Shortest distance between parabola and point 
Find the shortest distance between the parabola defined by $y^2 = 2x$ and a point $ E:= (1.5, 0)$.

I can't use the distance formula because I'm missing a set of points $(x, y)$ to plug into. So, instead, I have a normal that passes through the point $E$ from the parabola. Which is the definition of the shortest distance to a point.
$$y - y_1 = m(x - x_1)$$
The slope of the normal is $\frac{1}{y_1}$ by using implicit differentiation and that's where I'm stuck, because I plug the point E into it and I get
$$y_1^2=x_1-1.5$$
How do I prove the shortest distance is $\sqrt{2}$?

For a point $(x,y)$ on the parabola we have that
$$d^2=(y-0)^2+\left(x-\frac32\right)^2=y^2+x^2-3x+\frac94=x^2-x+\frac94$$
and the minimum is attained for $x=\frac12$ thus at the point $(\frac12,1)$ and therefore
$$d=\sqrt{1^2+1^2}=\sqrt 2$$
