Axis of a Parabola

The axis of a parabola is the line which divides the parabola into two symmetrically equal parts. The following figure shows this:
The axis divides the parabola symmetrically
The mirror image of each point on the parabola on one side of the axis lies on the other side of the axis. If you draw a line perpendicular to the axis meeting the parabola at two points then those two points are mirror images of each other.

The axis of a parabola is defined as the line perpendicular to the directrix and passing through its focus. This is shown in the following figure:
Axis of a parabola is perpendicular to the directrix and passes through its focus

The axis of a parabola can also be defined as the line passing through the vertex and perpendicular to the directrix.
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Ordinate And Double Ordinate of a Parabola

Ordinates of the parabola `y^2 = x`
Ordinate of a parabola is a line perpendicular to the axis of a parabola, starting from the axis and ending on a point on the parabola itself. Ordinates are always inside the parabola. The figure above shows a bunch of ordinates for the parabola `y^2 = x`.
Double ordinates for the parabola `y^2 = x`
A double ordinate, on the other hand, is the double of an ordinate. That is, it is a line perpendicular to the axis of the parabola and meeting the parabola at two points instead of just one. The figure above shows the double ordinates for the parabola `y^2 = x`.

Note: The double ordinate of a parabola passing through its focus are called its latus rectum.
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How to Find Equation of Directrix of a Parabola?

The equation of the directrix of a parabola can be determined in different ways depending on the given information about the parabola. These can be classified into three basic cases:
  1. You know the coordinates of vertex and focus
  2. You know the equation of the parabola
The method to determine the equation of directrix for each of these cases is described below:

1. You know the coordinates of vertex and focus

Suppose you know the coordinates of vertex and focus of a parabola. In order to find the equation of its directrix, we will use two properties of the directrix to find its equation:
  1. The directrix is perpendicular to the axis of the parabola
  2. The distance of any point on the parabola, and thus its vertex, from the directrix is equal to its distance from the focus
From the first property:

Since we already know the coordinates of focus and vertex of the parabola, and we know that the axis of a parabola passes through the focus and vertex, hence we can calculate the slope of the axis of the parabola from the given information.
Suppose the coordinates of vertex are `(1, 2)` and those of the focus are `(2, 2)`. Then, the slope of the axis is, by the formula for slope of a straight line,
`"Slope of axis" = (y_2 - y_1)/(x_2 - x_1) = (2 - 2)/(2 - 1) = 0`
Since the slope of the axis of the parabola is zero, this means that the axis of the parabola is parallel to the x-axis (all lines having slope zero are parallel to the x-axis). It follows that since the directrix is perpendicular to the axis of the parabola, hence the directrix is perpendicular to the x-axis, or parallel to the y-axis. We know that the equation of any straight line parallel to the y-axis is given by `x = a`,  where `a` is its intercept on the x-axis. Thus we assume that the equation of the directrix of the given parabola is `y = a`.

From the second property

 Now we will apply the second property of directrices: the distance of any point on the parabola, and particularly the vertex, from the directrix is equal to its distance from the focus. This means that the vertex is the midpoint between the focus and the point where the directrix of the parabola meets its axis. Thus, if `(x, y)` is the point where the axis and directrix intersect each other, then the vertex `(1, 2)` is the midpoint of `(x, y)` and `(2, 2)`. By using midpoint formula, we have
`(1, 2) = ((x + 1)/2, (y + 2)/2)`
Solving the above two equations (one for `x` and one for `y`):
`(x + 2)/2 = 1`
`x + 2 = 2`
`x = 2 - 2`
`x = 0`
and
`(y + 2)/2 = 2`
`y + 2 = 4`
`y = 4 - 2`
`y = 2`
Thus, the coordinates of the point of intersection of the directrix and axis are `(0, 2)`. This can be graphically represented as follows:
As you can see in the above image, the vertex `(1, 2)` is the midpoint of the focus `(2, 2)` and the point of intersection of the directrix and the axis `(0, 2)`.

