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"A parabola with vertex (0,1) is touching the x-axis , what is the minimum integral distance of the point of contact of parabola with the x-axis from the origin"

My attempt :

let the point of contact of parabola with x axis be (h,0) , and the slope of axis of the parabola be m So Equation of Axis is : y-1=mx

Now using reflection property of a parabola , a line with slope "-m" and passing through the point (h,0) would meet the axis at focus

So I found the co-ordinates of focus as ($h\over 2$-$1\over 2m$ , $mh\over 2$+$1\over 2$)

But I am stuck on how to go about finding a constraint in order to find the minimum positive value of h

Could someone please help me to proceed further ?

Thanks in advance

ARROW
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  • Hi and welcome the Mathematics Stackexchange! Did you try to solve this problem? What was the difficulty you encountered? Here's how I would start. All parabolas that touch the $x$-axis are of the form $$ f(x) = a(x-x_0)^2, \qquad a, x_0 \in \mathbb{R} $$ Then you only need to plug in the other condition... – Matti P. Apr 01 '20 at 11:01
  • … and note that the point of contact with the $x$ axis is $(x_0,0)$, so the condition that this is an integral distance from the origin tells you that $x_0$ must be an integer. – gandalf61 Apr 01 '20 at 11:14
  • This isn’t a “do my homework for me” site. If you’d like to try asking this question again without it’s being closed again, see How to ask a homework question for things that you should include in your next attempt. – amd Apr 01 '20 at 20:48
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    @MattiP. That’s not all of the parabolas that touch the $x$-axis. It’s only the ones with a vertical axis of symmetry. – amd Apr 01 '20 at 20:50
  • @amd thanks for the reference and guidance , I edited it , if I still missed something , any help would be appreciated , I did attempt the question before asking , I am new here so I didn't know what I should have included, I did try to format the question in order to show that it was taken up from somewhere , but I couldn't figure out how to format it that way , so I put it in quotes , that's all I could do – ARROW Apr 02 '20 at 04:19
  • Note that the question isn’t asking for just the minimum distance, but for the minimum integral distance. – amd Apr 03 '20 at 18:37

1 Answers1

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I’m not sure that knowing where the focus is really helps here. The key, I think, is the tangency condition.

If the equation of the axis is $mx-y+1=0$, then an equation of the tangent at the vertex $(0,1)$ is $x+my-m=0$, therefore an equation of a parabola with vertex at $(0,1)$ has the form $$(mx-y+1)^2=p(x+my-m).\tag1$$ We can get away with using a slope for the axis because a parabola with a vertical axis obviously isn’t possible here.

If $(h,0)$ lies on this parabola, we have $$(mh+1)^2=p(h-m).\tag2$$ The tangent to the parabola at this point has an equation of the form $$(mh+1)(mx-y+1)=\frac p2(x+my-2m+h)$$ which can be rearranged into $$\left(m+m^2h-\frac p2\right)x-\left(hm+\frac12mp+1\right)y+\left(hm+mp-\frac12hp+1\right)=0.\tag3$$ From equation (2) we have $p={(mh+1)^2\over h-m}$. Both the coefficient of $x$ and the constant term of equation (3) must vanish. Substituting into the former and factoring yields the condition $${(mh+1)(2m^2-mh+1)\over2(m-h)}=0.\tag4$$ Solving for $m$ in terms of $h$ and discarding the degenerate solution will give you a constraint on $h$. If you start instead with the constant term of equation (3) you will get one more factor than (4) in the resulting condition, but that additional factor has no real roots.

amd
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  • Got it ! Thanks a ton , and while reading this an idea came to my mind that foot of perpendicular to any tangent from the focus lies on the tangent at vertex so m = h/2 - 1/2m , which yields the same quadratic. I totally forgot about making use of tangent at vertex without reading this in my initial attempt , thanks again. – ARROW Apr 04 '20 at 03:41