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The question is this:

For the parabola: $$ (x-1)^2 + (y-1)^2 = \left(\dfrac{x+y} {\sqrt2}\right)^2 $$ what is the condition on the point $(h,h)$ (which lies on the axis of the parabola) that $3$ distinct normals can be drawn from it to the parabola?

Now, I have this doubt: from any point (inside the axis) on the axis we can draw three normals to the parabola (no concrete proof, just visualisation), right? So isn't this question wrong?

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  • $\begingroup$ What is your definition of "normal"? As far as I can see, it is possible to draw at most one normal on any given point on a plane curve, unless you're trying to embed the curve in a higher dimensional space... $\endgroup$ – DonAntonio Aug 25 '13 at 10:33
  • $\begingroup$ I am not talking about normal on/at a point. My question is about the number of normals that can be draw from that point. $\endgroup$ – Parth Thakkar Aug 25 '13 at 10:34
  • $\begingroup$ For instance, you cannot draw more than one normal from the those points of the axis which don't lie "inside" the parabola. $\endgroup$ – njguliyev Aug 25 '13 at 10:35
  • $\begingroup$ That's what I meant by from 'any point on the axis'. I'll edit the question. Thanks for pointing it out $\endgroup$ – Parth Thakkar Aug 25 '13 at 10:36
  • $\begingroup$ @ParthThakkar, that didn't help at all: what is "a normal from some point"? I'd say it is a straightline perpendicular (or orthogonal, or normal ) to the curve at that point, meaning: to that curve's tangent line at that point, so again: what is your definition of "normal from a point"? And normal...to what? $\endgroup$ – DonAntonio Aug 25 '13 at 10:37
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After all the discussion in the comments part we have, perhaps, the solution to the edited question:

From a point on the parabola's axis it is possible to draw three different normals to the parabola iff the point is contained ("inside") the parabola but it is not the parabola's vertex (which could be considered not to be "inside" the parabola, anyway...but it's better, imo, to be thorough here and mention this).

Added: Thanks to one of the last comments of Henning below this answer, it is possible to see that the $\,y-$intercept of any normal to $\,y=x^2\,$ is $\,\frac12+a^2\;$ , from where it follows that no point on the parabola's axis and inside the parabola can have $\,y-$intercept $\,\le\frac12\;$ and have any hope of being one of the wanted points from which three normals to the parabola can be drawn!

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    $\begingroup$ @ParthThakkar: The normal through a point ifinitesimally close to the vertex will intersect the axis at the focus. And the farther away the point on the parabola gets, the higher the intersection of its normal with the axis. There's no way to get a non-axis normal to intersect the axis below the focus. $\endgroup$ – hmakholm left over Monica Aug 25 '13 at 10:50
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    $\begingroup$ Oops, correction -- the critical point is not actually at the focus, but the center of the circle that osculates the parabola at the vertex. I thought this was the focus, but it's actually twice as far from the vertex as the focus is. $\endgroup$ – hmakholm left over Monica Aug 25 '13 at 10:58
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    $\begingroup$ Ah, too bad you realized that now, @HenningMakholm : I was almost done with a proof...hehe. +1 $\endgroup$ – DonAntonio Aug 25 '13 at 10:59
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    $\begingroup$ Indeed so, @ParthThakkar. I corrected the last line yet I forgot to do it on the equation itself. $\endgroup$ – DonAntonio Aug 25 '13 at 11:10
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    $\begingroup$ Leave it! Understood. Thanks a lot :D $\endgroup$ – Parth Thakkar Aug 25 '13 at 11:17
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This can be done in a more generic manner with all variables, in which case, from any point $P(h,k)$ there exist three normals to a parabola $y^2 = 4ax$ if $h> 2a$.

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  • $\begingroup$ This seems wrong to me. If three real normals can be drawn, then $h > 2a$, not the other way around. $\endgroup$ – Ashish Ahuja Nov 22 '20 at 14:50

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