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I have the parametric parabola: $$ y=f(x)=C(x-4)(x-5)+D $$ where $D$ is fixed.

I want to find for which value of $C$ the distance from the parabola to the point $(4,0)$ is exactly $\frac{1}{3}$ and the point $(4,0)$ lies at the left of the curve.

The staightforward way is to set up this equation: $$ \sqrt{(x-4)^2+(f(x))^2}=\frac{1}{3} $$ that by squaring becomes: $$ (C^2(x-5)^2+1)(x-4)^2+2CD(x-5)(x-4)+D^2-\frac{1}{9}=0 $$ and find the solutions with $x>4$. The value for $C$ that makes them coincide is the solution of our problem.

Is there another way I can use to solve this problem that does not use the full formula for quartic polynomials? I know that the parabola I am looking for has the same distance also from the point $(5,0)$ (we have a simmetry on the axis $x=\frac{9}{2}$), can I use that information to find a second order solving equation?

N74
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    Try using calculus instead of algebra. – Mikhail Katz Feb 03 '16 at 10:12
  • The point $x$ where the distance is minimized satisfies $((4,0)-(x,f(x))\perp f'(x)$ – Gregor de Cillia Feb 03 '16 at 10:13
  • @Gregor de Cillia It's not a minimization problem, it's the intersection of the parabola with another second-degree curve. – Justpassingby Feb 03 '16 at 10:17
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    @N74 May I suggest that you change the title into "Intersection of a parabola and a circle?" I was also misled into thinking we were looking for a minimum distance. – Justpassingby Feb 03 '16 at 10:18
  • @Justpassingby Whenever you are calculating the distance of a point $x$ and a set $A$ (in this case the graph of the parabola), you have to minimize. $d(x,A):=\min_{y\in A}|x-y|$ – Gregor de Cillia Feb 03 '16 at 10:21
  • @Gregor de Cillia The title suggests that we are looking for $d(x,A)$ but the actual question is about something different. – Justpassingby Feb 03 '16 at 10:23
  • @user72694 Finding a zero for $2(x-4)+2*f'(x)f(x)$ is no way easier than findind the solution of the quartic. And when I find that zero, still I have to substitute it into the equation to verify the distance. – N74 Feb 03 '16 at 10:23
  • @GregordeCillia Nice point, but still I need to solve a cubic polynomial. I feel there is something simpler. – N74 Feb 03 '16 at 10:25
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    @N74 The cubic is simpler than the quartic but irrelevant. All of these commenters including myself have been misled by the title into thinking it was about minimal distance. – Justpassingby Feb 03 '16 at 10:26
  • @Justpassingby, it is about the minimal distance, with a pair of parameters in addition. What is the distance to the parabola if not the minimal distance? – Mikhail Katz Feb 03 '16 at 10:31
  • @user72694 Read the question without paying attention to the title (which is wrong). – Justpassingby Feb 03 '16 at 10:37
  • @GregordeCillia I think you hint could lead to the solution. If you can assemble a full answer I will accept it. – N74 Feb 03 '16 at 10:42
  • @Justpassingby, I agree with you that the title is misleading, but the finding the minimal distance from parabola to a point is still solved using calculus (your solution) even if it depends on parameters! Afterwards one plugs in 1/3 to find the parameters. – Mikhail Katz Feb 03 '16 at 10:44
  • Not sure why the downvotes, though the title could be improved. – Mikhail Katz Feb 03 '16 at 10:45
  • @N74 Unfortunately, the equation resulting from the orthogonality condition $(4-x)-f(x)f'(x) = 0$ has no simple solution, see here. Therefore, I'm stuck at this point – Gregor de Cillia Feb 03 '16 at 11:40
  • The parabola with a given symmery axis has to touch a given circle. Can we get anywhere nearer by parametrizing the tangent point as a variable point on the circle? – Justpassingby Feb 03 '16 at 13:05
  • @GregordeCillia Yes, I made a mistake and thought I had to solve $(x-4)f'(x)-f(x)=0$ making the dot product with the normal vector. Too bad I have to make the dot product with the tangent vector... so still a 3rd order equation. – N74 Feb 03 '16 at 13:18
  • @Justpassingby I tried to substitute $x=1/3cos\theta+4$ but still I have a monster with all the powers of $cos\theta$ – N74 Feb 03 '16 at 13:43

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We are not necessarily looking for an explicit solution to the quartic equation in $x,$ as long as we find a nice relationship between $C$ and $D$ expressing the fact that the quartic equation has at least a double root.

That relationship is the condition that the discriminant of the quartic equation has to be zero. It can be obtained by eliminating $x$ from the system of two equations: (1) the quartic; and (2) its derivative.

The discriminant of a general quartic is a rather horrible homogeneous 6-th degree polynomial in the coefficients, with 16 terms.

In our particular case of a quartic with vanishing third-degree term there are only 6 terms, and they are all multiples of the leading coefficient so we are effectively left (after excluding the 'easy' solutions having $C=0,$ i.e., $|D|=1/3$) with a homogeneous 5-th degree polynomial. See https://en.wikipedia.org/wiki/Discriminant#Formulas_for_low_degrees

The result is an explicit polynomial relationship between $C$ and $D.$ It is at most of the 10th degree in $C$ (I have not verified it by machine but according to a quick pen-and-paper calculation the 10th degree coefficient is nonzero).

Justpassingby
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