Equation of a Parabola
Aligned to G-GPE.A.2 — Common Core State Standards for Mathematics.
What this lesson teaches
A parabola is the set of points equidistant from a fixed focus and a fixed line (the directrix). Setting the distance to the focus equal to the distance to the directrix and simplifying gives the equation. With focus (0, p) and directrix y = -p, the parabola is y = x^2/(4p).
Worked examples
Example 1
Find the equation with focus (0, 2) and directrix y = -2. A point (x, y) is equidistant from both. Distance to focus: sqrt(x^2 + (y - 2)^2). Distance to directrix: y + 2. Set equal and square: x^2 + (y - 2)^2 = (y + 2)^2. Expand: x^2 + y^2 - 4y + 4 = y^2 + 4y + 4. Simplify: x^2 = 8y, so y = x^2/8.
Example 2
Find the focus and directrix of y = x²/12. Compare to the standard form x² = 4py: here x² = 12y, so 4p = 12 and p = 3. The focus sits p units above the vertex at (0, 3), and the directrix is the horizontal line p units below, y = -3. Check the point (6, 3) on the curve: its distance to the focus is 6, and its distance to y = -3 is 3 + 3 = 6.
Example 3
Common mistake: confusing p with the coefficient. In y = x²/8 the number 8 is 4p, so p = 2 and the focus is at (0, 2) — not (0, 8) and not (0, 1/8). A useful sanity check: the larger p is, the WIDER the parabola opens, because the focus and directrix move farther apart. If your computed focus makes a narrow parabola look wide, re-check the 4.
Practice questions
- A parabola has focus (0, 3) and directrix y = -3. Set up the equal-distance equation.
- Derive the equation for focus (0, 1) and directrix y = -1.
- For y = x^2/12, identify the value of p and state the focus and directrix.
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Every lesson comes with a video taught in English and Spanish — the same video the QR code in the printed workbook opens.
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Ecuación de una parábola
Una parábola es el conjunto de puntos equidistantes de un foco fijo y una recta fija (la directriz). Igualar la distancia al foco con la distancia a la directriz y simplificar da la ecuación. Con foco (0, p) y directriz y = -p, la parábola es y = x^2/(4p).
Ejemplo 1: Halla la ecuación con foco (0, 2) y directriz y = -2. Un punto (x, y) equidista de ambos. Distancia al foco: raíz(x^2 + (y - 2)^2). Distancia a la directriz: y + 2. Iguala y eleva al cuadrado: x^2 + (y - 2)^2 = (y + 2)^2. Desarrolla: x^2 + y^2 - 4y + 4 = y^2 + 4y + 4. Simplifica: x^2 = 8y, así que y = x^2/8.
Ejemplo 2: Halla el foco y la directriz de y = x²/12. Compara con la forma estándar x² = 4py: aquí x² = 12y, así que 4p = 12 y p = 3. El foco queda p unidades arriba del vértice, en (0, 3), y la directriz es la recta horizontal p unidades abajo, y = -3. Comprueba el punto (6, 3) de la curva: su distancia al foco es 6, y su distancia a y = -3 es 3 + 3 = 6.
Ejemplo 3: Error común: confundir p con el coeficiente. En y = x²/8 el número 8 es 4p, así que p = 2 y el foco está en (0, 2) — no en (0, 8) ni en (0, 1/8). Una comprobación útil: cuanto mayor es p, MÁS ABIERTA es la parábola, porque el foco y la directriz se separan. Si tu foco calculado hace que una parábola angosta parezca ancha, revisa el 4.
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