= Solution
Parameterize the <unit circle> by $\zeta=e^{i\theta}$. Under the <Joukowski map>,
$$
z=e^{i\theta}+a^2e^{-i\theta}
=(1+a^2)\cos\theta+i(1-a^2)\sin\theta.
$$
Consequently its image is the <ellipse>
$$
\boxed{\frac{x^2}{(1+a^2)^2}+\frac{y^2}{(1-a^2)^2}=1,}
$$
with horizontal and vertical <semi-axes>
$$
\boxed{A=1+a^2,\qquad B=1-a^2.}
$$
Both are positive because $0<a<1$. The horizontal axis is the major axis. The derivative $dz/d\zeta=1-a^2/\zeta^2$ is nonzero for $|\zeta|\geq1$. Also, equality of two images implies $(\zeta_1-\zeta_2)(1-a^2/(\zeta_1\zeta_2))=0$; two distinct exterior points cannot satisfy the second factor since $|\zeta_1\zeta_2|>1>a^2$. Thus the exterior map is one-to-one, mapping the exterior of the <unit circle> conformally onto the exterior of the <ellipse>.
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