Question

Find the shortest distance of the point $\left(0,c\right)$ from the parabola $y=x^2$ where $0\le c\le 5$

Answer 1

Step 1.Find the critical point

Let the point on the parabola be $\left(x,y\right)$

Therefore The point is $\left(x,x^2\right)$

Therefore the distance of the point $\left(x,x^2\right)$ from $\left(0,c\right)$ is

$$\begin{gathered} \Rightarrow s\left(x\right)=\sqrt{{\left(x-0\right)}^2+{\left(x^2-c\right)}^2} \\ \Rightarrow s\left(x\right)=\sqrt{x^2+x^4-2x^{2}c+c^2} \\ \Rightarrow s^2\left(x\right)=x^2+x^4-2x^{2}c+c^2 \end{gathered}$$

For critical point $\frac{d{s}^2\left(x\right)}{dx}=0$

$$\begin{gathered} \therefore 2x+4x^3-4xc=0 \\ \Rightarrow 4x\left(x^2+\frac{1}{2}-c\right)=0 \\ \Rightarrow x=0,x=\pm \sqrt{c-\frac{1}{2}} \end{gathered}$$

Step 2. Find the point of minima

For minima $\frac{d^2{s}^2\left(x\right)}{d{x}^2}>0$

$\therefore \frac{d^2{s}^2\left(x\right)}{d{x}^2}=2x+12x^2-4c$

Case 1. For $x=0$

$$\begin{gathered} {\left[\frac{d^2{s}^2\left(x\right)}{d{x}^2}\right]}_{x=0}=-4c \\ <0\left[\because c\in \left[0,5\right]\right] \end{gathered}$$

Case 2. For $x=\sqrt{c-\frac{1}{2}}$

Clearly $c\ge \frac{1}{2}$

Therefore $c\in \left[\frac{1}{2},5\right]$

$$\begin{gathered} {\left[\frac{d^2{s}^2\left(x\right)}{d{x}^2}\right]}_{x=\sqrt{c-\frac{1}{2}}}=2\sqrt{c-\frac{1}{2}}+12\left(c-\frac{1}{2}\right)-4c \\ =2\sqrt{c-\frac{1}{2}}+8c-6 \\ >0\left[\because c\in \left[\frac{1}{2},5\right]\right] \end{gathered}$$

Case 3:For $x=-\sqrt{c-\frac{1}{2}}$

Clearly $c\ge \frac{1}{2}$

Therefore $c\in \left[\frac{1}{2},5\right]$

$$\begin{gathered} {\left[\frac{d^2{s}^2\left(x\right)}{d{x}^2}\right]}_{x=-\sqrt{c-\frac{1}{2}}}=-2\sqrt{c-\frac{1}{2}}+12\left(c-\frac{1}{2}\right)-4c \\ =-2\sqrt{c-\frac{1}{2}}+8c-6 \\ <0\left[\because c\in \left[\frac{1}{2},5\right]\right] \end{gathered}$$

Therefore the point of minimum is $\left(\sqrt{c-\frac{1}{2}},c-\frac{1}{2}\right)$

Step 3 Find the shortest distance

The shortest distance is

$$\begin{gathered} s\left(x\right)=\sqrt{{\left(x-0\right)}^2+{\left(y-c\right)}^2} \\ =\sqrt{{\left(\sqrt{c-\frac{1}{2}}-0\right)}^2+{\left(c-\frac{1}{2}-c\right)}^2} \\ =\sqrt{c-\frac{1}{2}+\frac{1}{4}} \end{gathered}$$

$=\sqrt{c-\frac{1}{4}}\forall c\in \left[\frac{1}{2},5\right]$

Hence, the shortest distance is $\sqrt{c-\frac{1}{4}}$