Lesson PARABOLAS

Source code of 'PARABOLAS'

This Lesson (PARABOLAS) was created by by Theo(3458)

This lesson covers a basic overview of PARABOLAS

Parabolas are structures that are the result of quadratic equations.  They have certain properties that will be discussed in more detail below.

To see more information relating to the standard form of quadratic equations and how to solve them, see the lesson on QUADRATIC EQUATIONS.

REFERENCES

<a href= "http://tutorial.math.lamar.edu/classes/alg/parabolas.aspx" target="_blank">Paul's Online Notes</a>
<a href= "http://www.intmath.com/Quadratic-equations/4_Graph-quadratic-function.php" target="_blank">Interactive Mathematics</a>
<a href= "http://www.purplemath.com/modules/parabola.htm" target="_blank">Purple Math</a>
<a href= "http://hotmath.com/hotmath_help/topics/parabolas.html" target="_blank">Hot Math</a>
<a href= "http://www.mathwarehouse.com/geometry/parabola/standard-and-vertex-form.php" target="_blank">Math Warehouse Quadratic and Vertex Equation Forms</a>
<a href= "http://jwilson.coe.uga.edu/EMT725/Class/Sarfaty/EMT669/InstructionalUnit/Parabolas/parabolas.html" target="_blank">Jack Sarfaty Wilson College</a>
<a href = "http://www.mathwords.com/d/directrix_parabola.htm" target = "_blank">Focus and Directrix of a Parabola (mathwords)</a>
<a href = "http://www.algebralab.org/lessons/lesson.aspx?file=Algebra_conics_directrix.xml" target = "_blank">Focusw and Directrix of a Parabola (algebralab)</a>
<a href = "http://www.intmath.com/Plane-analytic-geometry/4_Parabola.php" target = "_blank">Focus and Directrix of a Parabola (intmath)</a>
<a href = "http://www.mc.maricopa.edu/~steig/readings/conics/index.html" target = "_blank">Focus and Directrix of a Parabola (maricopa)</a>

STANDARD FORM FOR THE EQUATION OF A PARABOLA

The standard form for the equation of a parabola is the standard form of a quadratic equation.

The standard form of the quadratic equation is:

{{{y = ax^2 + bx + c}}}

a is the coefficient of the x^2 term.
b is the coefficient of the x term.
c is the constant term.

VERTEX OF A PARABOLA

The vertex of a parabola is the minimum / maximum point of the parabola.

If the parabola opens up, the vertex is the minimum point on the parabola.  If the parabola opens down, the parabola is the maximum point on the parabola.  If the parabola opens right, the vertex is the leftmost point on the parabola.  If the parabola opens left, the vertex is the rightmost point on the parabola.

ROOTS OF THE PARABOLA

In order to solve for the roots of the quadratic equation, you would set y = 0 and solve for x.

Note that x^2 and {{{x^2}}} are the same thing.  The first is the text version and the second is the algebra.com formula generator version.

An example of a parabola in standard form would be y = x^2 - 10x + 16 where:
a = 1
b = -10
c = 16

This parabola opens up.
That's because the "a" term in the equation of this parabola is positive.

Please note that when the parabola opens up, then the vertex of the parabola is the lowest point on the parabola.  This would also be the minimum point on the parabola because the value of y at this point is less than the value of y at any other point on the graph of this parabola.

Graph of y = x^2 - 10x + 16 is shown below.
{{{graph(300,300,-10,10,-10,10,x^2 - 10x + 16)}}}

To find the roots of this equation, set y = 0 and solve for x.

Factors for this equation would be in the general form of:
(dx+e)*(fx+g) = 0 where:
d*f = a
d*g + e*f = b
e*g = c

The factors for this equation of x^2 - 10x + 16 are:
(x-2)*(x-8) = 0
The standard form is:
(dx+e)*(fx+g) = 0
You can see that:
d*f = 1*1 = 1
d*g + e*f = 1*(-8) + (-2)*1 = -8-2 = -10
e*g = (-2)*(-8) = 16

The roots for this equation of x^2 - 10x + 16 are:
x = 2 and x = 8

If we make the "a" term in this equation negative, then the graph will open down.

