Parabola to a point

Let the fixed point be $Q(15, 20)$ and point $P(x,y)$ on the parabola where $PQ$ is the shortest. Then $PQ$ must be perpendicular to the tangent at point $P$ . Since the gradient at point $P$ is given by $\dfrac {dy}{dx} = \dfrac x{10}$ . Then the gradient of line $PQ$ is $-\dfrac {10}x$ and the following equation is true:

$\begin{aligned} \frac {\blue y-20}{x-15} & = - \frac {10}x & \small \blue{\text{Note that }y = \frac {x^2}{20}} \\ \frac {x^2}{20} - 20 & = \frac {150}x - 10 \\ \implies x^3 - 200x - 3000 & = 0 \end{aligned}$

We can use Cardano's method to solve the cubic equation as follows:

$\begin{aligned} x^3 - 200x - 3000 & = 0 & \small \blue{\text{Let }x = u+v} \\ u^3 + v^3 + (3uv-200)(u+v) - 3000 & = 0 & \small \blue{\text{Assume }3uv-200=0} \\ u^3 + v^3 & = 3000 & \small \blue{\implies u^3v^3 = \frac {8000000}{27}} \end{aligned}$

By Vieta's formula , $u^3$ and $v^3$ are roots of the following quadratic equation.

$\begin{aligned} w^2 - 3000 w + \frac {8000000}{27} & = 0 \end{aligned}$

$\implies w \approx \begin{cases} 2897.749514 & \implies u \approx 14.25674168 \\ 102.2504861 & \implies v \approx 4.676150286 \end{cases}$

$\implies x = u+v \approx 18.93289196 \implies y \approx 17.9227199$ and $PQ \approx \sqrt{(15-18.93289196)^2+(20-17.9227199)^2} \approx \boxed{4.45}$ .

Let us take a point $(h, \dfrac{h^2}{20})$ on the parabola. Then the square of the distance of this point from $(15,20)$ is $(h-15)^2+(\dfrac{h^2}{20}-20)^2$ . This attains a minimum when $h^3-200h-3000=0$ or $h=18.93289...$ . The minimum distance is $\boxed {4.4477782981...}$