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Is there an elementary proof for the diameter of an ellipse?
How to find the center of an (scaled) ellipse?Finding the eccentricity/focus/directrix of ellipses and hyperbolas under some rotationFinding the point on a rotated ellipse corresponding to a given tangential angleLongest parallel chord of an ellipseUsing the Pin-And-String Method to create parametric equation for an ellipseProperty of ellipses involving normals at the endpoints of a focal chord and the midpoint of that chordFit an ellipse with known semi-major-axis and pointsHow to find ellipse circumference using 5 points?Determining the normal of an ellipseGraphically locate the axes or foci of an ellipse from 5 arbitrary points on its perimeter.
$begingroup$
Let $A=(x,y) in mathbb R^2 , $ be an ellipse. ($a,b>0$). It is a known fact that the diameter of $A$, is $2maxa,b$. This can be proved via Lagrange multiplier's method.
Is there a more elementary proof? (not using Lagrange multiplier's method). Is there a proof without calculus at all?
I think that one can imagine the following process: Start with an arbitrary point $p_0$ on the ellipse, and connect it with a chord to its "antipodal point"- that is the point on the ellipse that maximizes the distance from $p_0$. (we can sort of know where it should be "visually", since the connecting chord should be orthogonal to the ellipse at that point).
Now, moving continuously the initial point $p_0$, it is rather intuitive that the longest chord should be the major axis.
Of course, this is not a rigorous proof.
multivariable-calculus optimization analytic-geometry conic-sections maxima-minima
$endgroup$
add a comment |
$begingroup$
Let $A=(x,y) in mathbb R^2 , $ be an ellipse. ($a,b>0$). It is a known fact that the diameter of $A$, is $2maxa,b$. This can be proved via Lagrange multiplier's method.
Is there a more elementary proof? (not using Lagrange multiplier's method). Is there a proof without calculus at all?
I think that one can imagine the following process: Start with an arbitrary point $p_0$ on the ellipse, and connect it with a chord to its "antipodal point"- that is the point on the ellipse that maximizes the distance from $p_0$. (we can sort of know where it should be "visually", since the connecting chord should be orthogonal to the ellipse at that point).
Now, moving continuously the initial point $p_0$, it is rather intuitive that the longest chord should be the major axis.
Of course, this is not a rigorous proof.
multivariable-calculus optimization analytic-geometry conic-sections maxima-minima
$endgroup$
add a comment |
$begingroup$
Let $A=(x,y) in mathbb R^2 , $ be an ellipse. ($a,b>0$). It is a known fact that the diameter of $A$, is $2maxa,b$. This can be proved via Lagrange multiplier's method.
Is there a more elementary proof? (not using Lagrange multiplier's method). Is there a proof without calculus at all?
I think that one can imagine the following process: Start with an arbitrary point $p_0$ on the ellipse, and connect it with a chord to its "antipodal point"- that is the point on the ellipse that maximizes the distance from $p_0$. (we can sort of know where it should be "visually", since the connecting chord should be orthogonal to the ellipse at that point).
Now, moving continuously the initial point $p_0$, it is rather intuitive that the longest chord should be the major axis.
Of course, this is not a rigorous proof.
multivariable-calculus optimization analytic-geometry conic-sections maxima-minima
$endgroup$
Let $A=(x,y) in mathbb R^2 , $ be an ellipse. ($a,b>0$). It is a known fact that the diameter of $A$, is $2maxa,b$. This can be proved via Lagrange multiplier's method.
Is there a more elementary proof? (not using Lagrange multiplier's method). Is there a proof without calculus at all?
I think that one can imagine the following process: Start with an arbitrary point $p_0$ on the ellipse, and connect it with a chord to its "antipodal point"- that is the point on the ellipse that maximizes the distance from $p_0$. (we can sort of know where it should be "visually", since the connecting chord should be orthogonal to the ellipse at that point).
Now, moving continuously the initial point $p_0$, it is rather intuitive that the longest chord should be the major axis.
