# Shortest distance of $(0,c)$ from a parabola

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

The point on the parabola $(x,x^2)$.

$\mathscr{l}(x)=\sqrt{x^2+(y-c)^2}=\sqrt{x^2+(x^2-c)^2}\\\mathscr{l}(y)=\sqrt{y+(y-c)^2}$

If I follow the expression $\mathscr{l}(y)=0$ then its fine and i'll get $\mathscr{l}(y)$ is minimum at $y=\frac{2c-1}{2}$ and $\mathscr{l}_{min}=\frac{\sqrt{4c-1}}{2}$. $$\frac{d\mathscr{l}(y)}{dy}=\frac{1+2(y-c)}{2\sqrt{y+(y-c)^2}}\\ \frac{d\mathscr{l}(y)}{dy}=0\implies y=\frac{2c-1}{2}\\ \mathscr{l}_{min}=\mathscr{l}(\frac{2c-1}{2})=\frac{\sqrt{4c-1}}{2}$$

My Attempt

But, if I consider $\mathscr{l}(x)$

$$\frac{d\mathscr{l}(x)}{dx}=\frac{2x+2(x^2-c).2x}{2\sqrt{x^2+(x^2-c)^2}}=\frac{x+2x(x^2-c)}{\sqrt{x^2+(x^2-c)^2}}=\frac{x\big[1+2(x^2-c)\big]}{\sqrt{x^2+(x^2-c)^2}}$$ $$\mathscr{l}'(x)=0\implies x\big[1+2(x^2-c)\big]=0\\ \implies x=0 \text{ or } x^2=\frac{2c-1}{2}\\ \implies x=0 \text{ or } x=\sqrt{\frac{2c-1}{2}}\text{ as }x\geq 0\\$$ How do I deal with the critical point $x=0$, and why do i not get this case if I use $\mathscr{l}(y)$ ?

Note: I have checked a similar problem link which does not completely address my post here.

To avoid confusion, I use $\ell$ to denote the distance. So, $\ell=\sqrt{x^2+(x^2-c)^2}=\sqrt{y+(y-c)^2}$.

$$\frac{d\ell}{dx}=\frac{d\ell}{dy}\cdot\frac{dy}{dx}=2x\frac{d\ell}{dy}$$

$\displaystyle \frac{d\ell}{dx}=0$ if and only if $\displaystyle \frac{d\ell}{dy}=0$ or $x=0$.

That's why we have one more critical point when differentiating w.r.t. $x$.

$$\frac{d\ell}{dx}=\frac{x\big[2x^2-(2c-1))\big]}{\sqrt{x^2+(x^2-c)^2}}$$

If $c>0.5$, $\displaystyle x=\pm\sqrt{\frac{2c-1}{2}}$ are local minimum points and $x=0$ is a local maximum point by the first derivative test.

If $0\le c\le 0.5$, $x=0$ is the only critical point which is a local minimum point by the first derivative test.

Note: When $0\le c<0.5$, $\displaystyle \frac{d\ell}{dy}>0$ for all $y\ge0$ and hence $\ell$ attains its least value when $y=0$.

Alternatively, we can use

$$\ell=\sqrt{x^4-(2c-1)x^2+c^2}=\sqrt{\left(x^2-\frac{2c-1}{2}\right)^2+\frac{4c-1}{4}}$$

If $c>0.5$, $\ell$ is the least if $\displaystyle x^2=\frac{2c-1}{2}$.

If $0\le c\le 0.5$, $\displaystyle \frac{2c-1}{2}\le0$ and $\ell$ is the least if $\displaystyle x=0$.