Now we know two things about the directrix. One is that it is a vertical line (since it is perpendicular to the axis) and second is that it passes through the point `(0, 2)`. The equation of the directrix is `x = a` where `a` is equal to 0 since it passes through `(0, 2)`.

Thus the equation of the directrix is `x = 0`

2. You know the equation of the parabola

The equation of the directrix can be determined from the equation of the parabola by comparing the equation of the parabola with one of the four standard forms. Since two equations can only be compared when they are written in the same way, hence if the equation of the parabola is not given in the standard form already, we will have to change it into the standard form by following the method described here. It involves mainly the completing the square method.

For example, you have the equation of the parabola `y^2 = 8x` We already know that the directrix of the standard parabola `y^2 = 4ax` is given by `x = -a`. Thus, comparing `y^2 = 8x` with `y^2 = 4ax`, we get
`4a = 8`, that is
`a = 2`
Therefore the equation of the directrix for the parabola `y^2 = 8x` is `x = -2`.
Similarly you can find the equation of the directrix for all equations that are directly comparable with one of the four standard forms.

But if the equation of the parabola is not in the standard form already, you can convert it into one of the four standard forms as described in this post. For example, if you have the equation `y = x^2 + 3x + 2`, you can convert it into the standard form by following the procedure described here. The steps of converting the equation into standard form are not given in this post for its length. However, after following the steps as described in the post linked above, you will get the following equation for `y = x^2 + 3x + 2`:
`(x + 3/2)^2 = (y + 1/4)`
We will now substitute `X` in place of `x + 3/2` and `Y` in place of `y + 1/4`, so the equation becomes
`X^2 = Y`
On comparing the above equation with
`x^2 = 4ay`
we get
`4a = 1` 
`a = 1/4` 
Since we know that the parabola `x^2 = 4ay` has directrix at `y = -a`, therefore this parbola with `a = 1/4` has directrix at
`X = -1/4`
Now we need to convert the capital `X` back to `x`. Since we had substituted `x + 3/2` in place of `X`, therefore we can write,
`x + 3/2 = X`
The equation of the directrix is `X = -1/4`. Thus substituting `X = -1/4` into the above equation we get
`x + 3/2 = -1/4`
`x = -1/4 - 3/2`
`x = -7/4`
Therefore the parabola `y = x^2 + 3x + 2` has the directrix at `x = -7/4`.
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Directrix of a Parabola

Definition

The directrix of a parabola is a fixed straight line not touching or intersecting the parabola such that the distance of any point on the parabola from directrix is equal to its distance from the focus.

The directrix is always perpendicular to the axis of the parabola and lies on the side opposite to which the parabola is facing (the parabola is said to be facing on the side where it opens - for example, the parabola below is right facing).

In the following graph of the parabola `y^2 = 4x`, the directrix is `x = -1` and focus is labelled S. A, B, and C are three points on the parabola. The distance of each point from the focus equals its distance from the directrix. This means that AP = AS, BR = BS, and CQ = CS.

Directrix of a Parabola
For right or left facing parabolas, like the one above, the directrix is a vertical line. For upright or down facing parabolas, their directrices are horizontal lines (Directrices is plural for directrix).

An interesting property of the directrix is that the vertex is the mid point of the line joining the focus and the directrix. This can be understood from the fact that the distance between any point on the parabola and the directrix is equal to the distance between that point and the focus. The vertex is a point on the parabola, hence its distance from the directrix is equal to its distance from the focus. Since both the focus and vertex lie on the same straight line (the axis), hence the vertex is the midpoint of the focus and the point of intersection of the directrix and the axis. Thus, knowing the coordinates of any two, say the vertex and focus, the coordinates of the third point can be be obtained  (the point of intersection in this case) by using the midpoint formula.