Of course, this is not a rigorous proof.
multivariable-calculus optimization analytic-geometry conic-sections maxima-minima
multivariable-calculus optimization analytic-geometry conic-sections maxima-minima
asked Mar 14 at 7:58
Asaf ShacharAsaf Shachar
5,77931144
5,77931144
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Geometrically, the map $xmapsto x/a$ stretches the $x$ axis by $a$, and similarly $ymapsto y/b$ stretches the $y$ axis by a factor of $b$. Thus the graph of $(x/a)^2+(y/b)^2=1$ is the graph of the circle, but stretched in the $x,y$ directions by a factor of $a,b$ respectively. The conclusion follows.
$endgroup$
$begingroup$
To tie all the loose ends, it is easy to show that for any two points $p,q$ at distance $d=|p-q|$, the transformed points $p',q'$ are at distance $d'=|p'-q'|$ with $bdle d'le ad$.
$endgroup$
– Rahul
Mar 14 at 8:59
$begingroup$
@Rahul Can you please elaborate on that? I do not see it...
$endgroup$
– Asaf Shachar
Mar 14 at 9:50
$begingroup$
Thank you for this answer. Can you please elaborate on how the conclusion follows from this geometric description? I don't see that immediately.
$endgroup$
– Asaf Shachar
Mar 14 at 9:51
1
$begingroup$
@Asaf: Let's say $p-q=(x,y)$. Then $d'=sqrt(ax)^2+(by)^2lesqrt(ax)^2+(ay)^2=asqrtx^2+y^2=ad$ (sorry, I assumed $age b$ in my comment). Similarly one can show $d'ge bd$.
$endgroup$
– Rahul
Mar 14 at 10:12
$begingroup$
@AsafShachar assuming $ageq b$, we know the antipodal points $(pm 1,0)$ have become $2a$ apart under the map. To see that it is the longest segment through the origin, note that $(x,y)$ with $x^2+y^2=1$ is mapped to $(ax,by)$ so that the distance from the origin to that point is $sqrta^2x^2+b^2y^2leqsqrta^2(x^2+y^2)=a$. So the segment on the $x$ axis maximises the length to the origin.
$endgroup$
– YiFan
Mar 14 at 14:16
add a comment |
$begingroup$
$(fracxa)^2 +(fracyb)^2=1$,
an ellipse centered at the origin.
Set $x= a cos t$; $y= b sin t$, $0 le t <2π$.
Note :
$x(t+π)=-x$, $y(t+π)= -y$;
the points $(x(t), y(t))$ and $(x(t+π),y(t+t))$ lie on the line $y= mx$ , $m= tan t$, on 'opposite' sides of the ellipse, and are equidistant from the origin.
Given the symmetry about the $y-$axis it is sufficient to consider the first quadrant, and find $r=(d/2).$
Distance $r$ from origin of a point $(x,y)$ on the ellipse:
$r^2=a^2cos ^2 t +b^2 sin^2 t.$
1) $a ge b$:
$r^2 = a^2- a^2sin^2 t +b^2 sin^2 t$;
$r^2= a^2 -(a^2-b^2)sin^2 t le a^2$.
$r_max^2 = a^2$, or $d=2r= 2a$;
Similarly:
2) $a lt b$:
$r^2 = b^2-(b^2-a^2)cos ^2 t le b^2$;
$r_max^2 = b^2$, or $d= 2r=2b$.
Hence $d=max (2a,2b)$.
$endgroup$
$begingroup$
Max.Thanks for suggestions.
$endgroup$
– Peter Szilas
Mar 14 at 9:45
add a comment |
$begingroup$
An ellipse $mathscrE$ is a smooth, convex, centrally-symmetric shape. Assume that $A,BinmathscrE$ are the endpoints of a diameter. Then $AB$ has to be perpendicular to the tangent at $A$ and the tangent at $B$, hence such tangents have to be parallel to each other. This implies that $AB$ goes through the center of the ellipse, hence $B$ is the symmetric of $A$ with respect to the center, and the determination of the longest chord in an ellipse is equivalent to the maximization problem
$$ max_fracx^2a^2+fracy^2b^2=1 (x^2+y^2) $$
or to the determination of the eigenvalues/eigenvectors of the symmetric matrix $beginpmatrixfrac1a&0\ 0&frac1bendpmatrix$, which are the trivial ones.