• Thanx. Could u pls explain a bit more on how you reached the conclusion that "if $c>0.5$, $x=\pm\frac{2c-1}{2}$ are local minimum and $x=0$ is a local maximum and if $0\leq c\leq 0.5$, $x=0$ is the only critical point which is local minimum" – ss1729 Apr 17 '18 at 16:53
• If $c<0.5$, it is impossible that $2x^2-(2c-1)$ as $-(2c-1)$ would be positive. So the derivative is zero only when $x=0$. If $x$ is slightly larger than $0$, then $\frac{2x^2-(2c-1)}{\sqrt{x^2+(x^2-c)^2}}$ is positive and hence $\frac{x[2x^2-(2c-1)]}{\sqrt{x^2+(x^2-c)^2}}$ is positive. If $x$ is slightly smaller than $0$, then $\frac{2x^2-(2c-1)}{\sqrt{x^2+(x^2-c)^2}}$ is positive and hence $\frac{x[2x^2-(2c-1)]}{\sqrt{x^2+(x^2-c)^2}}$ is negative. By first derivative test, $x=0$ is a minimum point. – CY Aries Apr 17 '18 at 16:59
• If $c>0.5$, $2x^2-(2c-1)=0$ if $x=\pm\sqrt{\frac{2c-1}{2}}$. When $x$ is slightly larger than $\sqrt{\frac{2c-1}{2}}$, $2x^2-(2c-1)>0$ $x>0$ and $\sqrt{x^2+(x^2-c)^2}>0$. So, the derivative is positive. When $x$ is slightly smaller than $\sqrt{\frac{2c-1}{2}}$, $2x^2-(2c-1)<0$ $x>0$ and $\sqrt{x^2+(x^2-c)^2}>0$. So, the derivative is negative. By the first derivative test, $x=\sqrt{\frac{2c-1}{2}}$ is a minimum point. The case for $x=-\sqrt{\frac{2c-1}{2}}$ can be similarly argued. – CY Aries Apr 17 '18 at 17:04
• When $c>0.5$, the derivative is also $0$ when $x=0$. When $x$ is slightly larger than $0$, $x>0$, $2x^2-(2c-1)<0$ and $\sqrt{x^2+(x^2-c)^2}>0$. So the derivative is negative. When $x$ is slightly smaller than $0$, $x<0$, $2x^2-(2c-1)<0$ and $\sqrt{x^2+(x^2-c)^2}>0$. So the derivative is positive. By the first derivative test, $x=0$ is a maximum point. – CY Aries Apr 17 '18 at 17:08
• $c>\frac{1}{2}\implies 2c>1\implies 2c-1>0$ and $x>0\implies 2x^2>0$. From there how do you conclude $2x^2-(2c-1)<0$ ? – ss1729 Apr 17 '18 at 17:18

A projection of $P(0,c)$ on the parabola is the point $Q(x,x^2)$ with $x\geq 0$ such that $x^2+(x^2-c)^2$ is minimized. This happens only if $\frac{d}{dx}\left[x^2+(x^2-c)^2\right]=0$. In particular, if $c\leq \frac{1}{2}$ the closest point is the vertex of the parabola$^{(*)}$ and the wanted distance is $c$. If $c>\frac{1}{2}$ the closest point in the first quadrant is $\left(\sqrt{c-\frac{1}{2}},c-\frac{1}{2}\right)$ and the wanted distance equals $\sqrt{c-\frac{1}{4}}$.

(*) It makes sense: the osculating circle at the vertex of the parabola has its center at $\left(0,\frac{1}{2}\right)$.

Remarkably, to project a generic point on a parabola is not task which can be solved by straightedge and compass only. Indeed, by assuming $c>0$, the projection of $P(c,0)$ on the parabola is the point $Q(x,x^2)$ with $x>0$ such that $(x-c)^2+x^4$ is minimized. This happens only if $$\frac{d}{dx}\left[(x-c)^2+x^4\right]=0\quad \Longleftrightarrow\quad 2x^3+x=c \quad \stackrel{x=\sqrt{\frac{2}{3}}z}{\Longleftrightarrow}\quad 4z^3+3z=3c\sqrt{\frac{3}{2}}$$ and since $\sinh(3\theta)=4\sinh^3(\theta)+3\sinh(\theta)$, the solution is given by $$x = \sqrt{\frac{2}{3}}\sinh\left(\frac{1}{3}\text{arcsinh}\left(3c\sqrt{\frac{3}{2}}\right)\right)$$ and the shortest distance is $$\sqrt{c^2+\frac{1}{3}\sinh^2\left(\frac{1}{3}\text{arcsinh}\left(3c\sqrt{\frac{3}{2}}\right)\right)-\sqrt{\frac{3}{2}}\sinh\left(\frac{1}{3}\text{arcsinh}\left(3c\sqrt{\frac{3}{2}}\right)\right)}$$ which over the interval $\left(0,\frac{1}{2}\right)$ is well-approximated by the simpler expression $\frac{c^2}{1+2c^2}$, over the interval $\left[\frac{1}{2},5\right)$ is well-approximated by the simpler expression $\sqrt{c^2+\frac{4}{3}}-1$.