Inner isoptics of a parabola

Abstract

For a given angle α, the α-isoptic curve of a parabola is the geometric locus of points through which passes a pair of tangents to the parabola making an angle equal to α. We explore the inner isoptics of parabolas: they are the envelopes of the lines joining the points of contact of the parabola with the tangents through points on a given isoptic. We show that the inner orthoptic (that is, the inner 90°-isoptic) of a parabola is a degenerate point (the focus), but in other cases the inner isoptic is an ellipse. Inner isoptics of ellipses have already been studied, at least partially, by the second author and Mozgawa, and by Naiman, Skrzypiec and Mozgawa. In our contribution we study a simpler case, the case of a parabola. Our result connects parabolas, their "outer" and "inner" isoptics (hyperbolas and ellipses, respectively), thus we present a remarkable connection among different kinds of conics. We give an overview on how our GeoGebra experiments turned into conjectures, and then into a statement, and how we found its analytical proof after implicitizing the problem via elimination and Gröbner bases.

Full text

5th Croatian Conference on Geometry and Graphics
Dubrovnik, September 4-8, 2022
Inner isoptics of a parabola
by Zoltán Kovács, Thierry Dana-Picard and Tomás Recio
Abstract
For a given angle α, the α-isoptic curve of a parabola is the geometric locus of points through which passes a pair of tangents to
the parabola making an angle equal to α. We explore the inner isoptics of parabolas: they are the envelopes of the lines joining
the points of contact of the parabola with the tangents through points on a given isoptic. We show that the inner orthoptic
(that is, the inner 90°-isoptic) of a parabola is a degenerate point (the focus), but in other cases the inner isoptic is an ellipse.
Inner isoptics of ellipses have already been studied, at least partially, by the second author and Mozgawa, and by Naiman,
Skrzypiec and Mozgawa. In our contribution we study a simpler case, the case of a parabola. Our result connects parabolas,
their “outer” and “inner” isoptics (hyperbolas and ellipses, respectively), thus we present a remarkable connection among
different kinds of conics. We give an overview on how our GeoGebra experiments turned into conjectures, and then into a
statement, and how we found its analytical proof after implicitizing the problem via elimination and Gröbner bases.
A collaboration of CAS and DGS
We think our contribution is an example of the fruitful collaboration of CAS and DGS, allowed by GeoGebra, a popular tool in
mathematics education and research, Giac/Xcas, a free CAS, and Maple, a powerful general tool for computer algebra.
What are isoptics?

For a given angle α, the α-isoptic curve of a circle is the geometric locus of points through which passes a pair of tangents to
the circle making an angle equal to α.
Clearly, for circles, isoptic curves are circles. For algebraic reasons we can usually see bisoptics (that is, two isoptic curves at
the same time).
For parabolas, for a given angle α, the α-isoptic is the geometric locus of points through which passes a pair of tangents to the
parabola making an angle equal to α.

It is well known that the isoptics of a parabola are hyperbolas (or the directrix for the case α=90°).
What are inner isoptics?
For a given α-isoptic of a curve, the inner isoptic is the envelope of the lines joining the points of contact of the curve with the
tangents through points on the given isoptic.

Clearly, for circles, inner isoptic curves are circles.
In our contribution we study the case of a parabola. Why?
The case of ellipses is complicated, but already solved in 2020 and 2022.
The case of hyperbolas is also complicated and yet unsolved, it is challenging computationally.
The case of parabolas is yet unsolved, and it seems feasible to solve.

We consider the parabola , defined by focus and directrix , and its -isoptic where (In this way, a
well-known algebraic formula can be used, and accordingly, a slider can be added to the applet.) Here the isoptic is the
hyperbola An arbitrary point and the tangents to through are labelled with and The tangent points are
and Helper objects circle and intersection points and allow us to analytically observe the situation. By denoting
by , the envelope of is the ellipse
From conjecture towards a proof
13 April 2022, 22:34 CET
Dear Tomás,
we are in a new kind of work with Noah.
As the Passover and the Easter holidays are approaching,
we have no time to deal with this issue in the next few days,
but I guess we need your help in this matter sooner or later. :-)
Usually, when we want to obtain a locus or envelope equation,
we get a curve in two variables. This is done by using elimination.
In some very rare cases the output is not an equation but two equations,
the first one is in variable x, the second one is in variable y.
(In this case the geometric answer is a set of points, a 0-dimensional set).
But now it seems there is another case. We get two curves for some interesting
geometric input (we want to obtain the inner isoptic of a parabola).
The first curve is a union of an ellipse and a line (a reducible cubic),
and the second one again the union of the same ellipse but a different line
(again, a reducible cubic). Is this case well-known to you?
In a next email I will send some more details to Noah, and I will CC the data
for you as well. I am unsure how to handle this case, maybe the union of
the ellipse and the two lines could be shown as graphical output...
Thanks in advance for your help.
Best, Zoltan
14 April 2022, 1:00 CET
Zoltan, I kind of understand your question.
Your definition of curve is “irreducible” curve…and this is a quite rare case.
Most locus are curves, yes, but reducible.
Tomas
14 April 2022, 20:57 CET
Zoltan,
I have made some computations and let me tell you that
the output of your elimination is
ONE ELLIPSE (2x^2+y^2-4*x-10*y-19=0)
PLUS
ONE POINT (point (1,3))
pF d
=
D
1
D
2
αt
= tan2
α
.
h
.
E
∈
h p E t
1
t
2.
T
1
T
2.
c H
1
H
2
T
1
T
2
sse
.

