 at 1:50
|
show 6 more comments
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$begingroup$
I just realized that my question might be rephrased as "Proof that the definition of an ellipse is unique".
$endgroup$
– Nighteen
Jan 18 at 1:03
$begingroup$
Do you have the distance formula at your disposal? In other words, do you have the Cartesian equation for ellipses? Or must you stick to the definition you've given?
$endgroup$
– Clayton
Jan 18 at 1:12
1
$begingroup$
The foci along with the distance $2a$ uniquely determine the major and minor axes of the ellipse.
$endgroup$
– Théophile
Jan 18 at 1:36
1
$begingroup$
@Théophile You would have to define major and minor axes as the longest and shortest diameter. You would also have to prove that if two ellipses have different axes then they are different (have one point P that is in one and not in the other ellipse).
$endgroup$
– Nighteen
Jan 18 at 1:43
$begingroup$
Isn't it immediate? If the axes are different, you can (very easily) determine a point on one but not the other.
$endgroup$
– Clayton
Jan 18 at 1:50