Find equation of non-concurrent curve A parabola and straight line (red)
$$ y-\frac{x^2}{2}-1=0,\quad y-\frac{x}{2}-2=0;\, \tag 1$$
are combined plotted and found to intersect at $ P(-1,1.5),Q(2,3)\;;$
Two more curves (blue) are manipulated to pass through $(P,Q)$ by setting their RHS $=1$ and multiplying the two to form a combined equation:
$$\left(y-\frac{x^2}{2}\right)\cdot\left(y-\frac{x}{2}-1\right)=1 \tag2$$

However only one blue curve passes through $(P,Q)$. What was missing?
How are separate equations of the two blue curves (2) determined?
cp2=ContourPlot[(y-x^2/2-1)(y-x/2-2)==0,{x,-4,4},{y,-8,6},ContourStyle->Red,GridLines->Automatic];
cp3=ContourPlot[(y-x^2/2)(y-x/2-1)==1,{x,-4,4},{y,-8,6},GridLines->Automatic,ContourStyle->{Thick,Blue}]
Show[{cp2,cp3},PlotRange->All]

 A: If you take two curves $C_f$ and $C_g$ defined by a homogeneous implicit equation
$$
\eqalign{
  & \left\{ \matrix{
  C_f  = \left\{ {\left( {x,y} \right)} \right\}\;:\;\;f_0 (x,y) = 0 \hfill \cr 
  C_g  = \left\{ {\left( {x,y} \right)} \right\}\;:\;\;g_0 (x,y) = 0 \hfill \cr 
  C_{fg}  = \left\{ {\left( {x,y} \right)} \right\}\;:\;\;f_0 (x,y)g_0 (x,y) = 0 \hfill \cr}  \right.\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad f_0 (x,y)g_0 (x,y) = 0\quad  \to   \cr 
  &  \to \quad \left( {f_0 (x,y) = 0 \wedge g_0 (x,y) = z} \right) \cup \left( {f_0 (x,y) = z \wedge g_0 (x,y) = 0} \right)\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad C_{fg}  = C_f  \cup C_g  \cr} 
$$
then the curve defined by the product of the two equations is
the union of the previous two curves.
If instead you express the implicit equations as equal to $1$, and multiply them, what you obtain is the intersection
of two surfaces
$$
\eqalign{
  & \left\{ \matrix{
  C'_f  = \left\{ {\left( {x,y} \right)} \right\}\;:\;\;f_1 (x,y) = 1 \hfill \cr 
  C'_g  = \left\{ {\left( {x,y} \right)} \right\}\;:\;\;g_1 (x,y) = 1 \hfill \cr 
  C'_{fg}  = \left\{ {\left( {x,y} \right)} \right\}\;:\;\;f_1 (x,y)g_1 (x,y) = 1 \hfill \cr}  \right.\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad f_1 (x,y)g_1 (x,y) = 1\quad  \to   \cr 
  &  \to \quad \left( {f_1 (x,y) = z \wedge g_1 (x,y) = 1/z} \right)  \cr 
  &  \to \quad \left( {F_0 (x,y,z) = 0 \wedge G_0 (x,y,z) = 0} \right)\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad C'_{fg}  = S_F  \cap S_G  \cr} 
$$
which of course will contain the point(s) where both $f_1 =g_1 = 1$, that is $f_0 =g_0 = 0$.
To tell it in another way
$$
\eqalign{
  & f_1 (x,y)g_1 (x,y) = 1\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left( {f_0 (x,y) + 1} \right)\left( {g_0 (x,y) + 1} \right) = 1\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad f_0 (x,y)g_0 (x,y) + f_0 (x,y) + g_0 (x,y) = 0 \cr} 
$$
versus
$$
f_0 (x,y)g_0 (x,y) = 0
$$
