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!