$endgroup$
add a comment |
$begingroup$
There sure is. Let $AB$ be a chord of the ellipse and let $ F_1$ and $F_2$ be the foci, as pictured in the diagram.
let $l$ be the focal length of the ellipse. Then by definition $AF_1+AF_2=l$ and $BF_1 +BF_2=l$. If $A$, $B$, $F_1$ and $F_2$ are not all collinear, then by the triangle inequality at least one of $ AB<AF_1+BF_1$ and $AB<AF_2+BF_2$ is true. Hence $$2l= AF_1 +AF_2 +BF_1+BF_2>2AB$$
$$l>AB$$
On the other hand if $A$, $B$, $F_1$ and $F_2$ are all collinear and $A ne B$
$$2l= AF_1 +AF_2 +BF_1+BF_2=2AB$$
$$l=AB$$
Placing coordinates so that the foci lie on the $x$ axis and the ellipse is $fracx^2a^2 + fracy^2b^2=1$. It is clear that the collinear case has $A$ and $B$ as the $x$ intercepts, $(a,0)$ and $(-a,0)$ , hence $l=2a$ and $AB < 2a$ whenever it does not pass through both foci.
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Geometrically, the map $xmapsto x/a$ stretches the $x$ axis by $a$, and similarly $ymapsto y/b$ stretches the $y$ axis by a factor of $b$. Thus the graph of $(x/a)^2+(y/b)^2=1$ is the graph of the circle, but stretched in the $x,y$ directions by a factor of $a,b$ respectively. The conclusion follows.
$endgroup$
$begingroup$
To tie all the loose ends, it is easy to show that for any two points $p,q$ at distance $d=|p-q|$, the transformed points $p',q'$ are at distance $d'=|p'-q'|$ with $bdle d'le ad$.
$endgroup$
– Rahul
Mar 14 at 8:59
$begingroup$
@Rahul Can you please elaborate on that? I do not see it...
$endgroup$
– Asaf Shachar
Mar 14 at 9:50
$begingroup$
Thank you for this answer. Can you please elaborate on how the conclusion follows from this geometric description? I don't see that immediately.
$endgroup$
– Asaf Shachar
Mar 14 at 9:51
1
$begingroup$
@Asaf: Let's say $p-q=(x,y)$. Then $d'=sqrt(ax)^2+(by)^2lesqrt(ax)^2+(ay)^2=asqrtx^2+y^2=ad$ (sorry, I assumed $age b$ in my comment). Similarly one can show $d'ge bd$.
$endgroup$
– Rahul
Mar 14 at 10:12
$begingroup$
@AsafShachar assuming $ageq b$, we know the antipodal points $(pm 1,0)$ have become $2a$ apart under the map. To see that it is the longest segment through the origin, note that $(x,y)$ with $x^2+y^2=1$ is mapped to $(ax,by)$ so that the distance from the origin to that point is $sqrta^2x^2+b^2y^2leqsqrta^2(x^2+y^2)=a$. So the segment on the $x$ axis maximises the length to the origin.
$endgroup$
– YiFan
Mar 14 at 14:16
add a comment |
$begingroup$
Geometrically, the map $xmapsto x/a$ stretches the $x$ axis by $a$, and similarly $ymapsto y/b$ stretches the $y$ axis by a factor of $b$. Thus the graph of $(x/a)^2+(y/b)^2=1$ is the graph of the circle, but stretched in the $x,y$ directions by a factor of $a,b$ respectively. The conclusion follows.
$endgroup$
$begingroup$
To tie all the loose ends, it is easy to show that for any two points $p,q$ at distance $d=|p-q|$, the transformed points $p',q'$ are at distance $d'=|p'-q'|$ with $bdle d'le ad$.
$endgroup$
– Rahul
Mar 14 at 8:59
$begingroup$
@Rahul Can you please elaborate on that? I do not see it...
$endgroup$
– Asaf Shachar
Mar 14 at 9:50
$begingroup$
Thank you for this answer. Can you please elaborate on how the conclusion follows from this geometric description? I don't see that immediately.
$endgroup$
– Asaf Shachar
Mar 14 at 9:51
1
$begingroup$
@Asaf: Let's say $p-q=(x,y)$. Then $d'=sqrt(ax)^2+(by)^2lesqrt(ax)^2+(ay)^2=asqrtx^2+y^2=ad$ (sorry, I assumed $age b$ in my comment). Similarly one can show $d'ge bd$.
