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 $\sqrt{2}$?

Given a point $P=\left(\frac{y^2}2,y\right)$ of your parabola, consider the line segment joining $P$ to $C=\left(\frac32,0\right)$. The slope of this line segment is $\frac{2y}{y^2-3}$. And the slope of the tangent to the parabola at $P$ is $\frac1y$. Since two lines are orthogonal if and only if one of them is horizontal and the other one is vertical or when the product of their slopes is $-1$, these lines are orthogonal if and only if $y=0$ or $\frac2{y^2-3}=-1$, which means that $y=0$ or that $y=\pm1$. Forget $0$: that's a local maximum. So, the distance from the parabola to $C$ is$$\left\|\left(\frac12,1\right)-\left(\frac32,0\right)\right\|=\sqrt2.$$