If we want to keep the roots the same so we have the equation in the same relative position on the graph only opening in the opposite direction, we multiply the whole equation by (-1) to get:
-x^2 + 10x - 16

Original equation is x^2 - 10x + 16
Revised equation is -x^2 + 10x - 16

It's the same equation only now it is opening down.

Please note that when the parabola opens down, then the vertex of the parabola is the highest point on the parabola.  This would also be the maximum point on the parabola because the value of y at this point is greater than the value of y at any other point on the graph of this parabola.

It has the same roots so it has essentially been pivoted on the x-axis.

Graph of y = -x^2 + 10x - 16 is shown below.
{{{graph(300,300,-10,10,-10,10,-x^2 + 10x - 16)}}}

PARABOLA THAT OPENS TO THE LEFT OR TO THE RIGHT

Let's take the original equation of y = x^2 - 10x + 16 which opens up.
Replace y with x and x with y to get x = y^2 - 10y + 16
This graph will open to the right because the "a" term is positive (coefficient of y^2).

Please note that when the parabola opens to the right, then the vertex of the parabola is the leftmost point on the parabola.  This would also be the minimum point on the parabola because the value of x at this point is less than the value of x at any other point on the graph of this parabola.

Graph of x = y^2 - 10y + 16 is shown below.
{{{graph(300,300,-10,10,-10,10,5+sqrt(x+9),5-sqrt(x+9))}}}

Now let's take the original equation of y = -x^2 + 10x - 16 which opens down.
Replace y with x and x with y to get x = -y^2 + 10y - 16
This graph will open to the left because the "a" term is negative (coefficient of y^2).

Please note that when the parabola opens to the left, then the vertex of the parabola is the rightmost point on the parabola.  Please note that this is also the maximum point of this parabola because the value of x at this point is greater than the value of x at any other point on the graph of this parabola.

Graph of x = -y^2 + 10y - 16 is shown below.
{{{graph(300,300,-10,10,-10,10,5+sqrt(-x+9),5-sqrt(-x+9))}}}

Note that we could not graph x = y^2 - 10y + 16 directly, nor could we graph x = -y^2 + 10y - 16 directly.

This is because the graphing software is fixed so that your equation has to be in the form of y = f(x) only.  It does not know how to handle x = f(y).

In order to graph these equations, we have to solve for x in order to convert the equation that is in the form of x = f(y) into the form of y = f(x).

I'll do one to show you how that was done since it's a good opportunity to show you how the completing the squares method can be helpful in solving equations.

Take the equation of x = y^2 - 10y + 16
Solve for y as follows:
Subtract 16 from both sides of this equation to get:
y^2 - 10y = x - 16
Complete the squares on the left hand side of this equation by taking one half of -10 (the b term) to get -5 and then squaring it to get 25 to get:
(y-5)^2 - 25 = x - 16

The -5 became part of the factor to be squared and the 25 is an adjustment of the result of (y-5)^2 to make it equal to y^2 - 10y rather than y^2 - 10y + 25.

Add 25 to both sides of this equation to get:
(y-5)^2 = x+9
Take the square root of both sides of this equation to get:
y-5 = +/- {{{sqrt(x+9)}}}
Add 5 to both sides of this equation to get:
y = 5 +/- {{{sqrt(x+9)}}}

You have now converted x = f(y) into the form of y = f(x).
The two forms of this equation are equivalent.  x = f(y) solves for x.  y = f(x) solves for y.

In a similar fashion, x = -y^2 + 10y - 16 was converted to:
y = 5 +/- {{{sqrt(-x+9)}}}

The equation of x = y^2 - 10x + 16 opens to the right.
The equivalent equation of y = 5 +/- {{{sqrt(x+9)}}} also opens to the right.
The coefficient of y^2 in the x = f(y) version of the equation is positive and the coefficient of x in the y = f(x) version of the same equation is also positive.  This will be generally true.

