1

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.

Gummy bears
  • 3,408
  • There is an infinite family of ellipses that circumscribe the rectangle. Are you asked to find the one with the smallest area? Or what? – TonyK Nov 27 '14 at 14:18
  • Oops, forgot to mention that it passes through the point (0, 4)) @TonyK – Gummy bears Nov 27 '14 at 14:56

2 Answers2

1

I've tried (only graphically - program Geogebra) - I would say, the ellipsis would have the same focus.

elipsis

Truth has user TonyK - there is an infinite family of ellipses that circumscribe the rectangle.

Edit - followed by:

$\frac{x^2}{a^2}+\frac{y^2}{16}=1,\quad (x=3, and \,y=2) \quad \Rightarrow a^2=12$

$\Rightarrow \frac{x^2}{12}+\frac{y^2}{16}=1$

georg
  • 2,749
  • Sorry, forgot to mention that it passes through the point (0,4). Also, when I found the answer using the assumption that they have the same focii, the answer came out to be incorrect. – Gummy bears Nov 27 '14 at 14:57
  • I just realized that the point given was on the y-axis and therefore gave the value of b. Thanks anyways. – Gummy bears Nov 27 '14 at 15:24
0

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 $a$.