Finding the Equation of a Parabola Given Vertex and Focus: A Comprehensive Guide

Parabolas are a fundamental part of conic sections and are widely used in various fields such as physics, engineering, and mathematics. In this guide, we will explore how to find the equation of a parabola when given its vertex and focus. We will provide step-by-step solutions, explain the underlying formulas, and delve into different methods to solve this problem.

Understanding the Parabola

A parabola is a U-shaped curve that is symmetric about its axis. The vertex of a parabola is its highest or lowest point, and the focus is a fixed point inside the parabola. Depending on the orientation of the parabola, it can open upwards, downwards, to the left, or to the right.

Given Data and Characteristics

Let's consider a parabola with the vertex at ( (5, 15) ) and the focus at ( (3, 13) ).

Vertex: ( (5, 15) ) Focus: ( (3, 13) )

Direction and Distance Calculation

Since the x-coordinates of the vertex and focus are the same, the parabola opens vertically. The focus is below the vertex, indicating that the parabola opens downwards. We can calculate the distance ( p ) between the vertex and the focus as follows:

p y_{text{vertex}} - y_{text{focus}} 15 - 13 2

Since the parabola opens downwards, ( p ) is negative:

p -2

Standard Form of a Vertical Parabola

The standard form of a parabola with a vertical axis of symmetry and vertex at ( (h, k) ) is:

y - k frac{1}{4p}(x - h)^2 quad text{or} quad y - k -frac{1}{4p}(x - h)^2 quad text{(for downward opening parabolas)}

Substituting the vertex ( (h, k) (5, 15) ) and ( p -2 ) into the formula, we get:

y - 15 -frac{1}{4(-2)}(x - 5)^2 quad Rightarrow quad y - 15 frac{1}{8}(x - 5)^2

Final Equation

To express the equation in a more standard form, we can expand and simplify:

y - 15 frac{1}{8}(x - 5)^2 quad Rightarrow quad y - 15 frac{1}{8}(x^2 - 1 25) quad Rightarrow quad y - 15 frac{1}{8}x^2 - frac{5}{4}x frac{25}{8}

y frac{1}{8}x^2 - frac{5}{4}x frac{25}{8} 15 quad Rightarrow quad y frac{1}{8}x^2 - frac{5}{4}x frac{25}{8} frac{120}{8} quad Rightarrow quad y frac{1}{8}x^2 - frac{5}{4}x frac{145}{8}

Alternative Methods

Let's explore two alternative methods to find the equation of a parabola given the vertex and focus.

Method 1: Using Directrix and Midpoint

The vertex of the parabola is the midpoint of the focus and the feet of the directrix. Let the feet of the directrix be ( (x_1, y_1) ).

x_1 frac{3 h}{2} quad Rightarrow quad x_1 frac{3 5}{2} 4 quad Rightarrow quad x_1 1 y_1 frac{13 k}{2} quad Rightarrow quad y_1 frac{13 15}{2} 14 quad Rightarrow quad y_1 7

The directrix is a horizontal line with equation ( y 7 ). Using the definition of a parabola ( PS PM ), where ( P ) is a moving point on the parabola, ( S ) is the focus, and ( M ) is the feet of the perpendicular from ( S ) to the directrix, we have:

x - 5^2 8(y - 15) quad Rightarrow quad (x - 5)^2 8(y - 15)

Method 2: Using Vertex and Focus Distance

The distance between the focus and the vertex is ( a ), where ( 4a ) is the length of the latus rectum. The axis of the parabola is parallel to the y-axis, and the vertex is at ( (5, 15) ). The length of the latus rectum is 8, so ( a 2 ). The equation of the parabola is:

(x - 5)^2 8(y - 15)

This is the same equation obtained in Method 1.

Conclusion

In conclusion, we have provided a detailed guide on how to find the equation of a parabola given its vertex and focus. We have demonstrated two methods—using the directrix and midpoint, and using the vertex and focus distance—both leading to the same final equation: