Parabola

Tags: math

Is a special plane formed from the cartesian equation \( y^2 = 4ax \) for some parameter \( a \). It can also be written through the parametric equation \( x = at^2 \) and \( y = 2at \). This plane has the special property that for all points on the parabola the distance from that point to another point, known as the focus \( (a, 0) \), is equivalent to the distance from that parabola to a special line known as the directrix formed at \( x = - a \).

\begin{figure}
  \centering
  \begin{tikzpicture}
    \draw[black, ultra thick, ->] (0, -4) -- (0, 4) node[left]{$y$};
    \draw[black, ultra thick, ->] (-3.5, 0) -- (4, 0) node[below]{$x$};
    \draw[dotted] (-3.5, -4) grid (4, 4);

    \draw[black, green!80!blue] plot[smooth, domain=-sqrt(8):sqrt(8)] (0.5 * \x * \x, \x);
    \fill[blue] (2, 2) circle (0.1);
    %\fill[red] (1, 4) circle (0.2) node[right, outer sep = 5pt, black]{$(1, 4)$};
    % \draw[red, dashed, thick] (1, 4) -- (2, 2) node[midway, left, outer sep = 5pt]{$d$};

    \draw[blue, line width=0.5mm] (-3, -4) -- (-3, 4) node[left, pos=0.9]{Directix}; % horizontal rule
    \draw[dashed, red, <->] (-3, 2) -- (2, 2) node[above, pos=0.25]{$d_1$};
    \draw[dashed, red, <->] (-3, 1.5) -- (0, 1.5) node[below, pos=0.5]{$a$};
    \fill[blue] (3, 0) circle (0.1) node[below, outer sep = 10pt, black]{$\text{focus} = (\alpha, 0)$};
    \draw[red, dashed, thick] (2, 2) -- (3, 0) node[midway, right, outer sep=5pt]{$d_2$};
    % \draw[dashed, red, <->] (2, 2) -- (3, 0) node[midway, left, outer, sep=5pt]{focus};

    % Draw grid elements
    \foreach \x in {-3, -2, -1, 1, 2, 3} \node[below] at (\x, 0){$\x$};
    \foreach \y in {0, 1} \node[left] at (0, \y){$\y$};
  \end{tikzpicture}
  \caption{Visualisation of the focus, directix, property of a parabola.}
  \label{fig:foci-parabella-prop}
\end{figure}

The proof for this is as follows. First lets clarify \( x = -a \) as the equation for the directrix, it follows that \( d_1 = a + x \) as the total distance \( d_1 \). Furthermore we can qualify the euclidean distance \( d_2 \) as \( d_2^2 = \Delta{y^2} + \Delta{x^2} \). The prior definitions can be substituted back into this to get \( d_2 = \sqrt{y^2 + (x-a)^2} \). Now we equate \( d_1 \) and \( d_2 \).

\begin{align*} d_2 &= \sqrt{y^2 + (x-a)^2} \
a+x &= \sqrt{y^2 + (x-a)^2} \
(a+x)^2 &= y^2 + (x-a)^2 \
x^2 + 2ax + a^2 &= y^2 + x^2 - 2ax + a^2 \
2ax &= y^2 - 2ax \
4ax &= y^2 \end{align*}

Which gives us back the original equation for the parabola. This shows that our initial assumptions were correct and the focus-directrix property holds.