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The tangent lines of a smooth plane conic form a smooth plane conic in the dual projective space. If the original equation is $v^{\mathsf{T}}Qv=0$, its tangent has coefficient vector $Qv$, giving the displayed equation. The matrix $Q$ is symmetric and invertible, so this correspondence is a projective linear isomorphism of the conics.

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

Write the projective plane as $\mathbb{P}(V)$ for a three-dimensional complex vector space $V$. A projective line is the zero set of a nonzero linear functional $\ell \in V^{\ast }$. Two such functionals give the same line exactly when they differ by a nonzero scalar. Hence the parameter space of lines is the dual projective space $\mathbb{P}(V^{\ast })$. A choice of basis identifies $V^{\ast }$ with ${\mathbb{C}}^3$, giving $\boxed{({\mathbb{P}}^2{)}^{\vee }\cong {\mathbb{P}}^2}$. The description as $\mathbb{P}(V^{\ast })$ is intrinsic; the identification with the original $\mathbb{P}(V)$ requires a choice.