The equation of x = -y^2 + 10x - 16 opens to the left.
The equivalent equation of y = 5 +/- {{{sqrt(-x+9)}}} also opens to the left.
The coefficient of y^2 in the x = f(y) version of the equation is negative and the coefficient of x in the y = f(x) version of the same equation is also negative.  This will be generally true.

FINDING THE VERTEX OF A PARABOLA WHEN THE EQUATION FOR THE PARABOLA IS IN STANDARD FORM AND THE PARABOLA OPENS UP OR DOWN

Please note that when the parabola opens up, then the vertex of the parabola is the lowest point on the parabola.  This is also called the minimum point on the parabola.  It is called the minimum point on the parabola because the value of y at that point is less than at any other point on the parabola.

Standard form of the equation of a parabola that opens up or down is y = ax^2 + bx + c

The x value of the vertex of the parabola can be found by the equation x = -b/2a

Once you find the x value, then you plug that value into the equation to find the y value.

The generalized form of the vertex of the parabola is (x,y) = (-b/2a,f(-b/2a)).

-b/2a is the x coordinate of the vertex.  f(-b/2a) is the y coordinate of the vertex.

You find the x coordinate of the vertex by using the formula x = -b/2a.  You find the y coordinate of the vertex by plugging the value of x into the equation and solving for y.  Since you equation is in the form of y = f(x), then you are solving for f(-b/2a) when you replace x with the value of -b/2a.

An example will help clarify this.

y = f(x) = x^2 - 10x + 16 is the standard form of the equation for a parabola that opens up.  The standard form of this equation is y = ax^2 + bx + c.  a is the coefficient of the x^2 term and is equal to 1.  The parabola is opening up because a = 1 is positive.  The vertex is the minimum point on the parabola.

a = 1
b = -10
c = 16

the equation for the x value of the vertex of this parabola is found by using the formula of x = -b/2a and substituting 1 for the a value and -10 for the b value to get:
x = -b/2a = -(-10)/2 = 10/2 = 5

The x value for the vertex of this parabola is 5.

To find the y value of the vertex of this parabola, replace x with 5 in the original equation and solve for y.
Note that y = f(x) = x^2 - 10x + 16 now becomes:
y = f(5) = (5)^2 - 10*(5) + 16 which becomes:
y = f(5) = 25 - 50 + 16 = -25 + 16 = -9

We have an x value of 5 and a y value of -9 for the vertex of the parabola that equals y = f(x) = x^2 - 10x + 16.

We found x = -b/2a = 5 and we found:
y = f(-b/2a) = -9

the coordinate point of (x,y) = (-b/2a,f(-b/2a)) becomes (5,-9).

If you look at the graph of the equation of y = x^2 - 10x + 16, you will see that the vertex of this equation is (5,-9) as shown below:
{{{graph(300,300,-10,10,-10,10,x^2 - 10x + 16,-9)}}}

I placed a horizontal line at y = -9 to help show this better.

You can also see that the vertex is the lowest / minimum point on this parabola.

FINDING THE VERTEX OF A PARABOLA WHEN THE EQUATION FOR THE PARABOLA IS IN STANDARD FORM AND THE PARABOLA OPENS LEFT OR RIGHT.

Please note that a parabola that opens left or right will be in the standard form of:
x = ay^2 + by + c while a parabola that opens up or down will be in the standard form of:
y = ax^2 + bx + c

In this section, we will be discussing the parabola that is in the form of:
x = ay^2 + by + c

Please note also that a parabola in the form of:
x = f(y) can also be shown in the form of:
y = f(x) but it will look a lot different.

It will look something like:
y = {{{(-b/2a)}}} +/- {{{sqrt(((x-c)/a)+(b/(2a))^2)}}}

Please note that if a is negative, then the {{{(x-c)/a}}} term under the square root sign will also be negative.

Bottom line is you can't miss it.  If it's an equation of a parabola that opens to the left or to the right, and it is in the form of y = f(x), then there is a square root on the right hand side of the equation and it definitely does not look like the standard form of the equation of a quadratic equation.

