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?
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).