Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
You know how when you construct an ellipse, you take a rope, fix it to 2 points, and stretch that rope?
When the rope is being stretched, let's call the part of the string attached to the first point d1, and the part of the string attached to the second point d2 (I know it's the same string be just bear with me).
My question is, why is the highest point of the ellipse formed when d1=d2? Why is this the case?
One approach which is somewhat roundabout but avoids calculus would be:
First prove that an ellipse is a squished circle. This shows that the ellipse has a single unique highest point (because certainly a circle does, and squishing the $y$-coordinates uniformly everywhere doesn't change which points are higher than others).
However, if the point with distances $d_1$ and $d_2$ is highest, then by symmetry the point with distances $d_2$ and $d_1$ is just as high. And if $d_1\ne d_2$ they are different points and can't both be highest, a contradiction.
The height of a point $P$ of the ellipse is given by the intersection of the two circles of equations $$\begin{cases} (x+d/2)^2 + y^2 = d_1^2\\ (x-d/2)^2 + y^2 = d_2^2 \end{cases}$$
where $(-d/2,0), (d/2,0)$ are the points where the strings are attached. And $d_1+d_2=a$ is a constant (the length of the string) larger than $d$. You're looking to the maximum of $y$ and want to prove that it is obtained when $x=0$.
So $$f(x,y)=\sqrt{(x+d/2)^2 + y^2} + \sqrt{(x-d/2)^2 + y^2}=a$$
$y$ is implicitely defined through $f$ as a function of $x$. You have $$\begin{cases} \frac{\partial f}{\partial x} = \frac{x+d/2}{\sqrt{(x+d/2)^2 + y^2}} + \frac{x-d/2}{\sqrt{(x-d/2)^2 + y^2}} \\ \frac{\partial f}{\partial y} = \frac{y}{\sqrt{(x+d/2)^2 + y^2}} + \frac{y}{\sqrt{(x-d/2)^2 + y^2}} \end{cases}$$
The General formula for derivative of implicit function tells you that $$y^\prime(x) = - \frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}$$ Therefore $y^\prime(x)$ vanishes when $\frac{\partial f}{\partial x}$ vanishes, that is for $x=0$ as desired. And in that case, $d_1^2=d_2^2=d^2/4+y^2 = a^2/4$.
Without calculus.
1) The "highest point" is reached when the tangent of the ellipse at that point is "horizontal", i.e. parallel to the line containing foci $A$ and $B$.
2) The line perpendicular to the tangent at $P$ (aka "normal") is the bisector of $\angle APB$ (this can be proved without calculus).
3) Hence the highest point $P$ is that with bisector of $APB$ perpendicular to $AB$, that is when $AP=BP$.
In an optimization problem ( calculus of variations) an extremization of $y$ coordinate
$$ y= b \sqrt{1- (x/a)^2} $$
is possible when subject to a constraint function $ (d_1+d_2)=$
$$ \sqrt{(x-c)^2 + y^2 }+ \sqrt{( (x+c)^2 + y^2 } $$
with $ a^2= b^2+ c^2 $ we see that happens when contributions of $d_1$ and $d_2$ are equal.
It is like saying that area of a rectangle of given perimeter is maximum when length equals breadth.
If a triangle with base vertices at the two foci $B,C$ of the ellipse is not isosceles then the angle bisector at the top vertex $A$ is different from the altitute at $A$. Therefore the line perpendicular to the angle bisector, which is the tangent line to the resulting ellipse, is not horizontal. Moving $A$ an infinitesimal amount along the tangent line will result in only a second order change in the sum of distances to $B$ and $C$, but it will result in a first-order change in the altitude. Therefore at a point of maximum, triangle $ABC$ has to be isosceles.
