Show that $\left(\frac{u}{a}\right)^2 + \left(\frac{v}{b}\right)^2 = 1$ is the image of a circle of radius 1 under a dillation $\mathcal Fa,b$ This is an exercise from "Basic Mathematics" by S. Lang (exercise 4a, p.178).
My problem:
Show that the set of points $(u, v)$ satisfying the equation
$${ \left(\frac{u}{a}\right)^2 + \left(\frac{v}{b}\right)^2 = 1 } \tag 1$$
is the image of the circle of radius 1, centered at O under the map $\mathcal Fa,b$.
The mapping $\mathcal Fa,b$ is such that, for every point $(u, v)$ we have:
$$\mathcal Fa,b(u, v) \mapsto (au, bv)$$
In other words, under such mapping, a point as it's $u$-coordinate stretched by $a$, and it's $v$-coordinate stretched by $b$.
My approach: The equation of the circle of radius 1 centered at O is
$$(u - 0)^2 + (v - 0)^2 = 1$$
$$\equiv u^2 + v^2 = 1 \tag 2$$
Under $\mathcal Fa,b$ every point $(u, v)$ gets mapped onto a point of coordinates $(au, bv)$, which means every point has its $u$ and $v$ coordinates stretched by $a$ and $b$ respectively.
Therefore, a circle of radius 1 would have it's radius stretched by $ab$.
In order to obtain the circle of radius 1 under such map, we have to divide the $u$-coordinates of a given set of points by $a$, and it's respective $v$-coordinates by $b$.
My question: Is my answer correct? Before arriving at this answer I tried to put $(1)$ and $(2)$ in an equation, i.e.
$$\left(\frac{u}{a}\right)^2 + \left(\frac{v}{b}\right)^2 = u^2 + v^2 $$
but quickly realized I didn't know where I was intending to head with it and proceeded to come up with the answer I posted above.
Thank you.
 A: Your approach has some good geometrical intuition.  I like it.  But it's not very formal.  Below I give two alternatives which might look a little nicer to a professor.
Also $$\left(\frac{u}{a}\right)^2 + \left(\frac{v}{b}\right)^2 = u^2 + v^2$$ is not really a true statement.  What the question here is hiding (which I make explicit in my answers) is that the $(u,v)$ on the left and the one on the right exist in two different spaces (or at least are given with respect to two different coordinate systems).  So it really doesn't make sense to compare them like this.

Take a point $(x,y)$ on the circle given by $x^2+y^2=1$.  Apply $\mathcal Fa,b$.
$$(x,y)\mapsto (ax,by)=(x',y')$$ 
Then, for every $(x,y)$
$$\left(\frac{ax}{a}\right)^2+\left(\frac{by}{b}\right)^2=\left(\frac{x'}{a}\right)^2+\left(\frac{y'}{b}\right)^2=1$$
So the circle in the $xy$-plane maps to an ellipse in the $x'y'$-plane.

Or more symbolically, let $S^1=\{(x,y) \mid x^2+y^2=1\}$.  Then $$\begin{align}\bbox[5px,border:2px solid red]{\mathcal Fa,b(S^1)} &= \{(x',y')\mid (x',y') = \mathcal Fa,b(x,y) \wedge (x,y)\in S^1\} \\ &= \{(x',y')\mid x'=ax \wedge y'=by \wedge x^2+y^2=1\} \\ &= \{(x',y')\mid x'=ax \wedge y'=by \wedge \left(\frac{ax}{a}\right)^2+\left(\frac{by}{b}\right)^2=1\} \\ &\bbox[5px,border:2px solid red]{= \{(x',y')\mid \left(\frac{x'}{a}\right)^2+\left(\frac{y'}{b}\right)^2=1\}}\end{align}$$