$endgroup$
– Rahul
Mar 14 at 10:12
$begingroup$
@AsafShachar assuming $ageq b$, we know the antipodal points $(pm 1,0)$ have become $2a$ apart under the map. To see that it is the longest segment through the origin, note that $(x,y)$ with $x^2+y^2=1$ is mapped to $(ax,by)$ so that the distance from the origin to that point is $sqrta^2x^2+b^2y^2leqsqrta^2(x^2+y^2)=a$. So the segment on the $x$ axis maximises the length to the origin.
$endgroup$
– YiFan
Mar 14 at 14:16
add a comment |
$begingroup$
Geometrically, the map $xmapsto x/a$ stretches the $x$ axis by $a$, and similarly $ymapsto y/b$ stretches the $y$ axis by a factor of $b$. Thus the graph of $(x/a)^2+(y/b)^2=1$ is the graph of the circle, but stretched in the $x,y$ directions by a factor of $a,b$ respectively. The conclusion follows.
$endgroup$
Geometrically, the map $xmapsto x/a$ stretches the $x$ axis by $a$, and similarly $ymapsto y/b$ stretches the $y$ axis by a factor of $b$. Thus the graph of $(x/a)^2+(y/b)^2=1$ is the graph of the circle, but stretched in the $x,y$ directions by a factor of $a,b$ respectively. The conclusion follows.
answered Mar 14 at 8:30
YiFanYiFan
4,7531727
4,7531727
$begingroup$
To tie all the loose ends, it is easy to show that for any two points $p,q$ at distance $d=|p-q|$, the transformed points $p',q'$ are at distance $d'=|p'-q'|$ with $bdle d'le ad$.
$endgroup$
– Rahul
Mar 14 at 8:59
$begingroup$
@Rahul Can you please elaborate on that? I do not see it...
$endgroup$
– Asaf Shachar
Mar 14 at 9:50
$begingroup$
Thank you for this answer. Can you please elaborate on how the conclusion follows from this geometric description? I don't see that immediately.
$endgroup$
– Asaf Shachar
Mar 14 at 9:51
1
$begingroup$
@Asaf: Let's say $p-q=(x,y)$. Then $d'=sqrt(ax)^2+(by)^2lesqrt(ax)^2+(ay)^2=asqrtx^2+y^2=ad$ (sorry, I assumed $age b$ in my comment). Similarly one can show $d'ge bd$.
$endgroup$
– Rahul
Mar 14 at 10:12
$begingroup$
@AsafShachar assuming $ageq b$, we know the antipodal points $(pm 1,0)$ have become $2a$ apart under the map. To see that it is the longest segment through the origin, note that $(x,y)$ with $x^2+y^2=1$ is mapped to $(ax,by)$ so that the distance from the origin to that point is $sqrta^2x^2+b^2y^2leqsqrta^2(x^2+y^2)=a$. So the segment on the $x$ axis maximises the length to the origin.
$endgroup$
– YiFan
Mar 14 at 14:16
add a comment |
$begingroup$
To tie all the loose ends, it is easy to show that for any two points $p,q$ at distance $d=|p-q|$, the transformed points $p',q'$ are at distance $d'=|p'-q'|$ with $bdle d'le ad$.
$endgroup$
– Rahul
Mar 14 at 8:59
$begingroup$
@Rahul Can you please elaborate on that? I do not see it...
$endgroup$
– Asaf Shachar
Mar 14 at 9:50
$begingroup$
Thank you for this answer. Can you please elaborate on how the conclusion follows from this geometric description? I don't see that immediately.
$endgroup$
– Asaf Shachar
Mar 14 at 9:51
1
$begingroup$
@Asaf: Let's say $p-q=(x,y)$. Then $d'=sqrt(ax)^2+(by)^2lesqrt(ax)^2+(ay)^2=asqrtx^2+y^2=ad$ (sorry, I assumed $age b$ in my comment). Similarly one can show $d'ge bd$.
$endgroup$
– Rahul
Mar 14 at 10:12
$begingroup$
@AsafShachar assuming $ageq b$, we know the antipodal points $(pm 1,0)$ have become $2a$ apart under the map. To see that it is the longest segment through the origin, note that $(x,y)$ with $x^2+y^2=1$ is mapped to $(ax,by)$ so that the distance from the origin to that point is $sqrta^2x^2+b^2y^2leqsqrta^2(x^2+y^2)=a$. So the segment on the $x$ axis maximises the length to the origin.
