Is there an elementary proof for the diameter of an ellipse?

Question

Let $A=\{(x,y) \in \mathbb R^2 \, | \, (\frac{x}{a})^2+(\frac{y}{b})^2=1\}$ be an ellipse. ($a,b>0$). It is a known fact that the diameter of $A$, is $2\max\{a,b\}$. 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.

Answer 1

Geometrically, the map $x\mapsto x/a$ stretches the $x$ axis by $a$, and similarly $y\mapsto y/b$ 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.

Answer 2

$(\frac{x}{a})^2 +(\frac{y}{b})^2=1$,

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:

$r^2=a^2\cos ^2 t +b^2 \sin^2 t.$

1) $a \ge b$:

$r^2 = a^2- a^2\sin^2 t +b^2 \sin^2 t$;

$r^2= a^2 -(a^2-b^2)\sin^2 t \le a^2$.

$r_{max}^2 = a^2$, 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 $d=\max (2a,2b)$.

Answer 3

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 $\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$. 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.

Answer 4

An ellipse $\mathscr{E}$ is a smooth, convex, centrally-symmetric shape. Assume that $A,B\in\mathscr{E}$ 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 $$ \max_{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1} (x^2+y^2) $$ or to the determination of the eigenvalues/eigenvectors of the symmetric matrix $\begin{pmatrix}\frac{1}{a}&0\\ 0&\frac{1}{b}\end{pmatrix}$, which are the trivial ones.