We'll continue to work with the x = f(y) form of the parabola that opens to the left or to the right.  We will, however, have to convert it to the y = f(x) form when we graph it.

Please note that if a is positive, then the parabola will open to the right, and if a is negative, then the parabola will open to the left.

Standard form of the equation of a parabola that opens to the left or to the right is:
x = ay^2 + by + c

The y value of the vertex of the parabola can be found by the formula y = -b/2a

Once you find the y value, then you plug that value into the equation to find the x value.

The generalized form of the vertex of a parabola that opens left or right would be (f(-b/2a),-b/2a)

An example will help clarify this.

x = f(y) = y^2 - 10y + 16 is the standard form of the equation for a specific parabola that opens to the right.  The general form of this equation is x = ay^2 + by + c.  a is the coefficient of the y^2 term and is equal to 1.  The parabola is opening to the right because a = 1 is positive.

a = 1
b = -10
c = 16

the equation for the y value of the vertex of this parabola is found by using the formula of y = -b/2a and substituting -10 for the b value and 1 for the a value to get:
y = -(-10)/2 = 10/2 = 5

The y value for the vertex of this parabola is 5.

To find the x value of the vertex of this parabola, replace y with 5 in the original equation and solve for x.

x = f(y) = y^2 - 10y + 16 now becomes:
x = f(5) = (5)^2 - 10*(5) + 16 which becomes:
x = f(5) = 25 - 50 + 16 = -25 + 16 = -9

We have a y value of 5 and an x value of -9 for the vertex of the parabola that equals x = f(y) = y^2 - 10y + 16.

We found y = -b/2a = 5 and we found x = f(-b/2a) = -9

the coordinate point of (f(-b/2a),-b/2a) becomes (-9,5).

If you look at the graph of the equation of x = y^2 - 10y + 16, you will see that the vertex of this equation is (-9,5) as shown below:
{{{graph(300,300,-10,10,-10,10,5+sqrt(x+9),5-sqrt(x+9))}}}
You will also see that the graph opens to the right because the "a" term is positive.

PIVOT POINTS ON A PARABOLA THAT OPENS UP OR DOWN

Before we go any further, I wish to point out something that may not be obvious.

I reversed the direction of the parabola that opens up by multiplying the whole equation by (-1).
ax^2 = bx + c became -ax^2 - bx - c.
This essentially pivoted the equation on the roots of that equation.

If you want to reverse the direction of the parabola at the vertex, then you have to do the following:

Assume your original equation for a parabola that opens up is 3x^2 - 6x + 5 where:
a = 3
b = -6
c = 5

Find the vertex at (-b/2a, f(-b/2a)) = (1,2)

this is the point you want to pivot at if your want your reverse parabola to open down from that same point.

Multiply your equation by (-1) to get the reverse direction equation.
3x^2 - 6x + 5 * (-1) = -3x^2 + 6x - 5

Your reverse equation is:
-3x^2 + 6x - 5

Remove the c term of -5 and replace it with the variable name of c because it is unknown at this time.

Your vertex from the original equation of 3x^2 - 6x + 5 is (x,y) = (1,2).

In your reverse equation of -3x^2 + 6x - c, replace x with the original x value of your vertex of 1 and set the equation equal to the original y value of your vertex of 2 and solve for c:

Your reverse equation of:
-3x^2 + 6x + c becomes:
-3(1)^2 + 6*(1) + c = 2 which becomes:
-3 + 6 + c = 2 which becomes:
c = -1

Your new c value is -1 making your reverse equation equal to:
-3x^2 + 6x - 1

The original equation is:
3x^2 - 6x + 5 
and the reverse equation is:
-3x^2 + 6x - 1

Graph of both the original and reverse equation is shown below:
{{{graph(300,300,-5,5,-5,5,3x^2 - 6x + 5, -3x^2 + 6x - 1)}}}

As you can see, the reverse equation has now been pivoted at the vertex.

AXIS OF SYMMETRY OF A PARABOLA

If the parabola points up or down, then the axis of symmetry for the parabola is the x value of the vertex of the parabola.