assuming v57 is x and v58 is y. See enclosed screenshot of my Maple computations.
I used YOUR output, changed v57 is x and v58 is y, then
considered the elimination ideal I called Ideal and asked
for its PrimeDecomposition. The result is a couple of ideals,
one is the ideal of the ellipse, the other is the ideal of the point.
So the elimination locus is the union of one curve and one point
(and I have verified the point is NOT in the ellipse). This is all…
Tomas
14 April 2022, 22:37 CET
Dear Tomás, thank you so much that you worked so hard on this!
I am starting to understand what happens. Finally I managed to separate
the input curves (a hyperbola plus a line) and if I use only one (the hyperbola),
the elimination ideal is just the ellipse. The newest version of GeoGebra
(not yet available publicly, I plan to do a release next week) plots the output
nicely and quite quickly -- even a moderately fast animation can be performed!
Now we can prove (well, in general not yet, but for many particular cases
it's clear symbolically) the following theorem:
Given a parabola p, an angle θ (0°<θ<180°) and the θ-isoptic i of p.
(It is well known that for θ=90°, i is a line: the directrix of p; otherwise
i is a hyperbola.) Let us consider the two tangents from a point X ∈ i to p,
and consider the tangent points P1 and P2. We construct the segment s=P1P2
and the envelope e of s, while X ∈ i. (That is, we consider e=Envelope(s,X).)
Now e is an ellipse.
This theorem seems like something new. It connects the three conic sections:
the parabola, the hyperbola and the ellipse, in a novel way. :-)
Best, Zoltán
27 April 2022, 18:10 CET
Dear Noah,
I am trying to prove that the non-90-isoptics of a parabola are always ellipses.
I am quite close to do that, but, for some reason, the elimination fails
if I construct the figure in a natural way.
On the other hand, I managed to make a conjecture.
Namely, we assume that we consider the parabola y=x^2/4. This can be done wlog.
According to Wikipedia (for example), it has the α-isoptic
x^2-tan^2(α) (y+1)^2-4y=0.
Denote tan^2(α) by t.
I found that the inner isoptic has the following equation in this case:
(2t+2)x^2 + (2t)y^2 -(4t+8)y + 2t=0
which can also be written as
x^2 + t/(t+1) y^2 - 2(t+2)/(t+1)y + t/(t+1) = 0.
I cannot prove this yet, but the conjecture is visually clear.
I attach a file where you can play with dragging D and t.
Best, Zoltan
A proof
Let us consider the parabola that has the directrix and focus . The -isoptic of this
parabola is
see Wikipedia. Let us denote by the number . Since , .
Consider a point with coordinates , that is, , thus
Let us consider the tangents and to . We know that the circle with center and radius intersects the directrix at
points and , that is,
and
Furthermore, the tangents and touch the parabola at points and .
Let us put a virtual locus point on the line :
Now we compute the determinant of the Jacobian matrix
c
:
x
2− 4
y
= 0
f
:
y
= −1
C
= (0, 1)
α a
Iα
(
x
,
y
) :=
x
2− tan2
α
(
y
+ 1)2− 4
y
= 0,
tan2
α t α
≠ 90o
t
> 0
L
∈
a L
= (
lx
,
ly
)
Iα
(
lx
,
ly
) = 0
e
1:
l
2
x
−
t
(
ly
+ 1)2− 4
ly
= 0.
k n c e L CL
M
= (
mx
, −1)
N
= (
nx
, −1)
e
2:
l
2
x
+ (
ly
− 1)2− (
mx
−
lx
)2− (
ly
+ 1)2= 0
e
3:
l
2
x
+ (
ly
− 1)2− (
nx
−
lx
)2− (
ly
+ 1)2= 0.
k n O
= (
mx
,
m
2
x
/4)
P
= (
nx
,
n
2
x
/4)
X
= (
x
,
y
)
OP
e
4: = 0.
∣
mxnxx
m
2
x
/4
n
2
x
/4
y
1 1 1
∣
D
∂ ∂ ∂ ∂

