The given curve is the parabola $y=x^2$ with point $(0,9)$.
We want to minimize $D^2=f(x,y)=x^2 + (y-9)^2$ with the constraint equation $g(x,y) =y-x^2=0$.
After finding the gradients of both $f$ and $g$ we get: $$2x = -2 \lambda x$$ $$2y-18=\lambda$$ Plugging in $\lambda$ from our $y$ equation into the $x$ equation yields that either $$x=0$$ $$y=\frac{19}{2}$$ When $x=0$ then $y=0$ and when $y=\frac{19}{2}$ then $x=\pm\sqrt{\frac{19}{2}}\space$,we cant to minimize the equation so we use the negative. Using these values of $x$ and $y$ in $D^2$ gives $$D^2=\sqrt{\frac{39}{4}}$$ $$D=\frac{\sqrt{39}}{2}$$ But this is where I run into a problem because $\frac{\sqrt{39}}{2}=9.75$ which is greater than my original point. So how do I find the lowest point then?
$f(x,y,\lambda) = x^2 + (y-9)^2 - \lambda (x^2 - y)\\ \frac {\partial f}{\partial x} = 2x (1-\lambda) = 0\\ \frac {\partial f}{\partial y} = 2(y-9) +\lambda = 0\\ \frac {\partial f}{\partial y} = y-x^2 = 0$
From the first equation, either $x = 0$ or $\lambda = 1$
If $x = 0$ then $(0,0)$ is a critical point. It looks like this is a local maximum.
If $\lambda = 1$
$2y - 17 =0\\ y = \frac {17}{2}$
This appears to be your error.
$(\sqrt \frac {17}{2} , \frac {17}{2})$ is a critical point.
Plugging this into:
$D^2 = x^2 + (y-9)^2\\ D^2 = \frac {17}{2} + (\frac {17}{2}-9)^2\\ D^2 = \frac {35}{4}$
$D$ is slightly less than $3$