If the parabola points left or right, then the axis of symmetry for the parabola is the y value of the vertex of the parabola.

The graph of the parabola that was pointing down with a vertex of (5,-9) is shown below.  The axis of symmetry is the vertical line at x = 5.
{{{graph(300,300,-10,10,-10,10,x^2 - 10x + 16)}}}

The graph of the parabola that was pointing left with a vertex of (-9,5) is shown below.  The axis of symmetry is the horizontal line at y = 5
{{{graph(300,300,-10,10,-10,10,5+sqrt(x+9),5-sqrt(x+9))}}}

To see a picture of vertical and horizontal Axis of Symmetry, click on the following hyperlink.
<a href = "http://theo.x10hosting.com/examples/Parabolas/ParabolaAxisOfSymmetry.html" target = "_blank">Axis Of Symmetry</a>

VERTEX FORM OF THE EQUATION FOR A PARABOLA THAT OPENS UP OR DOWN

The vertex form of the equation for a parabola that opens up or down is y = a(x-h)^2 + k where:

(h,k) is the vertex of the parabola.

h is the x coordinate
k is the y coordinate

a is the coefficient of the (x-h)^2 term.

When a is positive, the parabola opens up.
When a is negative, the parabola opens down.

Please note that when the parabola opens up, the vertex of the parabola is the lowest point on the parabola.  It is also the minimum point on the parabola because the value of y at that point is less than at any other point on the parabola.

Please note that when the parabola opens down, the vertex of the parabola is the highest point on the parabola.  It is also the maximum point on the parabola because the value of y at that point is greater than at any other point on the parabola.

VERTEX FORM OF THE EQUATION FOR A PARABOLA THAT OPENS LEFT OR RIGHT

The vertex form of the equation for a parabola that opens left or right is x = a(y-k)^2 + h where:

(h,k) is the vertex of the parabola.

h is the x coordinate
k is the y coordinate

a is the coefficient of the (y-k)^2 term.

When a is positive, the parabola opens to the right.
When a is negative, the parabola opens to the left.

Please note that when the parabola opens to the right, the vertex of the parabola is the leftmost point on the parabola.  It is also the minimum point on the parabola because the value of x at that point is less than at any other point on the parabola.

Please note that when the parabola opens to the left, the vertex of the parabola is the rightmost point on the parabola.  It is also the maximum point on the parabola because the value of x at that point is greater than at any other point on the parabola.

CONVERTING FROM THE STANDARD FORM OF A PARABOLA TO THE VERTEX FORM OF A PARABOLA WHEN THE PARABOLA OPENS UP OR DOWN

Standard form of the equation for a parabola opening up or down is y = ax^2 + bx + c
Vertex form of the equation for a parabola opening up or down is y = a(x-h)^2 + k

To show you how we convert from the standard form to the vertex form, we'll use an example.

Your equation in standard form is:
y = 5x^2 + 30x + 60 where:
a = 5
b = 30
c = 60

While this could be simplified further by dividing out by common factor of 5, we'll leave it the way it is for demonstration purposes because we want to keep the "a" term greater than 1 and we do not want to have to deal with fractions in the demonstration.

We factor out the "a" term of 5 to get:
{{{y = 5 * (x^2 + 6x + 12)}}}
We complete the square on the (x^2 + 6x) term to get:
{{{y = 5 * ((x+3)^2 - (3)^2 + 12)}}} which becomes:
{{{y = 5 * ((x+3)^2 - 9 + 12)}}} which becomes:
{{{y = 5 * ((x+3)^2 + 3)}}}
We remove the parentheses to get:
{{{y = 5 * (x+3)^2 + (3 * 5)}}} which becomes:
{{{y = 5*(x+3)^2 + 15}}}

This is now in vector form of {{{y = a * (x-h)^2 + k}}} where:
a = 5
h = (-3)
k = 15

The vertex of this equation is (-3,15)

The two equation of:
{{{y = 5x^2 + 30x + 60}}} and {{{y = 5(x+3)^2+15}}} are equivalent.
To prove it to yourself, simply choose any value for x and plug it in both equations and you will see that you get the same answer.
I chose x = 7 and got 515 for the first equation and 515 for the second equation.