$endgroup$
– YiFan
Mar 14 at 14:16
$begingroup$
To tie all the loose ends, it is easy to show that for any two points $p,q$ at distance $d=|p-q|$, the transformed points $p',q'$ are at distance $d'=|p'-q'|$ with $bdle d'le ad$.
$endgroup$
– Rahul
Mar 14 at 8:59
$begingroup$
To tie all the loose ends, it is easy to show that for any two points $p,q$ at distance $d=|p-q|$, the transformed points $p',q'$ are at distance $d'=|p'-q'|$ with $bdle d'le ad$.
$endgroup$
– Rahul
Mar 14 at 8:59
$begingroup$
@Rahul Can you please elaborate on that? I do not see it...
$endgroup$
– Asaf Shachar
Mar 14 at 9:50
$begingroup$
@Rahul Can you please elaborate on that? I do not see it...
$endgroup$
– Asaf Shachar
Mar 14 at 9:50
$begingroup$
Thank you for this answer. Can you please elaborate on how the conclusion follows from this geometric description? I don't see that immediately.
$endgroup$
– Asaf Shachar
Mar 14 at 9:51
$begingroup$
Thank you for this answer. Can you please elaborate on how the conclusion follows from this geometric description? I don't see that immediately.
$endgroup$
– Asaf Shachar
Mar 14 at 9:51
1
1
$begingroup$
@Asaf: Let's say $p-q=(x,y)$. Then $d'=sqrt(ax)^2+(by)^2lesqrt(ax)^2+(ay)^2=asqrtx^2+y^2=ad$ (sorry, I assumed $age b$ in my comment). Similarly one can show $d'ge bd$.
$endgroup$
– Rahul
Mar 14 at 10:12
$begingroup$
@Asaf: Let's say $p-q=(x,y)$. Then $d'=sqrt(ax)^2+(by)^2lesqrt(ax)^2+(ay)^2=asqrtx^2+y^2=ad$ (sorry, I assumed $age b$ in my comment). Similarly one can show $d'ge bd$.
$endgroup$
– Rahul
Mar 14 at 10:12
$begingroup$
@AsafShachar assuming $ageq b$, we know the antipodal points $(pm 1,0)$ have become $2a$ apart under the map. To see that it is the longest segment through the origin, note that $(x,y)$ with $x^2+y^2=1$ is mapped to $(ax,by)$ so that the distance from the origin to that point is $sqrta^2x^2+b^2y^2leqsqrta^2(x^2+y^2)=a$. So the segment on the $x$ axis maximises the length to the origin.
$endgroup$
– YiFan
Mar 14 at 14:16
$begingroup$
@AsafShachar assuming $ageq b$, we know the antipodal points $(pm 1,0)$ have become $2a$ apart under the map. To see that it is the longest segment through the origin, note that $(x,y)$ with $x^2+y^2=1$ is mapped to $(ax,by)$ so that the distance from the origin to that point is $sqrta^2x^2+b^2y^2leqsqrta^2(x^2+y^2)=a$. So the segment on the $x$ axis maximises the length to the origin.
$endgroup$
– YiFan
Mar 14 at 14:16
add a comment |
$begingroup$
$(fracxa)^2 +(fracyb)^2=1$,
an ellipse centered at the origin.
Set $x= a cos t$; $y= b sin t$, $0 le t <2π$.
Note :
$x(t+π)=-x$, $y(t+π)= -y$;
the points $(x(t), y(t))$ and $(x(t+π),y(t+t))$ lie on the line $y= mx$ , $m= tan t$, on 'opposite' sides of the ellipse, and are equidistant from the origin.
Given the symmetry about the $y-$axis it is sufficient to consider the first quadrant, and find $r=(d/2).$
Distance $r$ from origin of a point $(x,y)$ on the ellipse:
$r^2=a^2cos ^2 t +b^2 sin^2 t.$
1) $a ge b$:
$r^2 = a^2- a^2sin^2 t +b^2 sin^2 t$;
$r^2= a^2 -(a^2-b^2)sin^2 t le a^2$.
