Deriving the Equation of a Parabola: Step-by-Step Guide

Introduction to Parabolas

A parabola is a conic section defined as the locus of points that are equidistant from a fixed point, known as the focus, and a fixed straight line, known as the directrix. This unique property allows us to derive the equation of a parabola through various methods, including algebraic manipulation and geometric insights.

The Geometry of a Parabola

A mathematical definition of a parabola states that it is the set of all points in a plane that are equidistant from a point (the focus) and a line (the directrix). To understand this, let's consider a point F(xF, yF) as the focus and a line given by the equation ax by c 0 as the directrix.

Deriving the Basic Equation

To find the distance from any point (X, Y) on the parabola to the focus (xF, yF) and the directrix ax by c 0, we can use the distance formula and the condition of equidistance.

1. **Distance from a point to the focus**:

[d_{focus} sqrt{(X - x_{F})^2 (Y - y_{F})^2}]

2. **Distance from a point to the directrix**:

[d_{directrix} frac{|aX bY c|}{sqrt{a^2 b^2}}]

The equation of the parabola is derived from the condition that these distances are equal:

[sqrt{(X - x_{F})^2 (Y - y_{F})^2} frac{|aX bY c|}{sqrt{a^2 b^2}}]

Squaring both sides to eliminate the square root, we obtain:

[(X - x_{F})^2 (Y - y_{F})^2 frac{(aX bY c)^2}{a^2 b^2}]

This equation can be simplified and rearranged to yield the standard form of the parabola equation, which is useful for specific cases such as vertical or horizontal parabolas.

Special Cases: Vertical and Horizontal Parabolas

1. **Vertical Parabola**: If we choose the directrix to be ax by c 0 with b 0, the equation simplifies to:

[(X - x_{F})^2 (Y - y_{F})^2 frac{(aX c)^2}{a^2}]

By setting the focus coordinates to the origin and the directrix to x -1, we can derive the equation:

[x^2 4py]

Here, the vertex is at (0, 0) and the focus at (0, p), with p being the vertical distance from the vertex to the focus.

2. **Horizontal Parabola**: If we choose the directrix to be y 0 and the focus to be (a, 0), we derive:

[(X - a)^2 4pY]

Here, the vertex is at (a, 0) and the focus at (a, p), with p representing the horizontal distance from the vertex to the focus.

Lagrange Interpolation for Parabolas

For a more direct approach, we can use Lagrange interpolation to derive the equation of a parabola passing through three points. Let's denote these points as (a_1, b_1), (a_2, b_2), and (a_3, b_3). The function we seek is a quadratic polynomial f(x) Ax^2 Bx C that matches the points.

We start by defining the Lagrange basis polynomials:

[L_1(x) frac{(x - a_2)(x - a_3)}{(a_1 - a_2)(a_1 - a_3)}]

[L_2(x) frac{(x - a_1)(x - a_3)}{(a_2 - a_1)(a_2 - a_3)}]

[L_3(x) frac{(x - a_1)(x - a_2)}{(a_3 - a_1)(a_3 - a_2)}]

The function f(x) can then be expressed as:

[f(x) b_1L_1(x) b_2L_2(x) b_3L_3(x)]

Substituting the definitions of L_1(x), L_2(x), and L_3(x), we get:

[f(x) b_1 frac{(x - a_2)(x - a_3)}{(a_1 - a_2)(a_1 - a_3)} b_2 frac{(x - a_1)(x - a_3)}{(a_2 - a_1)(a_2 - a_3)} b_3 frac{(x - a_1)(x - a_2)}{(a_3 - a_1)(a_3 - a_2)}]

This formula provides a direct way to construct the equation of a parabola given three points without solving a system of equations.

Conclusion

The equations of parabolas can be derived using various methods, from geometric definitions to more advanced algebraic techniques like Lagrange interpolation. Understanding these methods helps in solving a wide range of problems involving parabolic shapes and their applications in physics, engineering, and mathematics.