Derive the equation of a parabola from its focus and directrix

Learning goal: Derive the equation of a parabola given a focus and a directrix, using the definition of a parabola as the set of points equidistant from the focus and the directrix, and move between the geometric description and the equation in both directions.

Scope note. G.GPE.2 is marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate a standard "for students who are planning on taking advanced courses". It is taught here in full as part of the Geometry course, as extension content for students on that path.

The geometric definition. A parabola is the set of all points in a plane that are the same distance from a fixed point, the focus, as from a fixed line, the directrix, where the focus is not on the directrix. The distance from a point to a line means the perpendicular distance. The vertex is the point of the parabola halfway between the focus and the directrix, and the axis of symmetry is the line through the focus perpendicular to the directrix. Until now a parabola was "the graph of a quadratic function"; this lesson shows that a parabola whose directrix is horizontal is exactly such a graph, by deriving the equation from the distance condition. (A parabola with a vertical or slanted directrix is still a parabola, but it is not the graph of a function of x.)

Setting up the derivation. Put the focus at F(0, p) and let the directrix be the horizontal line y = −p, with p > 0. The vertex is then the origin, halfway between them. Let P(x, y) be any point of the parabola. Its distance to the focus is √((x − 0)² + (y − p)²) by the distance formula. Its distance to the directrix is the vertical distance |y − (−p)| = |y + p|, because the directrix is horizontal, so the perpendicular from P to it is vertical. The defining condition is √(x² + (y − p)²) = |y + p|.

Doing the algebra. Square both sides (both are nonnegative, so no solutions are gained or lost): x² + (y − p)² = (y + p)². Expand: x² + y² − 2py + p² = y² + 2py + p². Subtract y² and p² from both sides: x² − 2py = 2py, so x² = 4py. Every point of the parabola satisfies x² = 4py, and reversing the steps shows every point satisfying x² = 4py is equidistant from F and the directrix. So the parabola with focus (0, p) and directrix y = −p has equation x² = 4py, or y = x²/(4p). This is a quadratic function, which is why its graph has the familiar U shape; p is the distance from the vertex to the focus, and also from the vertex to the directrix.

A worked model inside the lesson. Focus (0, 2), directrix y = −2. Here p = 2, so the equation is x² = 4·2·y = 8y, that is, y = x²/8. Check a point on it: (4, 2) satisfies 16 = 8·2. Its distance to the focus (0, 2) is √(4² + 0²) = 4; its distance to the directrix y = −2 is |2 − (−2)| = 4. Equal, as the definition demands. Check another: (8, 8) gives 64 = 8·8; distance to the focus is √(64 + 36) = √100 = 10; distance to the directrix is 8 + 2 = 10. And the vertex (0, 0) is 2 from the focus and 2 from the directrix. A point off the curve fails the test: (4, 3) has distance √(16 + 1) = √17 ≈ 4.12 to the focus but 5 to the directrix.

Shifting the vertex. If the vertex is at (h, k) instead of the origin, with the focus at (h, k + p) and the directrix y = k − p (for p > 0 the focus is p units above the vertex and the directrix p units below; for p < 0 they swap sides, each |p| units away), the same derivation with x − h and y − k in place of x and y gives (x − h)² = 4p(y − k). For focus (3, 1) and directrix y = −3: the vertex is halfway between, at (3, −1), and p = 2, so (x − 3)² = 8(y + 1). Check the point (7, 1): the left side is 16 and the right side is 8·2 = 16; its distance to the focus (3, 1) is 4 and to the line y = −3 is 1 − (−3) = 4. If the focus is below the directrix, p is negative in these formulas and the parabola opens downward.

Parabolas that open sideways. If the focus is (p, 0) and the directrix is the vertical line x = −p, swapping the roles of x and y in the derivation gives y² = 4px. For focus (2, 0) and directrix x = −2, the equation is y² = 8x; the point (2, 4) is on it (16 = 16), at distance 4 from the focus and 4 from the line x = −2. Such a parabola is not the graph of a function of x, but it is a parabola by the definition all the same.

