Why must an ellipsoid's R.H.S. be $1$?

Question

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?

Answer 1

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.

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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.

Answer 2

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

$$\frac{x^2}{\left(\frac{a}{2}\right)^2}+\frac{y^2}{\left(\frac{b}{2}\right)^2}=4$$