How can I create a function which describes the length between a parabola and a point on the y axis? 
This is what I'm thinking:
The distance of the line segment is between points (0,y) and$ (x,y)$ and the y coordinate of the parabola is $4-x^2
 so using he distance formula it is: 
$\sqrt{(x-0)^{2} +((4-x^2)-y)^{2}} = \sqrt{x^{2}+((4-x^2)-y)^2}$
Now what do I do? Do I take the derivative? Please help.
 A: Distance form point $(x,y)$ on the graph and $(0,2)$ can be parametrized by;
$L(x) = \sqrt{x^2 + (|y-2|)^2} = \sqrt{x^2 + (2-x^2)^2} = \sqrt{4 - 3 x^2 + x^4}$ (assuming $y>2$)
To find the minimum use the given point $\sqrt{3/2}$ and put it into $L(X)$
$L(\sqrt{3/2}) = \sqrt{4 - 3 (\sqrt{3/2})^2 + (\sqrt{3/2})^4} = \sqrt{4 - 3 \cdot 3/2 + 9/4} = \sqrt{ 16 - 18 + 9}/2 = \sqrt{7}/2$
A: Supposing that you don't know how (or don't want to) minimize the funtion $L(x)$, it is nice to get the same result by just looking for the radius of circle with center in $(0,2)$, and tangent to the given parabola.
The intersections between the generic circle with center in $(0,2)$ and radius $R$ and the parabola are given by the solution to
\begin{equation}
\begin{cases}
y = 4-x^2\\
x^2 + (y-2)^2 = R^2
\end{cases}
\end{equation}
Replacing $y$ into the second equation gives the biquadratic equation
\begin{equation}
x^4-3x^2+4-R^2=0.\tag{1}\label{biq}
\end{equation}
In order to have only two solutions this equation must have discriminant equal to $0$. That is
$$ \Delta = 9-16+4R^2=0,$$
which  implies $R=\frac{\sqrt{7}}{2}$, as in the shown Figure. Replacing than this value of $R$ in \eqref{biq}, gives the required abscissae.
