Find image $~u^2 + v^2 \leq 1, ~~x = au ,~~ y = bv$

Question

I'm trying to find the image of $~u^2 + v^2 \leq 1,~~ x = au ,~~ y = bv$

I'm pretty sure what I did is wrong, but I don't know why or how to know if the answer is correct.

I know $u^2 + v^2 \leq 1$ is a circle, $r = 1$.

$$ u \leq \sqrt{1-v^2}$$

$$x = a\sqrt{1-v^2}$$

$$v = \frac{b}{y}$$

$$\text{img} = x =a\sqrt{1-(\frac{b}{y})^2}$$

Answer 1

If you let $x = au$ and $y = bv$, the proposed relation becomes \begin{align*} \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}\leq 1 \end{align*}

which is the closure of an ellipse. Are you acquainted to such locus?