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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:

$$\left(y-\frac{x^2}{2}\right)\cdot\left(y-\frac{x}{2}-1\right)=1 \tag2$$

enter image description here

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]
Strictly_increasing
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Narasimham
  • 40,495

1 Answers1

1

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 $$ f_0 (x,y)g_0 (x,y) = 0 $$

G Cab
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