Definition 5.1.
A parabola is the set of points in a plane that are equidistant from a fixed point \(F\) (the focus) and a fixed line (the directrix).
Section 5.1 Parabola
The standard equation follows from that definition and nothing else. Choose coordinates that make the picture symmetric: put the focus at \(F(0,p)\) and the directrix at \(y = -p\text{,}\) so the two sit on opposite sides of the origin at equal distances. A point \(P(x,y)\) lies on the parabola exactly when its distance to \(F\) equals its distance to the directrix, and that second distance is measured straight down to the foot \((x,-p)\text{:}\)
\begin{equation*}
\sqrt{x^2 + (y-p)^2} = |y+p|.
\end{equation*}
Both sides are non-negative, so squaring loses nothing:
\begin{equation*}
x^2 + y^2 - 2py + p^2 = y^2 + 2py + p^2.
\end{equation*}
The \(y^2\) and \(p^2\) terms cancel from both sides, leaving \(x^2 = 4py\text{.}\)
So for the standard parabola \(x^2 = 4py\) the vertex is at the origin, the focus is at \((0,p)\text{,}\) and the directrix is the line \(y = -p\text{.}\) The vertex sits halfway between focus and directrix, which is why both are a distance \(|p|\) from it. If \(p \lt 0\) the same algebra applies unchanged and the curve opens downward; interchanging the roles of \(x\) and \(y\) gives the sideways parabola \(y^2 = 4px\text{.}\)
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Example 5.4 From equation to focus and directrix.
Find the focus and the directrix of \(y = -\frac{1}{2}x^2 + x - \frac{1}{2}\text{.}\)
Solution.
We begin by completing the square:
\begin{equation*}
y = -\frac{1}{2}x^2 + x - \frac{1}{2} = -\frac{1}{2}(x-1)^2,
\qquad\text{i.e.}\qquad (x-1)^2 = -2y.
\end{equation*}
Comparing with \(x^2 = 4py\) gives \(4p = -2\text{,}\) so \(p = -\frac{1}{2}\text{.}\) The graph is the standard parabola shifted right by one unit, so the vertex is at \((1,0)\text{.}\) Therefore the focus is at
\begin{equation*}
(1,\, p) = \left(1,\, -\tfrac{1}{2}\right),
\end{equation*}
and the directrix is the horizontal line \(y = -p = \tfrac{1}{2}\text{.}\)
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Example 5.6 From focus and directrix to equation (a sideways parabola).
Find the equation of the parabola whose focus is \((2,0)\) and whose directrix is the vertical line \(x = -2\text{.}\) In which direction does it open?
Solution.
Here we run Definition 5.1 in reverse. A point \((x,y)\) lies on the parabola exactly when its distance to the focus equals its distance to the directrix:
\begin{equation*}
\sqrt{(x-2)^2 + y^2} = |x+2|.
\end{equation*}
Squaring both sides and expanding,
\begin{equation*}
(x-2)^2 + y^2 = (x+2)^2
\;\Longrightarrow\; x^2 - 4x + 4 + y^2 = x^2 + 4x + 4.
\end{equation*}
The \(x^2\) and constant terms cancel, leaving
\begin{equation*}
y^2 = 8x.
\end{equation*}
Because the directrix is vertical and the focus lies to its right, the parabola opens to the right. Comparing with \(y^2 = 4px\) gives \(4p = 8\text{,}\) so \(p = 2\text{,}\) consistent with the focus at \((2,0)\text{.}\) This is the same equidistance idea as the previous example, but with a horizontal axis of symmetry.
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