How to identify which concics in $\mathbb R^2$ are compact and which conics are not?

Question

Which of the following are compact?

(a)The hyperbola $x^2 - y^2 = 1$ in $\mathbb R^2$.

(b)The parabola $y^2 = x$ in $\mathbb R^2$.

(c)The ellipse $x^2/a^2 +y^2/b^2 = 1$.

I am confused about how to show this. Now for (c) i can define a function from $(x,y)\rightarrow x^2 /a^2 +y^2/b^2$ where and $b$ are fixed and since $\{1\}$ is closed in $\mathbb R$ the ellipse is closed since the mapping is continuous and $$x^2+y^2 \leq a^2 x^2+b^2y^2=a^2b^2,$$

then the set is bounded so it is compact.

But for others two I can define such a function and show that the sets are closed. I think the other sets are not compact. But by the theorem if $A$ be a subset of $(X,d)$ is compact iff $A$ is closed in $X$. So is that enough to show that other sets are closed?

Answer 1

All of these sets are closed, so if you show one of them is bounded, then by Heine-Borel theorem, they're compact.

Now for c), the inequality you're written is wrong(for instance, it's false if $a<1$ and $b<1$). So if $(x,y) \in \Bbb R^2$ such that $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, then what can you say about $\|(x,y)\|$? I claim that in this case, $\|(x,y)\|\leq a^2+b^2$(you can easily check that this is the case). So the set in c) is compact. You can use a similar trick to show that the set in b) is NOT compact. For this it is enough to show that it's not bounded. For some $n \in \Bbb N$, let $p = (n,\sqrt{n})$. Then the point $p$ satisfies the equation $y = x^2$. But $\|p\| \geq n$. But this is true for every $n \in \Bbb N$. Therefore the set b) cannot be bounded and hence not compact. Now use a similar trick to show that the set in a) is not bounded(and hence not compact).

Answer 2

Consider the ellipse $(x/a)^2+(y/b)^2=1$. Then if you’re given a point $(x,y)$ on it, there exists $t\in[0,2\pi]$ such that $x/a=\cos t,y/a=\sin t$.

Conversely, any point $(a\cos t,b\sin t)$ belongs to the ellipse.

Thus the map $\varphi\colon[0,2\pi]\to\mathbb{R}^2$, $\varphi(t)=(a\cos t,b\sin t)$ is continuous and its range is the ellipse. Since $[0,2\pi]$ is compact, also the ellipse is compact.

The parabola and hyperbola are not bounded.

If $(t,t^2)$ is on the parabola, its distance from the origin is $\sqrt{t^2+t^4}=|t|\sqrt{1+t^2}\ge|t|$, so it can take arbitrarily large values.

For the hyperbola you can consider the intersections with lines of the form $x+y=k$, with $k>0$. You get $$ \begin{cases}x+y=k\\[4px] x-y=1/k\end{cases} $$ The distance of the intersection point from the origin can be easily computed as $$ \sqrt{\frac{1}{2}\Bigl(k^2+\frac{1}{k^2}\Bigr)}\ge\frac{k}{\sqrt{2}} $$