Edited to try to clarify OP's confusion:
The distance between $(x,y)$ and $(0,5)$ is indeed $\sqrt{x^2+(y-5)^2}$.
The distance between $(x,y)$ and the line $y=0$ is $|y|$ (if $y>0$, then it is above the $X$-axis, and the distance is just the $y$ coordinate; if $y<0$, then the distance is $-y = |y|$). So the points on the parabola are exactly the points for which the two distances are equal, that is, all $(x,y)$ for which:
$$\sqrt{x^2+(y-5)^2} = |y|.$$
Now square both sides, and you'll get the equation of the parabola you want.
How is this related to the formula you quote? It's really all just the same process. If the directrix is either horizontal (a line of the form $y=k$) or vertical (a line of the form $x=\ell$), then it is very easy to compute the distance from a point to the directrix: the point $(x,y)$ is $|y-k|$ away from the line $y=k$, and is $|x-\ell|$ away from the line $x=\ell$. If the focus of the parabola is at $(a,b)$, then you want the distance from $(x,y)$ to $(a,b)$ to equal the distance to the directrix, so you get:
\begin{align*}
\sqrt{(x-a)^2 + (y-b)^2} &= |y-k| &\qquad&\mbox{if the directrix is $y=k$,}\\
\sqrt{(x-a)^2 + (y-b)^2} &= |x-\ell| &&\mbox{if the directrix is $x=\ell$.}
\end{align*}
Then squaring both sides yields the equation of the parabola. Doing it correctly will cancel out the $y^2$ term when the directrix is horizontal, and the $x^2$ term when the directrix is vertical.
For more general parabolas, when the directrix is neither horizontal nor vertical but an arbitrary line $y=mx+b$, you need to work a bit harder, because the distance from $y=mx+b$ to $(x,y)$ is not so simple to compute. Not too hard, but not as simple. Once you have a formula for the distance, you set it equal to the distance to the focus, square both sides, and get the equation of the parabola.
I'm assuming you don't want any calculus involved here (too bad!), so let $\,(a,b)\,,\,(c,d)\,$ be the points on the parabola through which pass two tangent lines to it that are perpendicular:
$$\text{First tangent: we need to solve the system}\;\;\;\;y^2=4px\;\;,\;\;y-b=m(x-a)\Longrightarrow$$
$$(m(x-a)+b)^2=4px$$
$$\text{Second tangent: we need to solve the sytem}\;\;\;\;y^2=4px\;\;,\;\;y-d=-\frac{1}{m}(x-c)\Longrightarrow$$
$$\left(-\frac{1}{m}(x-c)+d\right)^2=4px$$
Of course, solving the above take into account that
$$b^2=4pa\;\;,\;\;d^2=4pc$$
since both points were chosen to be on the parabola.
Also, remember that two straight lines (none of which is horizontal/vertical) with slopes $\,m_1\,,\,m_2\,$ are perpendicular iff $\,m_1m_2=-1\,$, and this the reason we took the second tangent's slope to be $\,-1/m\,$
Best Answer
Yes, they are the same. In the textbook method, $2ydy=4pdx\rightarrow \frac{dy}{dx}=\frac{2p}{y}$. The slope $m=\frac{B}{A}=\frac{2p}{y}$. So $y=\frac{2p}{m}$.
Plugging this into the original equation $y^2=4px$ gives you $x=\frac{p}{m^2}$. Using the slope-intercept form of a line $y=mx+b$ would give you $b=\frac{p}{m}$.