An ellipse, whose equation is ${x^2\over9} + {y^2\over4} = 1$, is inscribed within a rectangle whose sides are parallel with the coordinate axes. Another ellipse is circumscribing the rectangle and passes through the point (0, 4). I am asked to find the eccentricity of the ellipse circumscribing the rectangle.
Is there any property which links the two ellipses? For example, I tried to check whether they would have the same focus, but that didn't come out to be true.
On
I've tried (only graphically - program Geogebra) - I would say, the ellipsis would have the same focus.
Truth has user TonyK - there is an infinite family of ellipses that circumscribe the rectangle.
Edit - followed by:
$\frac{x^2}{a^2}+\frac{y^2}{16}=1,\quad (x=3, and \,y=2) \quad \Rightarrow a^2=12$
$\Rightarrow \frac{x^2}{12}+\frac{y^2}{16}=1$
As you need a fifth point to determine the ellipse, the eccentricity is a degree of freedom and there is no useful relation to the inner ellipse.
Your red ellipse is $$\frac{x^2}{a^2}+\frac{y^2}{4^2}=1.$$ Plug the coordinates of a corner to determine $a$.