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Section 5.1 Parabola

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). 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 \(y = -p\text{.}\)
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Figure 5.1. The parabola \(x^2 = 4py\text{.}\) The point \(P\) is equidistant from the focus \((0,p)\) and the directrix \(y=-p\text{.}\)
Figure 5.2. Animation: as \(P\) moves along the parabola, its distance to the focus always equals its distance to the directrix.

Example 5.3. 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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Figure 5.4. The parabola \((x-1)^2 = -2y\text{,}\) opening downward with focus \(\left(1,-\tfrac12\right)\) and directrix \(y=\tfrac12\text{.}\)

Example 5.5. 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 the definition 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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Figure 5.6. The parabola \(y^2 = 8x\) opens to the right; \(P\) is equidistant from the focus \((2,0)\) and the directrix \(x=-2\text{.}\)
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