Tangent Perpendiculars to a Parabola
Tangent Perpendiculars to a Parabola
Paul Abbott
A recent question on the LinkedIn Mathematical Olympiads group asked: For a point inside the parabola defined by , given that the line segments , , and are perpendicular to the parabola, evaluate ++.
A(a,b)
y:=
2
x
AB
AC
AD
2
AB
2
AC
2
AD
Here is a diagram of the problem:
Out[]=
Equating the slope of the line through the points and to the slope of the perpendicular to the tangent line at requires that
(a,b)
(x,)
2
x
(x,)
2
x
2
x
x-a
1
2x
3
x
Compute and simplify ++, which is +, summed over the roots, , of this cubic via Viète’s formulas:
2
AB
2
AC
2
AD
2
(a-x)
2
(b-)
2
x
x
In[]:=
Factor/@RootSum[x2 -x (2 b-1)-a,x|->+]
3
x
2
(a-x)
2
(b-)
2
x
Out[]=
3(+)-
2
a
2
b
1
2
2
(2b-1)
So we have answered the question. But with a bit more work we can produce a nice visualization of the problem.
Compute the real roots of the cubic explicitly and determine the associated points on the parabola:
In[]:=
pts[{a_,b_}]={x,}/.Solve-b-,x,;
2
x
2
x
x-a
1
2x
Compute ++ for the points on the parabola satisfying the condition:
2
AB
2
AC
2
AD
In[]:=
(p(p-{a,b}).(p-{a,b}))/@pts[{a,b}]//Total//Simplify
Out[]=
This result is equivalent to that obtained above.
Compute the discriminant of the cubic and, for 3 real roots, require it to be positive:
In[]:=
Δ[a_,b_]=Discriminant[2 +(1-2 b) r-a,r]>0//FullSimplify
3
r
Out[]=
2>27
3
(2b-1)
2
a
Display the excluded region on a plot of the parabola:
In[]:=
=ShowPlot,{x,-2,2},,RegionPlot!Δ[a,b]&&b≥,{a,-2,2},{b,0,2},
2
x
2
a
Out[]=
Write a function to compute the tangent line segment at a point:
In[]:=
TangentLine[p_]:=Linep- Normalize[{1,2 First[p]}],p+ Normalize[{1,2 First[p]}]
1
2
1
2
Produce an interactive plot of the solutions constrained to the region where all solutions are real:
Out[]=
As noted in the comments on this problem, the boundary of the excluded region is the evolute of the parabola, which is the locus of all its centers of curvature defined by the tangent line .
AC