Rational curve and ellipse: exactly two intersections

There are no axial intercepts: the numerator never vanishes and $x=0$ is excluded. Asymptotes are $x=0,x=\mu,y=0$. Differentiate: $y^\prime=\dfrac{-2(2x-\mu)}{(x^2-\mu x)^2}$. Hence the sole stationary point is $(\mu/2,-8/\mu^2)$, a maximum on the negative middle branch. The exterior branches are positive. The graph is symmetric about $x=\mu/2$.

Divide by $a^2b$: $\dfrac{(x-a)^2}{a^2}+\dfrac{y^2}{b}=1$. Centre $(\mu/2,0)$, horizontal semi-axis $a=\mu/2$, vertical semi-axis $\sqrt b$. The ellipse occupies $0\le x\le\mu$, and the rational graph is negative there. The lower ellipse must extend below the rational maximum: $\sqrt b>8/\mu^2$, hence $b>64/\mu^4$. Equality gives one tangent intersection; a smaller $b$ gives none. To check that the strict condition gives exactly two, write $u=x-a$, $z=a^2-u^2>0$. Equating squared ordinates gives $bz/a^2=4/z^2$, or $z^3=4a^2/b$. There is a unique admissible $z<a^2$ exactly when $b>4/a^4=64/\mu^4$, and it yields two symmetric nonzero values of $u$.