find the coordinates of the area of the largest isosceles triangle that can be inscribed in an ellipsoid

Question

for the given ellipsoid $$x^2+ \frac{y^2}{4}=1$$ I have to find the coordinates of isosceles triangle that can be inscribed in this ellipsoid. the basis of the triangle is parallel to x axis

My attempt:
I am using Lagrange multipliers for solving this question.
using some Geometry , i noticed that , because I want to find the max area, if $(x,y)$ is a given point in the triangle, it should be on the ellipse.
So, if $(x,y)$ is one of the basis coordinates, then the other coordinate of the basis is $(-x,y)$ cause we want the max area.
Now,because we looking for isosceles triangle , the last coordinate is $(0,1)$
So, we get that the area of triangle is: $$x(1+y)$$ Then, I define $f(x,y)=x(1+y)$ and $g(x,y)=x^2+ \frac{y^2}{4}-1$
and for a given $(x,y)$ $x>0$ I can find the area of the triangle.
So i have the constraint $g(x,y)$ and the function $f$ ,using Lagrange multipliers, I can solve this problem.
My question is: Did I solve this question ? because I feel something isn't working well. Thanks very much!

Answer 1

$f(x,y)=x(1+y)$,

$g(x,y)=x^2+ \frac{y^2}{4}-1$

The point lies on the ellipse thus $ 0=x^2+ \frac{y^2}{4}-1$ $$y^2=4(1-x^2)$$ $$y=\pm2\sqrt{1-x^2}$$

put this value in $f(x,y)$

$$f(x,y)=x\left(1\pm2\sqrt{(1-x^2)}\right)$$

now can you maximize this?

Answer 2

Affine maps preserve the ratios of areas, hence in order to find the largest inscribed triangles we may just apply $\psi:(x,y)\mapsto (x,2y)$ to an equilateral triangle inscribed in the unit circle. Among these triangles with maximum area there clearly are isosceles triangles, for instance $\psi(ABC)$ where $ABC$ is an equilateral triangle inscribed in the unit circle and $AB$ is parallel to the $x$- or $y$- axis. The solution is clearly not unique, as well as the maximum area is clearly $\frac{3}{2}\sqrt{3}$.