= Solution
Apply a <projective transformation> to put the <cuspidal cubic> in the form $D=\{F=y^2z-x^3=0\}$. Since $L$ meets $D$ at only one smooth point $p$, the <Bézout theorem> forces intersection multiplicity three there. Hence $L$ is the <tangent line> at a <flex>.
The homogeneous <Hessian matrix> has determinant
$$
\det\begin{pmatrix}-6x&0&0\\0&2z&2y\\0&2y&0\end{pmatrix}=24xy^2.
$$
By the <Hessian criterion for a flex>, every smooth flex lies in $D\cap\{xy^2=0\}$. These intersections are $[0:0:1]$, the <cusp>, and $[0:1:0]$, which is smooth. Thus the <unique smooth flex of a cuspidal cubic> is $p=[0:1:0]$, with <tangent line> $L=\{z=0\}$. Conversely the pair in part (b) has exactly the required properties. Therefore \b[there is one projective-equivalence class], represented by
$$
\boxed{(C,D)=\bigl(3\{z=0\},\ \{y^2z=x^3\}\bigr).}
$$
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/iii/paper-25-cusp-and-flex.png]
{title=Cuspidal cubic in two projective charts: the cusp and its unique smooth flex}
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