Graph of this equation is shown below.
{{{graph (300,300,-15,15,-100,900,5x^2 + 30x + 60,5*(x+3)^2+15)}}}
I put both forms of the equation in the graph generator so that if there was a difference, you would see it.  If you only see one equation being graphed, then the two forms are equivalent.  This is another way of proving they are equivalent which is analagous to taking two objects and placing them on top of each other to see if they are equivalent.

CONVERTING FROM THE VERTEX FORM OF A PARABOLA TO THE STANDARD FORM OF A PARABOLA WHEN THE PARABOLA OPENS UP OR DOWN

To convert from the vertex form of the parabola to the standard form of the parabola, you simply reverse the process.

Our equation in vertex form is:
{{{y = 5(x+3)^2+15}}}
We square the (x+3) term and multiply it by 5 to get:
{{{y = 5 * (x^2 + 6x + 9) + 15}}} which becomes:
{{{y = 5x^2 + 30x + 45 + 15}}} which becomes:
{{{y = 5x^2 + 30x + 60}}} which is the same equation in standard form.

CONVERTING FROM THE STANDARD FORM OF A PARABOLA TO THE VERTEX FORM OF A PARABOLA WHEN THE PARABOLA OPENS LEFT OR RIGHT

Standard form of the equation for a parabola opening left or right is x = ay^2 + by + c
Vertex form of the equation for a parabola opening up or down is x = a(y-k)^2 + h

To show you how we convert from the standard form to the vertex form, we'll use an example.

Your equation in standard form is:
x = 5y^2 + 30y + 60 where:
a = 5
b = 30
c = 60

While this could be simplified further by dividing out by the common factor of 5, we'll leave it the way it is for demonstration purposes because we want to keep the "a" term greater than 1 and we do not want to have to deal with fractions in the demonstration.

We factor out the "a" term of 5 to get:
{{{x = 5 * (y^2 + 6y + 12)}}}
We complete the square on the (y^2 + 6y) term to get:
{{{x = 5 * ((y+3)^2 - (3)^2 + 12)}}} which becomes:
{{{x = 5 * ((y+3)^2 - 9 + 12)}}} which becomes:
{{{x = 5 * ((y+3)^2 + 3)}}}
We remove the parentheses to get:
{{{x = 5 * (y+3)^2 + (3 * 5)}}} which becomes:
{{{x = 5*(y+3)^2 + 15}}}

This is now in vector form of {{{x = a * (y-k)^2 + h}}} where:
a = 5
k = (-3)
h = 15

The vertex of this equation is (15,-3)

The two equations of:
{{{x = 5y^2 + 30y + 60}}} and {{{x = 5(y+3)^2+15}}} are equivalent.
To prove it to yourself, simply choose any value for y and plug it in both equations and you will see that you get the same answer.
I chose y = 7 and got 515 for the first equation and 515 for the second equation.

In order to graph this equation, we have to convert it from x = f(y) to y = f(x) by solving for y.
The equation of:
{{{x = 5*(y+3)^2 + 15}}} becomes:
y = {{{-3}}} +/- {{{sqrt((x-15)/5)}}}

Graph of this equation is shown below.
{{{graph (300,300,-100,900,-15,15,-3+sqrt((x-15)/5),-3-sqrt((x-15)/5))}}}

CONVERTING FROM THE VERTEX FORM OF A PARABOLA TO THE STANDARD FORM OF A PARABOLA WHEN THE PARABOLA OPENS LEFT OR RIGHT

To convert from the vertex form of the parabola to the standard form of the parabola, you simply reverse the process.

Our equation in vertex form is:
{{{x = 5(y+3)^2+15}}}
We square the (y+3) term and multiply it by 5 to get:
{{{x = 5 * (y^2 + 6y + 9) + 15}}} which becomes:
{{{x = 5y^2 + 30y + 45 + 15}}} which becomes:
{{{x = 5y^2 + 30y + 60}}} which is the same equation in standard form.

