Ellipse Diagonal's Length/Equation

Question

Excuse the vagueness of this question, but how can you find the equation and distance for the diagonal of any given ellipse, that is, the line from the most-northwestern point to the most southeastern point?

The crude drawing below helps clarify:

Answer 1

Assume the ellipse is given by $$E:\quad{x^2\over a^2}+{y^2\over b^2}=1\ .\tag{1}$$ There are two ways to define the most-northwestern and most-southeastern points. The simpler idea consists in intersecting the $45^\circ$ lines $y=\pm x$ with the ellipse. I'm sure you can do this yourself.

But most probably you want the points where the tangent to $E$ is $45^\circ$ ascending. Intuitively they are the points where a $45^\circ$ line translated towards $E$ from far away first hits the ellipse. In order to determine these points we note that $E$ can be viewed as level line of the function $$f(x,y):={x^2\over a^2}+{y^2\over b^2}\ ,$$ and that at each point ${\bf z}\in E$ the gradient $$\nabla f(x,y)=\left({2x\over a^2},\>{2y\over b^2}\right)$$ is orthogonal to the tangent there. In the points ${\bf z}$ we are after we therefore have $\nabla f({\bf z})\perp(1,1)$. Tthis can be expressed by $\nabla f({\bf z})\cdot(1,1)=0$, or $${2x\over a^2}+{2y\over b^2}=0\ .\tag{2}$$ Equations $(1)$ and $(2)$ together determine the two points ${\bf z}_1$, ${\bf z}_2$ in question.

Answer 2

Edit: Modified example

for example you have an equation for an ellipse like this: $k=2x^2+8y^2+1$.

k is now 9.

$9=2x^2+8y^2+5x-3y+5$

To find the endings of the axes you can use the Lagrangian Multiplier Method. The Funktion is here $\mathfrak L=x^2+y^2+\lambda (-4+2x^2+8y^2+5x-3y+5)$

If you max/min this function you will get the four coordinates for the endings of the two axis.

greetings,

calculus