From the equation back to the focus and directrix. Given y = x²/12, write it as x² = 12y and match with x² = 4py: 4p = 12, so p = 3. The vertex is (0, 0), the focus is (0, 3) and the directrix is y = −3. Given (x + 1)² = −4(y − 5): vertex (−1, 5), 4p = −4 so p = −1; the focus is (−1, 5 + (−1)) = (−1, 4), one unit below the vertex, and the directrix is y = 5 − (−1) = 6; the parabola opens downward.

The misconception: the focus is on the curve, or the vertex is the focus. The focus is inside the curve, never on it: the vertex is the point of the parabola nearest the focus, and it is |p| units away (p itself is a signed displacement, negative when the parabola opens downward). Students who put the focus at the vertex get p = 0, and the equation x² = 0 collapses to a line. A second misconception is measuring the distance to the directrix along a slanted line to some convenient point of it; the distance to a line is always perpendicular, which is why, for a horizontal directrix, it is simply the difference in y-coordinates.

Why the focus deserves its name. Every ray traveling parallel to the axis of a parabolic mirror reflects through the focus, and light from a source at the focus leaves as a parallel beam; that geometry follows from the equal-distance definition and is why many dish antennas and some lamp reflectors are made parabolic. The equation derived here is what an engineer uses to place the receiver: for a dish shaped like y = x²/8 in appropriate units, the receiver sits 2 units above the vertex.

Worked derivation from the definition: Focus (0, 3), directrix y = −3. For P(x, y): √(x² + (y − 3)²) = |y + 3|. Square: x² + y² − 6y + 9 = y² + 6y + 9. Simplify: x² = 12y. Check with (6, 3): 36 = 12·3; distance to the focus is 6, distance to the directrix is 3 + 3 = 6.

Worked shifted vertex: Focus (−2, 4), directrix y = 0. The vertex is midway, (−2, 2), and p = 2. Equation: (x + 2)² = 8(y − 2). Check the point (2, 4): left side 16, right side 8·2 = 16; distance to the focus √(16 + 0) = 4, distance to the directrix 4 − 0 = 4. Written as a function: y = (x + 2)²/8 + 2.

Worked downward-opening parabola: Focus (1, −1), directrix y = 5. The vertex is (1, 2), and the focus is 3 units below it, so p = −3: (x − 1)² = −12(y − 2). Check (7, −1): left 36, right −12·(−3) = 36; distance to the focus is 6, distance to the directrix is 5 − (−1) = 6. Equivalent function form: y = −(x − 1)²/12 + 2.

Worked equation to focus and directrix: For y² = −16x, compare with y² = 4px: 4p = −16, p = −4. The vertex is the origin, the focus is (−4, 0) and the directrix is x = 4; the parabola opens to the left. The point (−4, 8) is on it since 64 = −16·(−4); it is 8 from the focus and 8 from the line x = 4.

Definition-check record: For the parabola with focus (0, 2) and directrix y = −2, make a table of six points: three on the curve y = x²/8 (use x = 2, −6 and 8) and three not on it. For each, compute the distance to the focus with the distance formula and the distance to the directrix as a vertical distance, and record whether they are equal. Then sketch the focus, directrix, vertex and the six points.

Independent performance: Choose a focus and a horizontal directrix of your own with the vertex not at the origin, derive the equation from the distance condition showing every algebraic step, rewrite it in the form (x − h)² = 4p(y − k) and as y = a(x − h)² + k, and verify two points by the definition. Then take the equation y = x²/20, recover its focus and directrix, and explain where a receiver would sit if the curve were the cross-section of a dish. Success criteria: the derivation squares both sides only after noting both are nonnegative; p, the vertex, the focus and the directrix are all consistent with one another; each verification shows both distances; and the write-up states in one sentence what a parabola is.