FOCUS OF A PARABOLA

The focus of a parabola is a point on the axis of symmetry of the parabola that is a set distance from the vertex of the parabola.  The point is said to be inside the parabola.  If the parabola opens up, the point is above the vertex.  If the parabola opens down, the point is below the vertex.  If the parabola opens to the right, then the point is to the right of the vertex.  If the parabola opens to the left, then the point is to the left of the vertex.

DIRECTRIX OF A PARABOLA

The directrix of a parabola is a line (not a point) that is perpendicular to the axis of symmetry of the parabola and does not intersect with the parabola.  That line is said to be outside of the parabola.  If the parabola opens up, the directrix is below the vertex.  If the parabola opens down, the directrix is above the vertex.  If the parabola opens to the right, the directrix is to the left of the vertex.  If the parabola opens to the left, the directrix is to the right of the vertex.

RELATIONSHIP BETWEEN THE FOCUS OF A PARABOLA AND THE DIRECTRIX OF A PARABOLA

The focus is a point inside the parabola on the axis of symmetry of the parabola.
The directrix is a line outside the parabola that is perpendicular to the axis of symmetry of the parabola.

The distance between the Focus and the Vertex of the Parabola is the same as the distance between the Vertex and the Directrix at the point where the Directrix intersects with the Axis of Symmetry.

This is a defining characteristic of the Focus and Directrix of a Parabola.  They are in a fixed relationship with each other.

Let F = the Focus point on the axis of symmetry of the parabola.
Let V = the Vector point on the axis of symmetry of the parabola.
Let D = the point of intersection of the Directrix with the axis of symmetry of the parabola.

What we have is d = F-V = V-D.

You can't just pick any d.
There is only one d for each parabola that works.
This is because of the special relationship between the Focus and the Directrix and any other point on the parabola that will be discussed shortly.

In other words, there is a formula that determines what the distance between the Focus and the Vector, and the Vector and the Directrix, should be.

That formula is {{{d = 1/(4a)}}}

The "a" in the formula is the same "a" that you have been dealing with all along.

In the standard formula for a parabola of y = ax^2 + bx + c, that "a" is the coefficient of the x^2 term.

In the vertex formula for a parabola of y = a*(x-h)^2 + k, that "a" is the coefficient of the (x-h)^2 term.

If we know "a", then we can find "d" because of this fixed relationship between them given by the formula {{{d = 1/(4a)}}}.

EXAMPLE OF FINDING THE FOCUS AND THE POINT OF INTERSECTION OF THE DIRECTRIX WITH THE AXIS OF SYMMETRY OF A PARABOLA THAT IS OPENING UP USING THE STANDARD FORM OF THE EQUATION FOR THE PARABOLA

The formula we will be working with is in standard form of y = ax^2 + bx + c.
Our formula is:
y = 3x^2 + 24x + 43 where:
a = 3
b = 24
c = 43

The formula for the distance between the focus and the vector is:
d = 1/(4a)
Since a equals 3, this means that d = 1/(3*4) = 1/12.

The x value of the vector for this equation is -b/(2a) = -24/(2*3) = -24/6 = -4.
The y value of the vector for this equation is found by replacing x with 3 in the equation of y =:
3x^2 + 24x + 43 to get:
3*(-4)^2 + 24*(-4) + 43 = 48 - 96 + 43 = -5
The vertex for this parabola is:
V = (-4,-5)

Since this parabola opens up ("a" term is positive), the focus is above the vertex and the directrix is below the vertex.

Our parabola opens up, so the focus is above the vector.
It has the same x value as the vector.
It's y value is the y value of the vector plus d.
In this case, the  y value of the vector is -5 and d = 1/12.  Add 1/12 to -5 and you get -5 + 1/12 which is equivalent to -60/12 + 1/12 which is equal to -59/12.

The point of intersection of the directrix with the axis of symmetry of our parabola is below the vector and the same distance from it as the focus.
It has the same x value as the vector.
It's y value is the y value of the vector minus d.
In this case, the y value of the vector is -5.  Subtract 1/12 from -5 and you get -5 - 1/12 which is equivalent to -60/12 - 1/12 which is equal to -61/12.

