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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?

Parth Thakkar
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  • 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... – DonAntonio Aug 25 '13 at 10:33
  • 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. – Parth Thakkar Aug 25 '13 at 10:34
  • For instance, you cannot draw more than one normal from the those points of the axis which don't lie "inside" the parabola. – njguliyev Aug 25 '13 at 10:35
  • That's what I meant by from 'any point on the axis'. I'll edit the question. Thanks for pointing it out – Parth Thakkar Aug 25 '13 at 10:36
  • @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? – DonAntonio Aug 25 '13 at 10:37
  • Normal to the parabola. I think that is quite obvious as there is no other thing in the question to which normals can be drawn. – Parth Thakkar Aug 25 '13 at 10:39
  • Ok, if I now understood what you want, from any point on the axis of the parabola contained "inside" the parabola you can draw three normals: two to each of "the sides", one more to the vertex...except from the only point on the axis belonging to the parabola, namely: the parabola's vertex – DonAntonio Aug 25 '13 at 10:40
  • And again you didn't address the main parts of my questions, @ParthThakkar, yet perhaps I understood now mainly thanks to the comment of njguliyev – DonAntonio Aug 25 '13 at 10:41
  • That's what I was asking. I am sorry for all the confusion. I'll be extra careful while asking future questions. – Parth Thakkar Aug 25 '13 at 10:44
  • However, it is true that from any point 'inside' the parabola and lying on the axis, we can draw 3 lines which are normal to the curve - am I right? – Parth Thakkar Aug 25 '13 at 10:45

2 Answers2

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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!

DonAntonio
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  • Thank you for bearing with me despite the poor usage of words and answering the question :D As I said, I'll be more careful with future questions – Parth Thakkar Aug 25 '13 at 10:48
  • I'm not sure about that. For a point on the axis that is between the focus and the vertex, I think the axis itself is the only normal. – hmakholm left over Monica Aug 25 '13 at 10:48
  • @HenningMakholm, could you please elaborate? – Parth Thakkar Aug 25 '13 at 10:49
  • Why would a point between the vertex and the focus on the axis be different, @HenningMakholm ? – DonAntonio Aug 25 '13 at 10:50
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    @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. – hmakholm left over Monica Aug 25 '13 at 10:50
  • I'm not sure what is meant here by "infinitesimally close" to a point on the plane...For example, with $,y=x^2;$ , the focus is at $,\left(0,,,\frac14\right);$ . From any point $,(0,y);,;y>0;$ we can draw three normals to three different points on the parabola, one of these being the vertex $,(0,0),$...am I missing something here? – DonAntonio Aug 25 '13 at 10:53
  • @DonAntonio: I'm denying that you can draw three normals that intersect $(0,y)$ when $y<1/4$. Choose any point on the parabola except for $(0,0)$, and its normal will intersect the axis above $(0,1/4)$. – hmakholm left over Monica Aug 25 '13 at 10:55
  • @DonAntonio, the infintesimally close point HenningMakholm is talking about is the point infinitesimally close to the vertex on the parabola. The wording confused me too, but on reading 2-3 times, I understood what he says. That doesn't mean I am fully convinced about what he says. – Parth Thakkar Aug 25 '13 at 10:56
  • I'm not even sure what "intersecting the $,x-$axis above the point $,(0,,,1/4);$ " can possibly mean... – DonAntonio Aug 25 '13 at 10:57
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    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. – hmakholm left over Monica Aug 25 '13 at 10:58
  • @DonAntonio: Who said anything about intersecting the X-axis? The axis we're speaking about is the axis of the parabola (which also happens to be the $y$-axis of the coordinate system). – hmakholm left over Monica Aug 25 '13 at 10:59
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    Ah, too bad you realized that now, @HenningMakholm : I was almost done with a proof...hehe. +1 – DonAntonio Aug 25 '13 at 10:59
  • Yes @HenningMakholm, with all those x's I misread "axis" by "x-axis" – DonAntonio Aug 25 '13 at 11:00
  • @HenningMakholm, what?!? I don't understand what you mean by 'osculates the parabola at the vertex'. – Parth Thakkar Aug 25 '13 at 11:01
  • @DonAntonio, I'd like to see the proof if you've almost written it. – Parth Thakkar Aug 25 '13 at 11:01
  • @Parth, http://en.wikipedia.org/wiki/Osculating_circle – hmakholm left over Monica Aug 25 '13 at 11:02
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    Ah, got it. My dictionary (yes, I didn't google but dictionaried) called it 'intersecting at at least 3 points' - the source of all confusion. – Parth Thakkar Aug 25 '13 at 11:04
  • @ParthThakkar , a sketch :at any point $,(a,a^2),$ on $,y=x^2;$, the tangent line's given by $;y=2ax-a^2;$ and the normal there is $$y=-\frac1{2a}x+\frac12+a^2$$ . We can see the normal can intersect any point $,(0,b);,;b>0;$ since $,a+a^3=b ,,,b>0,$ has always a solution... – DonAntonio Aug 25 '13 at 11:05
  • Do you mean $y=-\frac1{2a}x+\frac12\large{+}a^2 $? – Parth Thakkar Aug 25 '13 at 11:09
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    Indeed so, @ParthThakkar. I corrected the last line yet I forgot to do it on the equation itself. – DonAntonio Aug 25 '13 at 11:10
  • I don't understand 2 parts: you say the normal can intersect at any point $(0,b)$ because $a + a^3 = b$ always has a solution. How did you get there? And how do I know that the equation has always got a solution? (I am more concerned about the 1st part) – Parth Thakkar Aug 25 '13 at 11:15
  • Perhaps the fact that the equation always has a solution for a positive $b$ follows from the intermediate value theorem, right? – Parth Thakkar Aug 25 '13 at 11:15
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    Leave it! Understood. Thanks a lot :D – Parth Thakkar Aug 25 '13 at 11:17
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    Just take the $,y-$intersect of the normal and equal it to $,b,$, @ParthThakkar ...About existence: differentiate wrt $,a,$ : $$(a^3+a)'=3a^2+1$$ and we get an increasing continuous function of $;a;$ with minimum at $,0,$ ... – DonAntonio Aug 25 '13 at 11:18
  • @DonAntonio: The $y$-intercept of $-\frac{1}{2a}x+\frac 12 + a^2$ is at $y=\frac 12 + a^2$ which is always at least $\frac 12$. So points $(0,y)$ with $y<\frac 12$ cannot be hit by normals from the side. – hmakholm left over Monica Aug 25 '13 at 11:26
  • It was meant the $,x-$intercept, @HenningMakholm...and you are right and that's exactly the condition Parth was looking for! – DonAntonio Aug 25 '13 at 11:32
  • I'm with Henning here. If a point $P$ on the axis is sufficiently close to the vertex (specifically, if the radius of curvature at the vertex is greater than the distance from $P$ to the vertex), there will only be one normal from $P$ to the parabola. – TonyK Aug 25 '13 at 11:32
  • After many... many confusions and misunderstandings, @TonyK , I added this last part of Henning's comments to the answer above and I think we can agree... almost ... on that. – DonAntonio Aug 25 '13 at 11:36
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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 $h> 2a$.

Glorfindel
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