The equation of the directrix of a parabola can be determined by following a few simple steps. These steps are explained in this post.
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Vertex of a Parabola

The vertex of a parabola is a point on the parabola itself. It is the point where the axis of the parabola intersects the parabola.

Vertex of a Parabola

For an upright parabola, vertex is the lowermost point on it:


For an upside down parabola, vertex is the uppermost point on it:


For a right facing parabola, vertex is the leftmost point on it:


And for a left facing parabola, the vertex is the rightmost point on it:

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Change Parabola Equation to Standard Form

The equation of a parabola has the following four simplest, standard, forms:

  • `y^2 = 4ax`
  • `y^2 = -4ax`
  • `x^2 = 4ay`
  • `x^2 = -4ay`
Properties of these standard forms are discussed here.

The equation of a parabola can be changed to match one of the standard forms given above. For example, 

`y = 2x^2 + 3x + 1`

We want to change this equation so that it matches `x^2 = 4ay`. This process can be broken down to three steps:

  1. Completing the square
  2. Rearranging the equation
  3. Substituting for x and y
These steps are discussed below with the example above.

1. Completing the square


This post describes in detail the method of completing the square, although the complete steps for that process are given below:
To complete the square, we first factor out the coefficient of x squared:
`y = 2(x^2 + 3/2x + 1/2)`
Now we take the coefficient of x, which is 3/2, divide it by 2, hence we get 3/4, and then square it, which gives us (3/4)^2. The result (3/4)^2 is then added and subtracted from the expression above, as follows:
`y = 2(x^2 + 3/2x + (3/4)^2 - (3/4)^2 + 1/2)`
Now compare the highlighted part above with the formula `a^2 + 2ab + b^2 = (a + b)^2` . That is, we take the part `x^2 + 3/2x + (3/4)^2` from the above equation, and, compare it with the formula `a^2 + 2ab + b^2`. It is clear that `a` corresponds with `x` and `b` corresponds with `3/4`. Thus, we can rewrite `x^2 + 3/2x + (3/4)^2` as `(x + 3/4)^2`. So the above equation can be rewritten as
`y = 2((x + 3/4)^2 - (3/4)^2 + 1/2)`
Now we just simplify the trailing fractions
`y = 2((x + 3/4)^2 - 1/16)`
If you didn't understand the above method of completing the square properly or want to see more examples, see this post.

2. Rearranging the equation

As you may notice, on comparing the last equation with the four standard forms described at the start, it is clear that the equation we have tends to match with `x^2 = 4ay`, although not completely. In order to make it more like the standard form, we are going to rearrange the equation a bit:

First we divide by 2 on both sides. We do this in order to move all the numbers in addition or multiplication on the side of `y` in the equation.
`y/2 = (x + 3/4)^2 - 1/16`
Next we move the fraction 1/16 to the left hand side. This is done by simply adding 1/16 to both sides of the equation:
`y/2 + 1/16 = (x + 3/4)^2`
Next we factor out the coefficient of `y` from the left side of the equation. Here the coefficient of y is 1/2, so we factor out 1/2 from the left side of the equation
`1/2(y + 1/8) = (x + 3/4)^2`
This is same as:

`(x + 3/4)^2 = 1/2(y + 1/8)`

3. Substituting for x and y

As you may notice in the last equation, if it is compared with the standard form `x^2 = 4ay`, there is `x + 3/4 `in place of `x` and `y + 1/8` in place of `y` in it.

This steps involves putting a capital `X`, or any other variable you may like, in place of whatever expression is in the equation in place of what should be there for `x` in the standard form. That is, we substitute `X` for `x + 3/4`. Likewise we substitute `Y` for `y - 1/8`. Thus we get
`X^2 = 1/2Y`
Now this equation is almost completely like the standard form `x^2 = 4ay` and can be compared with it to find its vertex, focus, directrix etc.
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How to find the length of latus rectum of a parabola?

