Chebyshev Filter Transfer Function
Note that the squared magnitude of the frequency response is given by
$$\big|H(j\omega)\big|^2=\frac{1}{1+\epsilon^2T^2_N\left(\frac{\omega}{\omega_c}\right)}\tag{1}$$
In the $s$-domain we have
$$H(s)H(-s)=\frac{1}{1+\epsilon^2T^2_N\left(\frac{s}{j\omega_c}\right)}\tag{2}$$
Computing the zeros of $(2)$ does not only result in the zeros of the filter's transfer function $H(s)$, but also in the zeros of $H(-s)$. The zeros of $(2)$ lie on an ellipse, and for each zero $s_0$ in the left half-plane, there's also a zero $-s_0$ in the right half-plane. Since we want a stable filter, we assign all zeros in the left half-plane to $H(s)$.