In[]:=
GeometricLocusMidpoints::usage="Find the equation of the geometric locus of the midpoints of the segment that joins any two points of the graph of the fuction y =
2
x
with abscises differentiated into two units. What kind of conic is it? References: Internet; Personal. ";
Development:
M is the midpoint of segment:
In[]:=
P
1
={a+2,
2
(a+2)
};
P
2
={a,
2
a
};M={x,y};​​M=
P
1
+
P
2
2
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
1
2
(2+2a),
1
2
(
2
a
+
2
(2+a)
)
In[]:=
{a+1,
2
a
+2a+2};​​x=a+1;a=x-1;​​y==
2
(x-1)
+2(x-1)+2;​​y=
2
x
+1;
It is a parabola.
Animation:
In[]:=
Manipulate[ζ[t_]:={t,
2
t
+1};
P
1
[t_]:={t+1,
2
(t+1)
};
P
2
[t_]:={t-1,
2
(t-1)
};​​OriginalCurve[t_]:={t,
2
t
};​​Show[{​​ParametricPlot[ζ[t1],{t1,-2,t},ColorFunction->"BlueGreenYellow"],​​ParametricPlot[Callout[OriginalCurve[t1],"y =
2
x
",{Scaled[0.75],Right},LabelStyle{14,Bold,Black}],{t1,-2,2},PlotStyle->{Blue,Thick}],​​Graphics[{Black,PointSize[0.015],​​Point[
P
1
[t]],Text[Style["
P
1
",14,Bold],
P
1
[t],{0,-1.2}],​​Point[
P
2
[t]],Text[Style["
P
2
",14,Bold],
P
2
[t],{0,-1.2}],​​Point[ζ[t]],Text[Style["M",14,Bold],ζ[t],{0,-1.2}],​​Thick,Line[{
P
1
[t],
P
2
[t]}]}]},AspectRatioAutomatic,​​AxesTrue,AxesOrigin{0,0},AxesLabel{x,y},PlotRange{{-2,2},{-0.5,4}},ImageSize500],Style["Geometric locus of the M midpoint of a segment that joins two points in a parabola.",Bold,Large],{{t,-2+0.00002,"Value (t)"},-2+0.000001,2,0.00001},ControlPlacementTop,SaveDefinitions->True]
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