Point on parabola which is at shortest distance? Find a point on parabola $y^2 =4x$ which is at shortest distance from $(1,0)$.
Answer given is $(0,0)$ 
But I am getting imaginary values.
How do i get $(0,0)$
 A: Note : $y^2=4x$ is a parabola, vertex $(0,0)$, symmetric about the $x-$axis.
We have $x  \ge 0.$
Distance$^2$: 
$d^2= y^2 + (x-1)^2 =4x +(x-1)^2$
$d^2=x^2+2x+1= (x+1)^2.$
$ \min  (d^2) =$
$ (0+1)^2 =1$ at  $x=0$ (why?)
(Recall $x \ge 0$)
The point on the parabola that minimizes $d^2$ to $(1,0)$ is (0,0)
Note : $d^2$ has been minimized, does this imply $d$ is minimal?
A: There are two ways you can understand this. 

First is the statement, that The vertex is the point on a parabola closest to the directrix. Since the eccentricity of a parabola is 1, the distance from any other point on the parabola to the directrix and focus is greater than the distance from the vertex to the focus.

And here $(1,0)$ is the focus and $(0,0)$ the vertex of this parabola.
The next way is actually finding the point at shortest distance from $(1,0)$
As, $x = \frac {y^2}{4}$
$\begin{align} D(y) &=  \sqrt{(x-1)^2 + (y-0)^2} \\ &=  \sqrt{(\frac {y^2}{4}-1)^2 + (y-0)^2} \end{align}$
For distance to be minimal,
$\begin{align} \frac{d}{dy}[(\frac {y^2}{4}-1)^2 + (y-0)^2] &= 0 \\ 2(\frac {y^2}{4}-1)y + 2(y-0) &= 0 \\ y^3 - 0 &= 0 \\ y &= 0\,\,\implies \,\,x  = 0 \end{align} $
A: Actually $(1,0)=F$ is a Focus of a parabola. Call $P$ the point on the parabola.  Then the ratio of the distance b/w $FP$ and the $M$ is $1$, where $M$ is the distance from $P$ to the directrix. That is $$\frac{FP}{M}=1$$ and so $FP=M$ 
We want a least $M$ 
Now the least distance from directrix to parabola is $1$, namely, the distance from $(-1,0)$ to $(0,0)$  , which is same as the distance from  $(0,0) $ to $ (1,0)$, so $(0,0)$ is a required point
For any other point $(h,k)$ on the parabola, the distance from the directrix  $x=-1$ to $(h,k)$ is  $>1$ 
