Trouble understanding equation of an ellipse I'm self-studying algebra by going through Sheldon Axler's algebra book. I'm a bit stuck on equation of an ellipse topic. Here's how it's explained in the book:

Example: Find an equation describing the ellipse in the xy-plane produced by stretching a circle of radius 1 centered at the origin horizontally by a factor of 5 and vertically by a factor of 3.
To find an equation describing ellipse, consider a typical point (u,v) on the circle of radius 1 centered at the origin. Thus $u^2 + v^2 = 1$. Stretching horizontally by a factor of 5 and stretching vertically by a factor of 3 transforms the point (u, v) to the point (5u, 3v). Rewrite the equation $u^2 + v^2 = 1$ in terms of this new point, getting $$\frac{(5u)^2}{25} + \frac{(3v)^2}{9} = 1$$

Shouldn't the resulting equation be $(5u)^2 + (3v)^2 = 1$? Where did 25 and 9 denominators come from?
The rest of the solution in the book:

Write the transformed point $(5u, 3v)$ as $(x,y)$, thus setting $x = 5u$ and $y = 3v$, getting $$\frac{x^2}{25} + \frac{y^2}{9} = 1,$$ which is the equation of the ellipse


What you quote is not the full solution, is it?
The point $(u,v)$ on the circle satisfies
$$u^2+v^2=1$$
which we just rewrite in equivalent form:
$$\frac{(5u)^2}{25}+\frac{(3v)^2}{9}=1$$
and from this we see that the point $(x,y)=(5u,3v)$, which belongs to the ellipse, satisfies
$$\frac{x^2}{25}+\frac{y^2}{9}=1$$
and so this is the equation of the ellipse.
It would be much more natural to think about it the other way around. Take the point $(x,y)$ on the ellipse, note that $(x/5, y/3)$ (squeezing inverse to your stretching) belongs to the circle, and therefore
$$\left(\frac{x}{5}\right)^2+\left(\frac{y}{3}\right)^2=1$$
which gives the same answer.
