Edited parabola question

Question

I am confused with the equation of a parabola. My teacher told me that it is in the form $$\text{(axis of parabola)}^2=4\text{(vertex of parabola)}$$ I feel that $\text{(the axis on which the vertex of parabola is)}^2$ should be $4\text{(axis of parabola)}$ For example: if the equation is $(y-a)^2=4(x-b)$ then is it in the form $$\text{(axis of parabola)}^2=4\text{(the axis on which vertex of parabola is)}$$ or the other way round. This is an edited version the question posted earlier.

Answer 1

The proper equation is: $$y^2=4ax$$ where $a$ is the distance of the focus point to the vertex.

In your terminology:

(distance from axis of parabola)${}^2$ = 4 (vertex-to-focus) (distance along axis of parabola)

$\LaTeX$ code:

\documentclass[border=10pt]{standalone}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{arrows}
\begin{document}

\begin{tikzpicture}
\draw[gray!50, very thin,-triangle 60] (-2,0) -- (4,-0); % x-axis
\draw[gray!50, very thin,-triangle 60] (0,-4) -- (0,4);  % y-axis
\draw[red,line width=2pt] (1,0) -- (1,2) node[right=1pt] {$\ell=2a$}; % semi latus rectum
\draw[rotate=-90,line width=2pt] (0,0) parabola (4,4);
\draw[rotate=-90,line width=2pt] (0,0) parabola (-4,4);
\node at (2.5,-2) {$y^2=4ax$ };
\node at (4,-3) {$r=\dfrac{2a}{1-\cos\theta}$ };
\node at (1.5,-3.5) {$(a t^2, 2at)$ };
\draw (-1,-4) -- (-1,3.5) node[above=1pt] {directrix}; % directrix
\node at (-0.5,3pt) {a};
\node at (0.5,3pt) {c=a};
\fill (1,0) circle (0.1); % focus
\end{tikzpicture}

\end{document}

Answer 2

It should be (Axis of parabola)$^2$ = (Length of latus rectum) * (Tangent at vertex$)$. For example, take: $y^2=4ax$. y=0 is the x-axis. Also, it forms the axis of the parabola. x=0 is the y-axis. Again, it forms the tangent at vertex. 4a is cleqrly the length of the latus rectum, i.e. the length of the focal chord perpendicular to the axis.