TITLE: Shortest distance between parabola and point
QUESTION [8 upvotes]: 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 $\sqrt{2}$?

REPLY [2 votes]: A reply to an old comment brought me back here, and I noticed that no one had offered a Lagrange multiplier solution.  You're looking for the minimum of
$$
    f(x,y) = (x-1.5)^2 +y^2
$$
subject to the constraint that $g(x,y) = 0$, where
$$
    g(x,y) = y^2-2x
$$
The critical points satisfy $\nabla f(x,y) = \lambda \nabla g(x,y)$ for some $\lambda$.  Or,
\begin{align*}
    2x-3 &= -2\lambda \\ 2y &= 2 \lambda y
\end{align*}
The second of these implies either $y=0$ or $\lambda = 1$.  If $y=0$, then $x=0$ and $f(0,0) = \frac{9}{4}$.  If $\lambda = 1$, we can substitute it into the first equation and get $x = \frac{1}{2}$.  If $x =\frac{1}{2}$, then $y = \pm 1$, and $f(\frac{1}{2},\pm1) = 2$.  So the minimum value of $f$ is at these two points, and the shortest distance is $\sqrt{2}$.