 Solution

Represent the smooth plane conic by $v^{\mathsf{T}}Qv=0$, with $Q$ an invertible symmetric $3\times 3$ 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 satisfiesConversely 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. The dual conic is therefore the smooth conic