that is,
i.e.,
To obtain the envelope equation that describes the inner isoptic, one needs to compute
This result is, however, the empty set. In fact, we need to exclude the degenerate case when . This is possible by
adding another equation:
by using Rabinowitsch's trick. Now
which is clearly the equation of a conic. (The computation took 10.04 seconds on a 4-years-old laptop.) It is well known (see
Wikipedia) that any conic given in the form is an ellipse if and only if
. In our case (because ), and this makes our statement proven.
Summary
When computing isoptics of ellipses and hyperbolas, toric sections appear. With parabolas, we remain within the framework of
conics. It is amazing that isoptics of a parabola are hyperbolas and the inner isoptics are ellipses, thus we have the complete set
of non-degenerate conics!
Acknowledgements
First and third author were partially supported by a grant PID2020-113192GB-I00 (Mathematical Visualization: Foundations,
Algorithms and Applications) from the Spanish MICINN. Géza Csima kindly verified our result and recommended some
additional ways to simplify the proof.
References
Csima, G., Szirmai J., Isoptic curves of conic sections in constant curvature geometries, Mathematical Communications
19(2) (2014), 277-290.
Dana-Picard, T., Mozgawa, W., Automated exploration of inner isoptics of an ellipse, Journal of Geometry 111 (2020),
1-10.
Naiman, A., Skrzypiec, M., Mozgawa, W., Implicit forms of inner isoptics of ellipses, Beiträge zur Algebra und
Geometrie (2022)
,
⎛
⎜
⎝
∂
e
1
∂
lx
∂
e
2
∂
lx
∂
e
3
∂
lx
∂
e
4
∂
lx
∂
e
1
∂
ly
∂
e
2
∂
ly
∂
e
3
∂
ly
∂
e
4
∂
ly
∂
e
1
∂
mx
∂
e
2
∂
mx
∂
e
3
∂
mx
∂
e
4
∂
mx
∂
e
1
∂
nx
∂
e
2
∂
nx
∂
n
3
∂
nx
∂
e
4
∂
nx
⎞
⎟
⎠
D
= ,
∣
2
lx
2
mx
2
nx
0
−2
lyt
− 2
t
− 4 −4 −4 0
0 2
lx
− 2
mx
0−2
mxnx
+2
mxx
+
n
2
x
−4
y
4
0 0 2
lx
− 2
nx
−
m
2
x
+2
mxnx
−2
nxx
+4
y
4
∣
D
= − 4
l
2
xm
2
x
+ 8
l
2
xmxx
+ 4
l
2
xn
2
x
− 8
l
2
xnxx
+ 4
lxm
3
x
+ 6
lxm
2
xnxtly
+ 6
lxm
2
xnxt
+ 4
lxm
2
xnx
− 4
lxm
2
xtlyx
− 4
lxm
2
xtx
− 8
lxm
2
xx
− 6
lxmxn
2
xtly
− 6
lxmxn
2
xt
− 4
lxmxn
2
x
+ 8
lxmxtlyy
+ 8
lxmxty
− 4
lxn
3
x
+ 4
lxn
2
xtlyx
+ 4
lxn
2
xtx
+ 8
lxn
2
xx
− 8
lxnxtlyy
− 8
lxnxty
− 2
m
3
xnxtly
− 2
m
3
xnxt
− 4
m
3
xnx
+ 4
m
2
xnxtlyx
+ 4
m
2
xnxtx
+ 8
m
2
xnxx
+ 2
mxn
3
xtly
+ 2
mxn
3
xt
+ 4
mxn
3
x
− 4
mxn
2
xtlyx
− 4
mxn
2
xtx
− 8
mxn
2
xx
.
⟨
e
1,
e
2,
e
3,
e
4,
D
⟩
∩
R
[
x
,
y
].
mx
=
nx
e
5: (
mx
−
nx
)
u
− 1 = 0
⟨
e
1,
e
2,
e
3,
e
4,
e
5,
D
⟩
∩
R
[
x
,
y
] =
tx
2+
ty
2+
x
2− 2
ty
+
t
− 4
y
AX
2+
BXY
+
CY
2+
DX
+
EY
+
F
= 0
B
2− 4
AC
< 0
B
2− 4
AC
= 0 − 4(
t
+ 1)
t
< 0
t
> 0