Show that figure is "confined"

Question

Show that figure defined by equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is confined.

I could not find definition in english of "confined figure", but it is such figure such there exists circle including that figure. For example, square is confined-unlike parabola.

My idea was to show that circle defined by $x^2+y^2=r^2$ includes that elipse for some $r$, but mixing inequalities with equalities is nonsense.

Similar question: Show that figure defined by equation $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is not confined.

Answer 1

The word that you're after is “bounded”.

If $\lvert a\rvert\geqslant\lvert b\rvert$, then$$1=\frac{x^2}{a^2}+\frac{y^2}{b^2}\leqslant\frac{x^2+y^2}{b^2}$$and therefore $x^2+y^2\leqslant b^2$. So, that figure is contained in the disk centered at $(0,0)$ with radius $b$. And, of course, if $\lvert a\rvert\leqslant\lvert b\rvert$, then that figure is contained in the disk centered at $(0,0)$ with radius $a$.

Can you deal with the other figure now?

Answer 2

"Bounded" is a more common term.

For $ \sqrt {1- \dfrac{y^2}{b^2}}$ to be real $y$ is bounded in a figure of ellipse in interval $$ (-b<y<+b), $$

and similarly $x$ is bounded.

For $ \sqrt {1+ \dfrac{x^2}{a^2}}$ to be real $y$ is not bounded in a figure of hyperbola, so the open interval is $$ (-\infty<y<+\infty). $$