Question

Find the shortest distance of the point $(0,c)$ from the parabola $y=x^2,$ where $\frac{1}{2}\le c\le 5.$

Answer 1

Given parabola is $y=x^2$
Let the point on the parabola be $(h,k)$
Distance between $(h,k)$ and $(0,c)$
$D=\sqrt{(h-0{)}^2+(k-c{)}^2}$
$\Rightarrow D=\sqrt{h^2+(k-c{)}^2}$
As $(h,k)$ lies on $y=x^2,$
so $k=h^2$
Now,
$D=\sqrt{k+(k-c{)}^2}$
We need the minimum value of $D.$
Differentiating w.r.t. $k,$

$D^{\prime }\left(k\right)=\frac{1+2(k-c)}{2\sqrt{k+(k-c{)}^2}}$
$\Rightarrow D^{\prime }\left(k\right)=\frac{[2k-2c+1]}{2\sqrt{k+(k-c{)}^2}}$
$\Rightarrow D^{\prime }\left(k\right)=\frac{[k-\left(\frac{2c-1}{2}\right)]}{\sqrt{k+(k-c{)}^2}}$
Putting $D^{\prime }\left(k\right)=0,$ we get
$\frac{[k-\left(\frac{2c-1}{2}\right)]}{\sqrt{k+(k-c{)}^2}}=0$

$\Rightarrow k=\frac{2c-1}{2}$
Using first derivatives test,
$k<\frac{2c-1}{2}\Rightarrow D^{\prime }\left(k\right)<0$
$k>\frac{2c-1}{2}\Rightarrow D^{\prime }\left(k\right)>0$
So,
$k=\frac{2c-1}{2}$ is a point of local minima.
Therefore, the minimum distance is $D=\sqrt{k+(k-c{)}^2}$

$\Rightarrow D=\sqrt{\frac{2c-1}{2}+{(\frac{2c-1}{2}-c)}^2}$
$\Rightarrow D=\sqrt{\frac{2c-1}{2}+\frac{1}{4}}$
$\Rightarrow D=\sqrt{\frac{4c-2+1}{4}}$
$\therefore D=\sqrt{\frac{4c-1}{4}}$
Hence, the shortest distance is $\frac{\sqrt{4c-1}}{2}$