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d ( F , P ) = d ( P , Q ) ( 0 x ) 2 + ( p y ) 2 = ( x x ) 2 + ( p y ) 2 .

Squaring both sides and simplifying yields

x 2 + ( p y ) 2 = 0 2 + ( p y ) 2 x 2 + p 2 2 p y + y 2 = p 2 + 2 p y + y 2 x 2 2 p y = 2 p y x 2 = 4 p y .
A parabola is drawn with vertex at the origin and opening up. A focus is drawn as F at (0, p). A point P is marked on the line at coordinates (x, y), and the distance from the focus to P is marked d. A line marked the directrix is drawn, and it is y = − p. The distance from P to the directrix at (x, −p) is marked d.
A typical parabola in which the distance from the focus to the vertex is represented by the variable p .

Now suppose we want to relocate the vertex. We use the variables ( h , k ) to denote the coordinates of the vertex. Then if the focus is directly above the vertex, it has coordinates ( h , k + p ) and the directrix has the equation y = k p . Going through the same derivation yields the formula ( x h ) 2 = 4 p ( y k ) . Solving this equation for y leads to the following theorem.

Equations for parabolas

Given a parabola opening upward with vertex located at ( h , k ) and focus located at ( h , k + p ) , where p is a constant, the equation for the parabola is given by

y = 1 4 p ( x h ) 2 + k .

This is the standard form    of a parabola.

We can also study the cases when the parabola opens down or to the left or the right. The equation for each of these cases can also be written in standard form as shown in the following graphs.

This figure has four figures, each a parabola facing a different way. In the first figure, a parabola is drawn opening up with equation y = (1/(4p))(x − h)2 + k. The vertex is given as (h, k), the focus is drawn at (h, k + p), and the directrix is drawn as y = k − p. In the second figure, a parabola is drawn opening down with equation y = −(1/(4p))(x − h)2 + k. The vertex is given as (h, k), the focus is drawn at (h, k − p), and the directrix is drawn as y = k + p. In the third figure, a parabola is drawn opening to the right with equation x = (1/(4p))(y − k)2 + h. The vertex is given as (h, k), the focus is drawn at (h + p, k), and the directrix is drawn as x = h − p. In the fourth figure, a parabola is drawn opening left with equation x = −(1/(4p))(y − k)2 + h. The vertex is given as (h, k), the focus is drawn at (h – p, k), and the directrix is drawn as x = h + p.
Four parabolas, opening in various directions, along with their equations in standard form.

In addition, the equation of a parabola can be written in the general form    , though in this form the values of h , k , and p are not immediately recognizable. The general form of a parabola is written as

a x 2 + b x + c y + d = 0 or a y 2 + b x + c y + d = 0 .

The first equation represents a parabola that opens either up or down. The second equation represents a parabola that opens either to the left or to the right. To put the equation into standard form, use the method of completing the square.

Converting the equation of a parabola from general into standard form

Put the equation x 2 4 x 8 y + 12 = 0 into standard form and graph the resulting parabola.

Since y is not squared in this equation, we know that the parabola opens either upward or downward. Therefore we need to solve this equation for y, which will put the equation into standard form. To do that, first add 8 y to both sides of the equation:

8 y = x 2 4 x + 12 .

The next step is to complete the square on the right-hand side. Start by grouping the first two terms on the right-hand side using parentheses:

8 y = ( x 2 4 x ) + 12 .

Next determine the constant that, when added inside the parentheses, makes the quantity inside the parentheses a perfect square trinomial. To do this, take half the coefficient of x and square it. This gives ( −4 2 ) 2 = 4 . Add 4 inside the parentheses and subtract 4 outside the parentheses, so the value of the equation is not changed:

8 y = ( x 2 4 x + 4 ) + 12 4 .

Now combine like terms and factor the quantity inside the parentheses:

8 y = ( x 2 ) 2 + 8 .

Finally, divide by 8:

y = 1 8 ( x 2 ) 2 + 1 .

This equation is now in standard form. Comparing this to [link] gives h = 2 , k = 1 , and p = 2 . The parabola opens up, with vertex at ( 2 , 1 ) , focus at ( 2 , 3 ) , and directrix y = −1 . The graph of this parabola appears as follows.

A parabola is drawn with vertex at (2, 1) and opening up with equation x2 – 4x – 8y + 12 = 0. The focus is drawn at (1, 3). The directrix is drawn at y = − 1.
The parabola in [link] .
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Put the equation 2 y 2 x + 12 y + 16 = 0 into standard form and graph the resulting parabola.

x = 2 ( y + 3 ) 2 2
A parabola is drawn with vertex at (−2, −3) and opening to the right with equation x = 2(y + 3)2 – 2. The focus is drawn at (0, −3). The directrix is drawn at x = −4.

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The axis of symmetry of a vertical (opening up or down) parabola is a vertical line passing through the vertex. The parabola has an interesting reflective property. Suppose we have a satellite dish with a parabolic cross section. If a beam of electromagnetic waves, such as light or radio waves, comes into the dish in a straight line from a satellite (parallel to the axis of symmetry), then the waves reflect off the dish and collect at the focus of the parabola as shown.

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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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