Shortest distance between parabola and point

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

$$d=\sqrt{1^2+1^2}=\sqrt 2$$

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$$\left\|\left(\frac12,1\right)-\left(\frac32,0\right)\right\|=\sqrt2.$$

Any point of the parabola is of the type $(x,Y)$ with $y^{2}=2x$. The distance between $(x,y)$ and $(1.5,0)$ is $\sqrt (x-1.5)^{2}+y^{2}=(x-1.5)^{2}+2x$. You have to find the minimum value of this quantity over all $x \geq 0$. [$x=(y^{2} /2) \geq 0$ on the curve]. Equivalently, you can mimimize teh square of teh distance. Differentiate and set the derivative equal to 0. You will find that the minimum is attained at $x=/2$. The minumum value is $\sqrt 2$.