Question

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

Answer 1

Let $P(x,y)$ be any point on the parabola and $Q(0,c)$ be the given point. Then,
$P{Q}^2=x^2+(y-c{)}^2$
$\Rightarrow P{Q}^2=x^2+(x^2-c{)}^2\Rightarrow P{Q}^2=x^4+x^2(1-2c)+c^2$

Clearly, $PQ$ will be minimum when $P{Q}^2$ is minimum.
Let $z=P{Q}^2$. Then,
$z=x^4+x^2(1-2c)+c^2$
$\frac{dZ}{dx}=4x^3+2x(1-2c)$ and
$\frac{d^{2}Z}{d{x}^2}=12x^2+2(1-2c)$

The critical numbers of $Z$ are given by $\frac{dZ}{dx}=0$
$\Rightarrow 4x^3+2x(1-2c)=0$
$\Rightarrow 2x[2x^2+(1-2c\left)\right]=0$
$\Rightarrow x=0,x=\pm \alpha$, where $\alpha =\sqrt{\frac{2c-1}{2}}$

Now, at $x=0,$
$\frac{d^{2}Z}{d{x}^2}{|}_{x=0}=2(1-2c)>0$ when $0<c<\frac{1}{2}$

So, in this case i.e., when $0<c<\frac{1}{2},$ $x=0$ gives the shortest distance and is equal to $\sqrt{c^2}=c$

And at $x=\pm \alpha$,
$\frac{d^{2}Z}{d{x}^2}{|}_{x=\pm \alpha }=12\left(\frac{2c-1}{2}\right)+2(1-2c)=4(2c-1)>0$ when $c>\frac{1}{2}$

So, in this case $Z$ is minimum when $c>\frac{1}{2}$ and the minimum value of $PQ$ is given by
$P{Q}^2={\alpha }^2+({\alpha }^2-c^2{)}^2$
$\Rightarrow P{Q}^2=\frac{2c-1}{2}+{(\frac{2c-1}{2}-c)}^2=\frac{4c-1}{4}$
$\Rightarrow PQ=\frac{\sqrt{4c-1}}{2}$
Therefore, the shortest distance is $c$ when $0<c<\frac{1}{2}$ and $\frac{\sqrt{4c-1}}{2}$ when $c>\frac{1}{2}$