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I know that from any point a maximum of three normals could be drawn to a parabola because the equation of normal is cubic. But I want to know the condition on the point for the number of normals

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  • $\begingroup$ In two dimensions, there are only two normal vectors to a parabola at any given point. In three dimensions, there are infinitely many normal vectors to a parabola at any given point... Perhaps I am misunderstanding the question. $\endgroup$ – pseudoeuclidean Feb 7 '17 at 18:42
  • $\begingroup$ No buddy I'm talking about normals being drawn on a parabola from any point (not necessarily on the parabola) $\endgroup$ – Palash gupta Feb 7 '17 at 18:45
  • $\begingroup$ I see. You mean to say "normals drawn from a parabola that contain a given point." Interesting query. What have you tried so far? $\endgroup$ – pseudoeuclidean Feb 7 '17 at 18:49
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    $\begingroup$ From that point three normals can't be drawn $\endgroup$ – Palash gupta Feb 8 '17 at 2:15
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    $\begingroup$ Yup it seems good $\endgroup$ – Palash gupta Feb 8 '17 at 2:25
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If $(p_x,p_y)$ are the coordinates of a point off the parabola and we'd like the line from $(p_x,p_y)$ to the point $(x,x^2)$ on the parabola to be normal to the parabola, then we need:

$$ \begin{pmatrix}p_x \\ p_y\end{pmatrix} = \begin{pmatrix}x \\ x^2\end{pmatrix} + t \begin{pmatrix}-2x \\ 1\end{pmatrix}. $$

This defines a pair of equation from which we can eliminate $x$ to obtain a cubic in $t$:

$$ 0 = 4t^3 - 4(p_y+1)t^2 + (4 p_y+1) t + p_x^2 - p_y = 0 $$

The discriminant of a cubic equation tells how many roots there are:

$$ 0 = ax^3+bx^2+cx+d\\ \Delta=18abcd-4b^{3}d+b^{2}c^{2}-4ac^{3}-27a^{2}d^{2} $$

Then $\Delta > 0$ corresponds to 3 real roots, $\Delta < 0$ to 1 real root and 2 complex roots, and $\Delta = 0$ to multiple root with all roots real.

I don't know for the case $\Delta = 0$ whether there are two distinct roots (one with multiplicity 2) or one root with multiplicity 3. Maybe both could occur.

I plotted the sign of the discriminant, with the parabola and a grid of unit size overlayed. The grey regions correspond to positive discriminant (3 real roots), white regions to negative discriminant (1 real root).

parabola normal count

Here is the GLSL source code, for use with Fragmentarium:

#include "Progressive2D.frag"

vec3 color(vec2 p)
{
  float s = length(vec4(dFdx(p), dFdy(p)));
  if (p.y - s < p.x * p.x && p.x * p.x < p.y + s) return vec3(0.0);
  if (abs(mod(p.x + 0.5, 1.0) -0.5) < 0.5 * s) return vec3(0.5);
  if (abs(mod(p.y + 0.5, 1.0) -0.5) < 0.5 * s) return vec3(0.5);
  if ((abs(p.x) - s) * (abs(p.x) - s) < p.y &&
      p.y < (abs(p.x) + s) * (abs(p.x) + s)) return vec3(0.0);
  float a = 4.0;
  float b = -4.0 * (p.y + 1.0);
  float c = 4.0 * p.y + 1.0;
  float d = p.x * p.x - p.y;
  float discriminant
    = 18.0 * a * b * c * d
    - 4.0 * b * b * b * d
    + b * b * c * c
    - 4.0 * a * c * c * c
    - 27.0 * a * a * d * d;
  if (discriminant > 0.0) return vec3(0.7);
  if (discriminant < 0.0) return vec3(1.0);
                          return vec3(0.2);
}
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  • $\begingroup$ I like this answer a lot but had to think the first bit through to follow it. I hope you don't mind my edit. Also, I believe that there's one normal at the cusp and two normals elsewhere. I'm going to post an answer illustrating what the normals look like $\endgroup$ – Mark McClure Feb 8 '17 at 4:11
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    $\begingroup$ When you are on the boundary curve, called the absolute off the parabola, you could get either two roots with one doubly degenerate, or one trippy degenerate root. The latter occurs when you choose the sharp tip or cusp of the evolute; elsewhere along that curve you get two roots with one doubly degenerate. The degenerate normal in this latter case is always the one that crosses the axis. $\endgroup$ – Oscar Lanzi May 25 at 15:19
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Claude's technique can be extended slightly to find the the points and normal lines given a particular point off of the parabola. Using this, I obtained an animation of some of the normals.

enter image description here

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Only two perpendicular tangents can be drawn from a point P on a parabola directrix ( standard result, need not be again proved). Taking this as given we complete the rectangle by drawing parallels to arrive at a unique opposite point inside point Q ... from which we conclude accordingly only 2 unique normals can be drawn from an arbitrary but inside point like Q.

enter image description here

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  • $\begingroup$ there is a third normal to the parabola (heading to the right) from your point marked Q, see mathr.co.uk/tmp/ZdYXF-mod.png its existence can be seen from the marked right hand normal crossing above Q, from the normal line moving below Q as x increases, and from the smoothness of the parabola $\endgroup$ – Claude Feb 7 '17 at 21:02
  • $\begingroup$ @Claude I have more carefully drawn it by Geogebra. Only two normals in this particular case.as at first stated, no third point. $\endgroup$ – Narasimham Feb 7 '17 at 22:28
  • $\begingroup$ @Narasinham your new diagram has $Q$ at around $(-0.7, -2.7)$, your old diagram had $Q$ at around $(-1.1, -3.7)$. $\endgroup$ – Claude Feb 7 '17 at 23:17
  • $\begingroup$ This is just wrong. $\endgroup$ – Mark McClure Feb 8 '17 at 4:12
  • $\begingroup$ Agreed clearly wrong. I fail to "properly" connect it to the third possibility, shall delete until it is found. $\endgroup$ – Narasimham Feb 8 '17 at 13:28

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