Why must an ellipsoid's R.H.S. be $1$? 
Why must the ellipsoid equation or ellipse equation have $1$ as the R.H.S?

The standard equation of an ellipse is:
$$\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2=1$$
And similar one for the ellipsoid: 
$$\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2+\left(\frac{z}{c}\right)^2=1$$
Whereas, the equation for the circle (respectively the sphere) is:
$$(x-a)^2+(y-b)^2=r^2;\quad (x-a)^2+(y-b)^2+(z-c)^2=r^2$$
Looking at this, one question naturally comes, why can't there be something else than $1$ as the R.H.S. for the ellipsoid/ellipse?
 A: I assume you're asking about the standard equation for an ellipsoid: 
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$$
It is designed this way for convenience, so that the numbers $a,b,c$ represent half the lengths of the principal axes. 

Let's suppose you want the right hand side to be $r$ instead of $1$, and suppose $r$ is positive. 
Then we can divide both sides by $r$ and rewrite the equation in the standard form:
$$\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = r$$
$$\frac{x^2}{a^2r} + \frac{y^2}{b^2r} + \frac{z^2}{c^2r} = 1$$
$$\frac{x^2}{\left(a\sqrt{r}\right)^2} + \frac{y^2}{\left(b\sqrt{r}\right)^2} + \frac{z^2}{\left(c\sqrt{r}\right)^2} = 1$$
As you can see, we still have an ellipsoid, but the lengths of all three principal axes have been scaled by a factor of $\sqrt{r}$. 
So yes, the right hand side can be something other than $1$, and you will still have an ellipsoid as long as that something is a positive number. However, it's more convenient to rearrange the equation into standard form. 
A: It's convention. The right hand side can also be something else, as long as it's positive, since you can always divide by that number and get $1$ on that side.
The thing is that every ellipse can be uniquely represented by the equation $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
Of course, for every $\alpha>0$, the equation
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=\alpha$$
also represents an ellipse, but you lose uniqueness, since
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
is the same ellipse as
$$\frac{x^2}{\left(\frac{a}{2}\right)^2}+\frac{y^2}{\left(\frac{b}{2}\right)^2}=4$$