$r_max^2 = a^2$, or $d=2r= 2a$;
Similarly:
2) $a lt b$:
$r^2 = b^2-(b^2-a^2)cos ^2 t le b^2$;
$r_max^2 = b^2$, or $d= 2r=2b$.
Hence $d=max (2a,2b)$.
$endgroup$
$begingroup$
Max.Thanks for suggestions.
$endgroup$
– Peter Szilas
Mar 14 at 9:45
add a comment |
$begingroup$
$(fracxa)^2 +(fracyb)^2=1$,
an ellipse centered at the origin.
Set $x= a cos t$; $y= b sin t$, $0 le t <2π$.
Note :
$x(t+π)=-x$, $y(t+π)= -y$;
the points $(x(t), y(t))$ and $(x(t+π),y(t+t))$ lie on the line $y= mx$ , $m= tan t$, on 'opposite' sides of the ellipse, and are equidistant from the origin.
Given the symmetry about the $y-$axis it is sufficient to consider the first quadrant, and find $r=(d/2).$
Distance $r$ from origin of a point $(x,y)$ on the ellipse:
$r^2=a^2cos ^2 t +b^2 sin^2 t.$
1) $a ge b$:
$r^2 = a^2- a^2sin^2 t +b^2 sin^2 t$;
$r^2= a^2 -(a^2-b^2)sin^2 t le a^2$.
$r_max^2 = a^2$, or $d=2r= 2a$;
Similarly:
2) $a lt b$:
$r^2 = b^2-(b^2-a^2)cos ^2 t le b^2$;
$r_max^2 = b^2$, or $d= 2r=2b$.
Hence $d=max (2a,2b)$.
$endgroup$
$begingroup$
Max.Thanks for suggestions.
$endgroup$
– Peter Szilas
Mar 14 at 9:45
add a comment |
$begingroup$
$(fracxa)^2 +(fracyb)^2=1$,
an ellipse centered at the origin.
Set $x= a cos t$; $y= b sin t$, $0 le t <2π$.
Note :
$x(t+π)=-x$, $y(t+π)= -y$;
the points $(x(t), y(t))$ and $(x(t+π),y(t+t))$ lie on the line $y= mx$ , $m= tan t$, on 'opposite' sides of the ellipse, and are equidistant from the origin.
Given the symmetry about the $y-$axis it is sufficient to consider the first quadrant, and find $r=(d/2).$
Distance $r$ from origin of a point $(x,y)$ on the ellipse:
$r^2=a^2cos ^2 t +b^2 sin^2 t.$
1) $a ge b$:
$r^2 = a^2- a^2sin^2 t +b^2 sin^2 t$;
$r^2= a^2 -(a^2-b^2)sin^2 t le a^2$.
$r_max^2 = a^2$, or $d=2r= 2a$;
Similarly:
2) $a lt b$:
$r^2 = b^2-(b^2-a^2)cos ^2 t le b^2$;
$r_max^2 = b^2$, or $d= 2r=2b$.
Hence $d=max (2a,2b)$.
$endgroup$
$(fracxa)^2 +(fracyb)^2=1$,
an ellipse centered at the origin.
Set $x= a cos t$; $y= b sin t$, $0 le t <2π$.
Note :
$x(t+π)=-x$, $y(t+π)= -y$;
the points $(x(t), y(t))$ and $(x(t+π),y(t+t))$ lie on the line $y= mx$ , $m= tan t$, on 'opposite' sides of the ellipse, and are equidistant from the origin.
Given the symmetry about the $y-$axis it is sufficient to consider the first quadrant, and find $r=(d/2).$
Distance $r$ from origin of a point $(x,y)$ on the ellipse:
$r^2=a^2cos ^2 t +b^2 sin^2 t.$
1) $a ge b$:
$r^2 = a^2- a^2sin^2 t +b^2 sin^2 t$;
$r^2= a^2 -(a^2-b^2)sin^2 t le a^2$.
$r_max^2 = a^2$, or $d=2r= 2a$;
Similarly:
2) $a lt b$:
$r^2 = b^2-(b^2-a^2)cos ^2 t le b^2$;
$r_max^2 = b^2$, or $d= 2r=2b$.
Hence $d=max (2a,2b)$.
edited Mar 14 at 10:11
Max
9011318
9011318
answered Mar 14 at 8:50
Peter SzilasPeter Szilas
11.6k2822
11.6k2822
$begingroup$
Max.Thanks for suggestions.