The length of the latus rectum is equal to the distance between the focus and vertex of the parabola. It is also equal to the perpendicular distance of the vertex from the directrix of the parabola.

Hence if you know the coordinates of the vertex and focus of a parabola, you can find the length of its latus rectum by using the distance formula as explained in the following example:

Example: A parabola has vertex (0, 3) and focus (0, 6).

By using distance formula, distance between the focus and vertex is:
`= sqrt((6 - 3)^2 + (0 - 0)^2)`
`= 3` 
Hence length of the latus rectum is equal to 3.

Simple. Isn't it?

But suppose you are given the equation of the parabola, then the method for finding the latus rectum becomes a bit different. Of course, you can find the vertex and focus coordinates from the equation and then use the distance formula, as above, to find the length of the latus rectum, but the following procedure describes how you would just find the length of the latus rectum from the equation of the parabola:

The length of the latus rectum is equal to four times the distance between the focus and the vertex. Thus in order to find the length of the latus rectum, first you have to determine the distance between the focus and the vertex of the parabola.

Thus if you know the coordinates of the vertex and focus of a parabola, you can simply use the distance formula to find the distance between the two points, then multiply it by four to get the length of the latus rectum.

On the other hand, if the equation of the parabola is given, you have to follow the following steps to find the length of the latus rectum:

Example: Suppose we have the equation `x^2 = 4ay`. 

It's graph is as follows:


In this equation 'a' is a constant value and is equal to the distance between the focus and vertex of the parabola.

Find Coordinates of Focus of a Parabola Method 2

The coordinates of the focus of a parabola can be determined from its equation. Generally all equations of parabolas can be converted into one of its standard forms.

Since we already know the foci of the standard forms, it becomes quite easy to find the focus from the equations reduced to match the standard forms.

Let us take some examples:

Example 1: `y^2 = 2x`

This equation is already in the standard form `y^2 = 4ax` (If you haven't read about standard forms then read it here).

Comparing the coefficients of 'x' in both we get
`4a = 2`
so
`a = 1/2`
We already know that the focus of the standard parabola `y^2 = 4ax` is `(a, 0)`. (If you don't know how we got that, see this post.)

Putting `a = 1/2` in it, we get
`(1/2, 0)`
Thus, the focus of the parabola `y^2 = 2x` is `(1/2, 0)`.

Example 2: `x^2 = -8y`

Again this equation is already in the standard form `x^2 = -4ay` (If you haven't read about standard forms then read it here).

Comparing the coefficients of 'x' in both we get
`4a = 8`
so
`a = 2`
We already know that the focus of the standard parabola `x^2 = -4ay` is `(0, -a)`. (If you don't know how we got that, see this post.)

Putting `a = 2` in it, we get
`(2, 0)`
Thus, the focus of the parabola `x^2 = -8y` is `(0, -2)`.

Example 3: `y = 2x^2 + 3x + 1`

This equation does not match with any of the four standard forms. We are going to follow some simple steps to convert it to standard form. These steps are discussed in detail in this post: Change Parabola Equation to Standard Form

The steps to convert it to a standard form are given below:

First, convert the equation to match a standard form by the method of completing the square. This works as follows:
Factor out coefficient of x squared
`y = 2(x^2 + 3/2x + 1)`
Take the coefficient of 'x', divide it by 2, square it and add and subtract it from the equation
`y = 2(x^2 + 3/2x + (3/4)^2 - (3/4)^2 + 1)`
Take the part containing x squared, x, and the squared term you added above and compare it with the formula `a^2 + 2ab + b^2 = (a + b)^2`. Then you get
`y = 2((x + 3/4)^2 - (3/4)^2 + 1)`
Simplify the numerical part:
`y = 2((x + 3/4)^2 - 13/4)`
Move the constants over to the right side:
`y/2 = (x + 3/4)^2 - 13/4`
`y/2 + 13/4 = (x + 3/4)^2`
Again, factor out the coefficient of y from the right hand side,
`1/2(y + 13/2) = (x + 3/4)^2`
Or
`(x + 3/4)^2 = 1/2(y + 13/2)`
Now we can change this equation to resemble the standard equation `x^2 = 4ay`. In order to do so, we will replace x + 3/4 with X and y + 13/2 with Y. That is,
`X^2 = 1/2Y`
...where `X = x + 3/4` and `Y = y + 13/2`
Note: If you are having trouble with what we did above then read this:
Change Parabola Equation to Standard Form
 Compare the above equation with the standard form `x^2 = 4ay`. Thus we have
`4a = 1/2`
`a = 1/8`
We already know that the focus of the parabola `x^2 = 4ay` lies on (0, a). Hence, the focus of the parabola above lies on (0, 1/8). But here we have taken
`X = x + 3/4`
Which means that
`x = X - 3/4`
And
`Y =  y + 13/2`
Which means that
`y = Y - 13/2`
That is, the actual values of x and y coordinates of any point are `(X - 3/4, Y - 13/2)`. Thus, since the focus is (0, 1/8) hence the coordinates of the actual focus are
`(-3/4, 1/8 - 13/2)`
That is,
`(-3/4, -51/8)`
Hence the coordinates of the focus of the parabola `y = 2x^2 + 3x + 1` are `(-3/4, -51/8)`.

Focus of a Parabola

The focus of a parabola is a fixed point inside the parabola. The distance of any point of the parabola from its focus is equal to the distance of that point from a fixed straight line called the directrix of the parabola.

The focus of a parabola always lies on its axis, that is, on the line about which the parabola is symmetric and passing through the vertex. The focus never lies on or outside the parabola.
Focus of a parabola
Each parabola has its own focus. The coordinates of the focus can be determined from the equation of a parabola as explained in this post.

Latus Rectum of a Parabola

The latus rectum of a parabola is a line passing through its focus and perpendicular to its axis.
Latus rectum is the line passing through the focus and perpendicular to the axis of a parabola
The length of the latus rectum of a parabola is always equal to four times the distance of focus from the vertex of the parabola. This post explains how to find the length of the latus rectum of a parabola.

The four standard forms of a parabola

As discussed earlier, there are four standard forms of a parabola. In short, they represent
  • A right handed parabola: `y^2 = 4ax`
  • A left handed parabola: `y^2 = -4ax`
  • An up facing parabola: `x^2 = 4ay`
  • A down facing parabola: `x^2 = -4ay`
In each of these equations, `x` and `y` are variables and `a` is any number.

Now as you may know, a parabola has some characteristic properties, namely the vertex, focus, directrix, axis, latus rectum, etc. Each of the four standard forms have their own set of properties. These are discussed below: 

`y^2 = 4ax` or Right Handed Parabola

  • Focus: (a, 0)
  • Directrix: x = -a
  • Vertex: (0, 0)
  • Axis: y = 0 (that is, the x-axis)
  • Length of latus rectum: 4a
For example, for the equation `y^2 = 2x`, comparing it with standard form `y^2 = 4ax`, we get a 4a = 2, that is, a = 1/2. Therefore the focus, directrix, etc. for `y^2 = 2x` are as follows:
  • Focus: (1/2, 0)
  • Directrix: x = -1/2
  • Vertex: (0, 0)
  • Axis: y = 0 (that is, the x-axis)
  • Length of latus rectum: 4(1/2) = 1/2
Graph of y^2 = 2x
`y^2 = 2x` and its characteristics

`y^2 = -4ax` or Left Handed Parabola

  • Focus: (-a, 0)
  • Directrix: (a, 0)
  • Vertex: (0, 0)
  • Axis: y = 0 (that is, the x-axis)
  • Length of latus rectum: 4a
For example, for the equation `y^2 = -2x`, comparing it with standard form `y^2 = -4ax`, we get a 4a = 2, that is, a = 1/2. Therefore the focus, directrix, etc. for `y^2 = -2x` are as follows:
  • Focus: (-1/2, 0)
  • Directrix: x = 1/2
  • Vertex: (0, 0)
  • Axis: y = 0 (that is, the x-axis)
  • Length of latus rectum: 4(1/2) = 1/2

