= Solution
Represent the <smooth plane conic> by $v^{\mathsf T}Qv=0$, with $Q$ an invertible symmetric $3\times3$ matrix. The tangent line at $[v]$ has coefficients $\ell=Qv$, because the differential is $2v^{\mathsf T}Q$. Thus its image in the <dual projective space> satisfies
$$
\ell^{\mathsf T}Q^{-1}\ell=0.
$$
Conversely any nonzero $\ell$ satisfying this equation gives $v=Q^{-1}\ell$, a point of the original <smooth plane conic> whose tangent is $[\ell]$. This correspondence is the restriction of an invertible projective linear transformation. \b[The <dual conic> is therefore the smooth conic]
$$
\boxed{C_1^\vee:\ell^{\mathsf T}Q^{-1}\ell=0.}
$$
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