Our Focus is at (-4,-59/12)
Our Vector is at (-4,-60/12)
Our Directrix Point of Intersection with the Axis of Symmetry is at (-4,-61/12)

EXAMPLE OF FINDING THE FOCUS AND THE POINT OF INTERSECTION OF THE DIRECTRIX WITH THE AXIS OF SYMMETRY OF A PARABOLA THAT IS OPENING UP USING THE VECTOR FORM OF THE EQUATION FOR THE PARABOLA

The standard form of our equation is y = ax^2 = bx + c which equals:
y = 3x^2 + 24x + 43
The vector form of our equation is y = a*(x-h)^2 + k which equals:
y = 3*(x+4)^2 - 5

Our vector is (-4,-5)
d = 1/(4a) = (1/12)

Once we have the vector and the distance between the focus and the vector, all other calculations are the same.

Since d is a distance, it will always be positive.

If the parabola opens up or to the right, then F = V + d
If the parabola opens down or to the left, then F = V - d

If the parabola opens up or down, then the y value of F and V are changing, and the x value remains the same (on the Axis of Symmetry).
If the parabola opens right or left, then the x value of F and V are changing, and the y value remains the same (on the Axis of symmetry).

RELATIONSHIP BETWEEN ANY POINT ON A PARABOLA AND THE DISTANCE BETWEEN THAT POINT AND THE FOCUS OF THAT PARABOLA, AND THE DISTANCE BETWEEN THAT POINT AND THE DIRECTRIX OF THAT PARABOLA

The distance between the focus and the Vertex is equal to the distance between the Vertex and the point of intersection between the Directrix and the Axis of Symmetry.
In the picture of the graph below, d1a is the distance between the Focus and the Vector, and d1b is the distance between the Vector and the intersection between the Directrix and the Axis of Symmetry.

The distance between any point on the Graph of the Parabola and the Focus is equal to the distance between that point and the Directrix.

The distance between that point and the focus is a straight line between those 2 points.

the distance between that point and the directrix is a vertical straight line between that point and the directrix where the intersection of that line with the directrix forms a right angle.  In other words, that vertical straight line is prependicular to the directrix at the intersection point.

In the picture of the graph below, d2a is the distance between any point on the graph of the Parabola, and d2b is the distance between that point on the Parabola and the intersecting point on the Directrix from a perpendicular line dropped from that point.

The following details relating to the graph below are displayed below.  They are also shown in the picture of the graph below.

Standard Form of Equation of Parabola: 
     y = .125*x^2 + .125*x - .375
Vertex Form of Equation of Parabola:
     y = .125*(x+.5)^2 – .40625
Vertex of Parabola:
     (x,y) = (-.5,-.40625)
Coefficient of x^2 term and (x+.5)^2 term:
     a = .125
Distance from Focus to Vertex:
     d1a = 1/(4a) = (1/.5) = 2
Distance from Vertex to Directrix:
     d1b = 1/(4a) = (1/.5) = 2
Axis of Symmetry:
     x = -.5
Focus of Parabola:
     (x,y) = (-.5, 1.59375)
Directrix of Parabola:
     y = -2.40625
Point on Parabola:
     (x,y) = (3, 1.125)
Distance from Point to Focus:
     d2a = sqrt((3.5)^2 + (-.46875)^2) = sqrt(12.46972656)
Distance from Point to Directrix:
     d2b = sqrt(0^2 + (3.53125)^2) = sqrt(12.46972656)

This is the graph of this parabola.
{{{graph(600,600,-4,4,-4,4,.125*x^2+.125*x-.375)}}}

To see the picture of this graph with comments and descriptive information, click on the following hyperlink.

<a href = "http://theo.x10hosting.com/examples/Parabolas/ParabolaFocusDirectrix.html" target = "_blank">Focus and Directrix of a Parabola</a>

You are encouraged to visit the references to view the many examples and problems they have up there.

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