`x^2 = 4ay` or Up Facing Parabola


  • Focus: (0, a)
  • Directrix: (0, -a)
  • Vertex: (0, 0)
  • Axis: x = 0 (that is, the y-axis)
  • Length of latus rectum: 4a
For example, for the equation `x^2 = 2y`, comparing it with standard form `x^2 = 4ay`, we get a 4a = 2, that is, a = 1/2. Therefore the focus, directrix, etc. for `x^2 = 2y` are as follows:
  • Focus: (0, 1/2)
  • Directrix: y = -1/2
  • Vertex: (0, 0)
  • Axis: x = 0 (that is, the x-axis)
  • Length of latus rectum: 4(1/2) = 1/2
Graph of x^2 = 2y
`x^2 = 2y` and its characteristics

`x^2 = -4ay` or Down Facing Parabola

  • Focus: (0, -a)
  • Directrix: (0, a)
  • Vertex: (0, 0)
  • Axis: x = 0 (that is, the y-axis)
  • Length of latus rectum: 4a
For example, for the equation `x^2 = -2y`, comparing it with standard form `x^2 = -4ay`, we get a 4a = 2, that is, a = 1/2. Therefore the focus, directrix, etc. for `x^2 = -2y` are as follows:
  • Focus: (0, -1/2)
  • Directrix: y = +1/2
  • Vertex: (0, 0)
  • Axis: x = 0 (that is, the x-axis)
  • Length of latus rectum: 4(1/2) = 1/2
`x^2 = -2y` and its characteristics

Standards form of a parabola

There are four simplest equations of a parabola which are called its standard forms. They are:
  • `y^2 = 4ax`
  • `y^2 = -4ax`
  • `x^2 = 4ay`
  • `x^2 = -4ay`
These are called the standard forms because they represent the simplest possible graphs of parabolas. In each of these equations, `x` and `y` represents a variable and `a` can be any number.

For example, `y^2 = 4ax` represents a right handed parabola as follows:
`y^2 = 4ax`
`y^2 = -4ax` represents a left handed parabola:
`y^2 = -4ax`
`x^2 = 4ay` represents a upward facing parabola:
`x^2 = 4ay`
`x^2 = -4ay` represents a upside-down or inverted parabola:
`x^2 = -4ay`
As you can see these are the simplest kinds of graphs of a parabola possible. So they are termed as standard forms.

These standard forms help determine the various properties of a parabola. For example, for `y = 2x^2 + 3x + 1`, though it does not match with any of the standard forms above, can be easily converted to match any one of them (It can be rewritten as `y = 2(x + 3/4)^2 - 1/8`, which is comparable with the standard form `x^2 = 4ay` - you can learn more about converting a parabola to the above four standard forms here). Its characteristics can be determined by comparing it to the standard form.

Each standard form of a parabola has its own set of properties or characteristics, such as the coordinates of focus, vertex, the equation of directrix, etc. These are discussed in this post.

Differential Equations

If you take the equation of a circle, `x^2 + y^2 = 25`, and differentiate it with respect to x, you get the following equation:
`2x + 2y dy/dx = 0`
or `x + y dy/dx = 0`      ... (i)
Equation (i) above is called a differential equation and it represents all those circles which have center at (0, 0) and any radius. All these circles form a family of circles.

Likewise, a differential equation represents a family of equations which correspond to some geometrical figure.

For example, the differential equation `dy/dx = 4x + 6` represents a family of all parabolas because it is obtained by differentiating the equation of a parabola `y = 2x^2 + 6x + 11`, and the differential equation `dy/dx = 1` represents a family of straight lines having slope 1.
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