Showing values of x fall into a range or domain, when y is also part of the equation I am told to consider the relation: 
$$\frac{x^2}{\:20}+\frac{y^2}{5} = 1$$
(i) Show that the values of x for which the relationship is defined are given by:$-\sqrt{20}<=\:x\:<=\:\sqrt{20}$
(ii) Similarly, find the values of y for which the relation is defined.
Do I need to figure out what y is algebraically first and then try values of x that fall into the domain provided? And the next part it is saying.....I have no idea.
 A: Your equation is
$$\frac{x^2}{20} + \frac{y^2}{5} = 1 \implies \frac{x^2}{20} = 1 - \frac{y^2}{5} \tag{1}\label{eq1A}$$
You have $\frac{y^2}{5} \ge 0 \implies -\frac{y^2}{5} \le 0 \implies 1 - \frac{y^2}{5} \le 1$. Thus, $\frac{x^2}{20} \le 1 \implies x^2 \le 20 \implies -\sqrt{20} \le x \le \sqrt{20}$. Similarly, since $\frac{x^2}{20} \ge 0 \implies 1 - \frac{x^2}{20} \le 1$, you have that $\frac{y^2}{5} \le 1 \implies y^2 \le 5 \implies -\sqrt{5} \le y \le \sqrt{5}$.
A: Firstly, let's make $y^2$ the subject of the equation:
$$\frac{y^2}{5}=1-\frac{x^2}{20}$$
$$y^2=5-\frac{x^2}{4}=\frac{20-x^2}{4}$$
Now for this to be defined (real), $y^2\geq 0$.
$$\frac{20-x^2}{4}\geq 0$$
$$x^2\leq 20$$
$$-\sqrt{20}\leq x\leq \sqrt{20}$$, as desired.
For part $(ii)$ it is similar - rearrange and solve the inequality
A: Ellipse:
$ \frac{x^2}{(\sqrt{20})^2}+\frac{y^2}{(√5)^2};$
Set $x'=\frac{x}{\sqrt{20}}$, and $y'=\frac{y}{√5}.$
Circle:
$(x')^2+(y')^2=1$, describes a circle center $(0,0)$ radius$=1.$
Then
$-1\le x' \le 1$; and $-1\le y'\le 1$.
Now express $x',y'$ in terms of original coordinates $x,y$.
