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Derive the Equation
of a Parabola

Learn how every point on a parabola is equidistant from a fixed point (focus) and a fixed line (directrix), then turn that geometric fact into an algebraic equation.

Geometry

What Is a Parabola?

A parabola is the set of all points in a plane that are equidistant from a fixed point called the focus and a fixed line called the directrix.

Directrix Focus Vertex P d₁ d₂ d₁ = d₂

Key Vocabulary

Focus
The fixed point inside the parabola. Denoted \( F \). All points on the curve are equidistant from it and the directrix.
Directrix
The fixed line outside the parabola. Every point on the curve has the same distance to this line as to the focus.
Vertex
The point on the parabola closest to the directrix, located exactly halfway between the focus and directrix.

Why Does This Matter?

Parabolas appear everywhere: satellite dishes focus signals to a receiver at the focus, headlight reflectors send light outward in parallel rays, and projectile paths trace parabolic arcs. Deriving the equation from the focus and directrix connects the geometric definition to the algebraic formula you already know.

Geometry to Algebra
Translate the equal-distance condition into \( (x - h)^2 = 4p(y - k) \).
Real-World Design
Engineers use the focus-directrix relationship to design reflectors and antennas.

The Equidistance Condition

Place the vertex at the origin. Let the focus be at \( (0, p) \) and the directrix be the line \( y = -p \). For any point \( (x, y) \) on the parabola:

x y y = -p F(0, p) Vertex (0, 0) P(x, y) p p

Deriving the Standard Form

1
Start with the equidistance condition: \( \sqrt{x^2 + (y - p)^2} = |y + p| \)
2
Square both sides: \( x^2 + (y - p)^2 = (y + p)^2 \)
3
Expand: \( x^2 + y^2 - 2py + p^2 = y^2 + 2py + p^2 \)
4
Cancel \( y^2 \) and \( p^2 \): \( x^2 - 2py = 2py \)
5
Solve: \( x^2 = 4py \)

Four Orientations

Depending on whether the parabola opens up, down, right, or left, the equation takes a different form. In every case the vertex is at the origin and \( p > 0 \).

Opens Equation Focus Directrix
Up \( x^2 = 4py \) \( (0, p) \) \( y = -p \)
Down \( x^2 = -4py \) \( (0, -p) \) \( y = p \)
Right \( y^2 = 4px \) \( (p, 0) \) \( x = -p \)
Left \( y^2 = -4px \) \( (-p, 0) \) \( x = p \)

General Vertex Form

When the vertex is at \( (h, k) \) instead of the origin, shift both coordinates:

Vertical Axis
\( (x - h)^2 = 4p(y - k) \)
Focus: \( (h,\, k + p) \)
Directrix: \( y = k - p \)
Horizontal Axis
\( (y - k)^2 = 4p(x - h) \)
Focus: \( (h + p,\, k) \)
Directrix: \( x = h - p \)

Parabola vs. Other Conics

Feature Parabola Ellipse / Circle
Shape Open curve, one branch Closed curve
Foci Exactly 1 focus 2 foci (circle: coincident)
Directrix 1 directrix line Not typically used
Eccentricity \( e = 1 \) always \( 0 \leq e < 1 \)
Squared terms Only one variable squared Both variables squared

How to Derive the Equation

1
Identify the focus and directrix from the problem.
2
Find the vertex: midpoint between the focus and directrix.
3
Determine \( p \): distance from vertex to focus (positive value).
4
Choose the correct form based on orientation (vertical or horizontal axis).
5
Substitute \( h, k, p \) into the general vertex form and simplify.

Example 1: Vertical Parabola

Given: Focus \( (0, 3) \), directrix \( y = -3 \). Derive the equation.

x y F(0, 3) y = -3 V(0, 0) p = 3

Example 2: Horizontal Parabola

Given: Focus \( (5, 0) \), directrix \( x = -5 \). Derive the equation.

x y x = -5 F(5, 0) V(0, 0) p = 5

Example 3: Shifted Vertex

Given: Focus \( (2, 5) \), directrix \( y = 1 \). Derive the equation.

1
Find the vertex: midpoint between focus and directrix. The focus y-coordinate is 5 and the directrix is \( y = 1 \), so vertex: \( \left(2,\, \frac{5+1}{2}\right) = (2, 3) \).
2
Find \( p \): distance from vertex to focus = \( 5 - 3 = 2 \). Since the focus is above the vertex, the parabola opens upward.
3
Substitute into \( (x - h)^2 = 4p(y - k) \) with \( h = 2,\, k = 3,\, p = 2 \).
4
Result: \( (x - 2)^2 = 8(y - 3) \).

Example 4: Opens Downward

Given: Focus \( (0, -2) \), directrix \( y = 2 \). Derive the equation.

x y F(0, -2) y = 2 V(0, 0) p = 2

Common Mistakes

✕ Confusing \( p \) with \( 4p \)
The coefficient of \( y \) (or \( x \)) in the standard form is \( 4p \), not \( p \). If the focus is at distance 3, write \( 4(3) = 12 \), not 3.
✕ Wrong sign for direction
A parabola opening downward or left uses a negative coefficient. Forgetting the negative flips the curve the wrong way.
✕ Mixing up the axis
Use \( x^2 = \ldots \) for vertical axis (up/down) and \( y^2 = \ldots \) for horizontal axis (right/left). Writing the wrong squared variable gives the wrong orientation.

Quick Reference Summary

Step Action
1 Find vertex: midpoint of focus and directrix
2 Find \( p \): distance from vertex to focus
3 Determine direction (up/down/left/right)
4 Vertical axis: \( (x-h)^2 = \pm 4p(y-k) \)
5 Horizontal axis: \( (y-k)^2 = \pm 4p(x-h) \)
6 Substitute \( h,\, k,\, p \) and simplify

Question 1 of 5

A parabola has focus \( (0, 4) \) and directrix \( y = -4 \). What is its equation?

Question 2 of 5

A parabola has focus \( (-3, 0) \) and directrix \( x = 3 \). Which equation represents this parabola?

Question 3 of 5

A parabola has vertex \( (1, 2) \) and focus \( (1, 5) \). What is the equation?

Question 4 of 5

What is the value of \( p \) for the parabola \( x^2 = 20y \)?

Question 5 of 5

A parabola has focus \( (4, -1) \) and directrix \( x = 0 \). What is its equation?