Let $b,p \in \mathbb{R}$, $p>0$. Find the distance from the point $(0,b)$ to the curve $x^2-4py=0$.

Question

Let $b,p \in \mathbb{R}$, $p>0$. Find the distance from the point $(0,b)$ to the curve $x^2-4py=0$.

The given curve can be identified by a parabola and in $\mathbb{R}^2$ the distance from a point $P$ to set $A$ is $$inf_{Q \in A}d(P,Q)$$ where $d$ is the euclidean metric in $\mathbb{R}^2$. My idea is to find a function in terms of the parabola $x^2-4py=0$ lets say $P(x)=(x, ?)$ leading to calculate the values $x$ and $b$ where $d(x)=d(b,P(x))$ takes its minimum values.

Answer 1

You're trying to minimize the distance $$\sqrt{(0-x)^2+(b-y)^2}$$ between $(0,b)$ and the parabola consisting of points $(x,y)$. You can substitute in $y=\frac{x^2}{4p}$ since $p>0$ and then the above distance is just a function of $x$ (which you can find the minimum of with routine calculus). In fact, you can omit the square root when doing the minimizing calculations (since the distance is minimized simultaneously with the square distance).

Answer 2

By the method of Lagrange multipliers:

Let $\displaystyle L=x^2+(b-y)^2 - \lambda (x^2 -4py)$.

$\displaystyle 0=\frac{1}{2} \frac{\partial L}{\partial x} = x-\lambda$

$\displaystyle 0=\frac{1}{2} \frac{\partial L}{\partial y} = b - y + 2p \lambda$

$\displaystyle 0= -\frac{\partial L}{\partial \lambda} = x^2-4py$

Combine the first two equations to get a linear relation between $x$ and $y$ and substitute into the third equation. Then use the quadratic formula or complete-the-square.