Finding The Equation Of A Parabola

Finding the Equation of a Parabola

Basic Definition

A parabola is a set of all points in a plane that are equidistant from a fixed point and a fixed line. The fixed point is called the “focus” and the fixed line is called the “directrix”. The focus may not be on the directrix.

The line through the focus and perpendicular to the directrix is called the “axis of symmetry” or just the axis of the parabola. The line segment through the focus and perpendicular to the axis of symmetry which is cut by the parabola is called the focal chord or “latus rectum” and its length is the focal width. The point of intersection of the axis of symmetry and the parabola is called the vertex. It then follows by the definition that the vertex is equidistant from the focus and the directrix.

Standard Forms of the Equations of a Parabola

An equation of the parabola with its vertex at (h, k) and focus at ( h+p. k) is :

( y – k ) ^2 = 4p ( x – h ).

The axis of symmetry of this parabola is parallel to the X-axis and the graph opens to the right if p > 0 and to the left if p < 0. The equation of the axis of symmetry is y = k

With directrix x = h – p.

An equation of the parabola with the vertex at (h, k) and focus at (h, k+p) is:

(x – h ) ^2 = 4p( y – k).

The axis of symmetry of this parabola is parallel to the Y-axis and the graph opens up if p > 0 and down if p < 0. The equation of the axis of symmetry is x = h with directrix

y = k – p.

If the vertex(h, k) is at the origin then h = k = 0, and the equation has the following forms :

Y ^ 2 = 4px

X ^ 2 = 4py.

The focal width of the parabola is 4p. The distance between vertex and focus is I p I.


Sample Problem Number One :

Find the equation and focal width of the parabola with vertex at (5, 1) and focus at (7, 1).

Solution :

In this problem, the vertex and focus are of the same distance from the X-axis ; therefore the line y = 1 is the axis of symmetry . Hence the form of the equation is ( y – k )^2 = 4p ( x – h ) with h = 5 and k = 1. Since the focus is two units to the right of the vertex , p = 2

4p = 4(2) = 8

Consequently, the equation of the parabola is :

(y – 1)^2 = 8 ( x – 5 ) and the focal width is absolute value of 8 = 8.


Sample Problem Number Two :

Find the equation of the parabola given vertex at (5, 0) and y = -8 as dirctrix.

The axis of symmetry is parallel to the Y –axis. The equation for the directrix is y = k – p. Substituting k = 0 and y = -8 to the equation :

-8 = 0 – p

-8 = -p

8 = p or p = 8

Therefore the equation of the parabola we are looking for is :

(x – 5 ) ^ 2 = 8y


Sample Problem Number Three :

Find the equation of the parabola given directrix y = 3 and focus at (5, -1 ).

The axis of symmetry of this parabola is parallel to the Y-axis. The parabola opens downward. The vertical distance from the dirctrix to the vertex is equal to p which is equal to the vertical distance from the vertex to the focus.Therefore vertical distance from the directrix to the focus is 2p. Get the directed distance from the dirctrix to the focus.

Absolute value of ( -1 – 3 ) = 4

2p = 4

p = 2 thus 4p = 8.

Therefore the vertex is at (5, -1 + 2 ) or at ( 5, 1 ).

The equation of the parabola we are looking for is: (x – 5)^2 = 8 (y – 1 ).


Sample Problem Number Four :

Find the equation of the parabola with vertex at (-2, 3) axis parallel to the X-axis and passing through (4, 9).

The form of the equation must be : (y – 3 ) ^2 = 4p (x + 2 ).

Substituting x = 4 and y = 9 gives 36 = 4p (6) and so 4p = 6. The equation thus is :

(y – 3 ) ^ 2 = 6 ( x + 2 ).


Sample Problem Number Five :

Find the equation of the parabola given vertex at (5, 2) and ends of the focal chord at (3, 6) and (3, -2 ).

The axis of symmetry is parallel to the X-axis. The parabola opens to the left .

4p = length of the focal chord.

Get the vertical or directed distance between the two ends of the focal chord.

Absolute value of ( -2-6) = 8

4p = 8

Thus the equation of the parabola we are looking for is :

(y – 2 )^2 = -8 ( x – 5 ).

SOURCE:

COLLEGE ALGEBRA By Rees Sparks Rees

More by this Author

  • Solving Investment Problems
    5

    Solving Investment Problems One of the most important applications of linear equations is found in solving investment problems. Investment problems use the Simple Interest formula I = Prt, where P =...

  • Solving Word Problems Involving Quadratic Functions
    18

    Solving Word Problems Involving Quadratic Function Some of the most important functions in applications are the quadratic functions. The quadratic functions are one of the simplest type of polynomial...

  • Solving Word Problems Involving Angular Velocity
    21

      Solving Problems Involving Angular Velocity   Among the challenging problems I encounter  in Trigonometry are problems involving angular velocity. In this hub, I presented several problems...


Comments 1 comment

JeanieR profile image

JeanieR 5 years ago from Sequoia National Forest, CA

I'm gonna have to wake up my grandson to help me with this...out of my league!

    Sign in or sign up and post using a HubPages Network account.

    0 of 8192 characters used
    Post Comment

    No HTML is allowed in comments, but URLs will be hyperlinked. Comments are not for promoting your articles or other sites.


    Click to Rate This Article
    working