$endgroup$
– Peter Szilas
Mar 14 at 9:45
add a comment |
$begingroup$
Max.Thanks for suggestions.
$endgroup$
– Peter Szilas
Mar 14 at 9:45
$begingroup$
Max.Thanks for suggestions.
$endgroup$
– Peter Szilas
Mar 14 at 9:45
$begingroup$
Max.Thanks for suggestions.
$endgroup$
– Peter Szilas
Mar 14 at 9:45
add a comment |
$begingroup$
An ellipse $mathscrE$ is a smooth, convex, centrally-symmetric shape. Assume that $A,BinmathscrE$ are the endpoints of a diameter. Then $AB$ has to be perpendicular to the tangent at $A$ and the tangent at $B$, hence such tangents have to be parallel to each other. This implies that $AB$ goes through the center of the ellipse, hence $B$ is the symmetric of $A$ with respect to the center, and the determination of the longest chord in an ellipse is equivalent to the maximization problem
$$ max_fracx^2a^2+fracy^2b^2=1 (x^2+y^2) $$
or to the determination of the eigenvalues/eigenvectors of the symmetric matrix $beginpmatrixfrac1a&0\ 0&frac1bendpmatrix$, which are the trivial ones.
$endgroup$
add a comment |
$begingroup$
An ellipse $mathscrE$ is a smooth, convex, centrally-symmetric shape. Assume that $A,BinmathscrE$ are the endpoints of a diameter. Then $AB$ has to be perpendicular to the tangent at $A$ and the tangent at $B$, hence such tangents have to be parallel to each other. This implies that $AB$ goes through the center of the ellipse, hence $B$ is the symmetric of $A$ with respect to the center, and the determination of the longest chord in an ellipse is equivalent to the maximization problem
$$ max_fracx^2a^2+fracy^2b^2=1 (x^2+y^2) $$
or to the determination of the eigenvalues/eigenvectors of the symmetric matrix $beginpmatrixfrac1a&0\ 0&frac1bendpmatrix$, which are the trivial ones.
$endgroup$
add a comment |
$begingroup$
An ellipse $mathscrE$ is a smooth, convex, centrally-symmetric shape. Assume that $A,BinmathscrE$ are the endpoints of a diameter. Then $AB$ has to be perpendicular to the tangent at $A$ and the tangent at $B$, hence such tangents have to be parallel to each other. This implies that $AB$ goes through the center of the ellipse, hence $B$ is the symmetric of $A$ with respect to the center, and the determination of the longest chord in an ellipse is equivalent to the maximization problem
$$ max_fracx^2a^2+fracy^2b^2=1 (x^2+y^2) $$
or to the determination of the eigenvalues/eigenvectors of the symmetric matrix $beginpmatrixfrac1a&0\ 0&frac1bendpmatrix$, which are the trivial ones.
$endgroup$
An ellipse $mathscrE$ is a smooth, convex, centrally-symmetric shape. Assume that $A,BinmathscrE$ are the endpoints of a diameter. Then $AB$ has to be perpendicular to the tangent at $A$ and the tangent at $B$, hence such tangents have to be parallel to each other. This implies that $AB$ goes through the center of the ellipse, hence $B$ is the symmetric of $A$ with respect to the center, and the determination of the longest chord in an ellipse is equivalent to the maximization problem
$$ max_fracx^2a^2+fracy^2b^2=1 (x^2+y^2) $$
or to the determination of the eigenvalues/eigenvectors of the symmetric matrix $beginpmatrixfrac1a&0\ 0&frac1bendpmatrix$, which are the trivial ones.
answered Mar 14 at 13:50
Jack D'AurizioJack D'Aurizio
291k33284669
291k33284669
add a comment |
add a comment |
$begingroup$
There sure is. Let $AB$ be a chord of the ellipse and let $ F_1$ and $F_2$ be the foci, as pictured in the diagram.
let $l$ be the focal length of the ellipse. Then by definition $AF_1+AF_2=l$ and $BF_1 +BF_2=l$. If $A$, $B$, $F_1$ and $F_2$ are not all collinear, then by the triangle inequality at least one of $ AB<AF_1+BF_1$ and $AB<AF_2+BF_2$ is true. Hence $$2l= AF_1 +AF_2 +BF_1+BF_2>2AB$$
$$l>AB$$
On the other hand if $A$, $B$, $F_1$ and $F_2$ are all collinear and $A ne B$
$$2l= AF_1 +AF_2 +BF_1+BF_2=2AB$$
$$l=AB$$
Placing coordinates so that the foci lie on the $x$ axis and the ellipse is $fracx^2a^2 + fracy^2b^2=1$. It is clear that the collinear case has $A$ and $B$ as the $x$ intercepts, $(a,0)$ and $(-a,0)$ , hence $l=2a$ and $AB < 2a$ whenever it does not pass through both foci.
$endgroup$
add a comment |
$begingroup$
There sure is. Let $AB$ be a chord of the ellipse and let $ F_1$ and $F_2$ be the foci, as pictured in the diagram.
let $l$ be the focal length of the ellipse. Then by definition $AF_1+AF_2=l$ and $BF_1 +BF_2=l$. If $A$, $B$, $F_1$ and $F_2$ are not all collinear, then by the triangle inequality at least one of $ AB<AF_1+BF_1$ and $AB<AF_2+BF_2$ is true. Hence $$2l= AF_1 +AF_2 +BF_1+BF_2>2AB$$
$$l>AB$$
On the other hand if $A$, $B$, $F_1$ and $F_2$ are all collinear and $A ne B$
$$2l= AF_1 +AF_2 +BF_1+BF_2=2AB$$
$$l=AB$$
Placing coordinates so that the foci lie on the $x$ axis and the ellipse is $fracx^2a^2 + fracy^2b^2=1$. It is clear that the collinear case has $A$ and $B$ as the $x$ intercepts, $(a,0)$ and $(-a,0)$ , hence $l=2a$ and $AB < 2a$ whenever it does not pass through both foci.
$endgroup$
add a comment |
$begingroup$
There sure is. Let $AB$ be a chord of the ellipse and let $ F_1$ and $F_2$ be the foci, as pictured in the diagram.
let $l$ be the focal length of the ellipse. Then by definition $AF_1+AF_2=l$ and $BF_1 +BF_2=l$. If $A$, $B$, $F_1$ and $F_2$ are not all collinear, then by the triangle inequality at least one of $ AB<AF_1+BF_1$ and $AB<AF_2+BF_2$ is true. Hence $$2l= AF_1 +AF_2 +BF_1+BF_2>2AB$$
$$l>AB$$
On the other hand if $A$, $B$, $F_1$ and $F_2$ are all collinear and $A ne B$
$$2l= AF_1 +AF_2 +BF_1+BF_2=2AB$$
$$l=AB$$
Placing coordinates so that the foci lie on the $x$ axis and the ellipse is $fracx^2a^2 + fracy^2b^2=1$. It is clear that the collinear case has $A$ and $B$ as the $x$ intercepts, $(a,0)$ and $(-a,0)$ , hence $l=2a$ and $AB < 2a$ whenever it does not pass through both foci.
$endgroup$
There sure is. Let $AB$ be a chord of the ellipse and let $ F_1$ and $F_2$ be the foci, as pictured in the diagram.
let $l$ be the focal length of the ellipse. Then by definition $AF_1+AF_2=l$ and $BF_1 +BF_2=l$. If $A$, $B$, $F_1$ and $F_2$ are not all collinear, then by the triangle inequality at least one of $ AB<AF_1+BF_1$ and $AB<AF_2+BF_2$ is true. Hence $$2l= AF_1 +AF_2 +BF_1+BF_2>2AB$$
$$l>AB$$
On the other hand if $A$, $B$, $F_1$ and $F_2$ are all collinear and $A ne B$
$$2l= AF_1 +AF_2 +BF_1+BF_2=2AB$$
$$l=AB$$
Placing coordinates so that the foci lie on the $x$ axis and the ellipse is $fracx^2a^2 + fracy^2b^2=1$. It is clear that the collinear case has $A$ and $B$ as the $x$ intercepts, $(a,0)$ and $(-a,0)$ , hence $l=2a$ and $AB < 2a$ whenever it does not pass through both foci.
edited Mar 15 at 9:46
answered Mar 14 at 9:00
B.MartinB.Martin
4121210
4121210
add a comment |
add a comment |
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