Parabola Equations and Graphs, Directrix and Focus and How to Find Roots of Quadratic Equations
The Parabola, a Mathematical Function
In this tutorial you'll learn about a mathematical function called the parabola. We'll cover the definition of the parabola first and how it relates to the solid shape called the cone. Next we'll explore different ways in which the equation of a parabola can be expressed. Also covered will be how to work out the maxima and minima of a parabola and how to find the intersection with the x and y axes. Finally we'll discover what a quadratic equation is and how you can solve it.
Definition of a Parabola
"A locus is a curve or other figure formed by all the points satisfying a particular equation."
One way we can define a parabola is that it is the locus of points that are equidistant from both a line called the directrix and a point called the focus. So each point P on the parabola is the same distance from the focus as it is from the directrix as you can see in the animation below.
We notice also that when x is 0, the distance from P to the vertex equals the distance from the vertex to the directrix. So the focus and directrix are equidistant from the vertex.
Definition of a Parabola
A parabola is a locus of points equidistant from a line called the directrix and point called the focus.
A Parabola is a Conic Section
Another way of defining a parabola
When a plane intersects a cone, we get different shapes or conic sections where the plane intersects the outer surface of the cone. If the plane is parallel to the bottom of the cone, we just get a circle. As the angle A in the animation below changes, it eventually becomes equal to B and the conic section is a parabola.
Equations of Parabolas
There are several ways we can express the equation of a parabola:
 As a quadratic function
 Vertex form
 Focus form
We'll explore these later, but first let's look at the simplest parabola.
The Simplest Parabola y = x²
^{The simplest parabola with the vertex at the origin, point (0,0) on the graph, has the equation y = x². }
^{The value of y is simply the value of x multiplied by itself.}
x
 y = x²


1
 1

2
 4

3
 9

4
 16

5
 25

Graph of y = x²  The Simplest Parabola
Let's Give x a Coefficient!
The simplest parabola is y = x^{2} but if we give x a coefficient, we can generate an infinite number of parabolas with different "widths" depending on the value of the coefficient ɑ.
So lets make y = ɑx^{2}
In the graph below, ɑ has various values. Notice that when ɑ is negative, the parabola is "upside down". We'll discover more about this later. Remember the y = ɑx^{2} form of the equation of a parabola is when its vertex is at the origin.
Making ɑ smaller results in a "wider" parabola. If we make ɑ bigger, the parabola gets narrower.
Turning the Simplest Parabola on Its Side
If we turn the parabola y = x^{2} on its side, we get a new function y^{2} = x or x = y^{2}. This just means we can think of y as being the independent variable and squaring it gives us the corresponding value for x.
So:
When y = 2, x = y^{2} = 4
when y = 3, x = y^{2} = 9
when y = 4, x = y^{2} = 16
and so on...
Just like the case of the vertical parabola, we can again add a coefficient to y^{2.}
So we have x = ɑy^{2}
Vertex Form of a Parabola Parallel to Y Axis
One way we can express the equation of a parabola is in terms of the coordinates of the vertex. The equation depends on whether the axis of the parabola is parallel to the x or y axis, but in both cases, the vertex is located at the coordinates (h,k). In the equations, ɑ is a coefficient and can have any value.
When the axis is parallel to y axis:
y = ɑ(x  h)^{2} + k
if ɑ = 1 and (h,k) is the origin (0,0) we get the simple parabola we saw at the start of the tutorial:
y = 1(x  0)^{2} + 0 = x^{2}
When the axis is parallel to the x axis:
x = ɑ(y  h)^{2} + k
Notice that this doesn't give us any information about the location of the focus or directrix.
Equation a Parabola in Terms of the Coordinates of the Focus
Another way of expressing the equation of a parabola is in terms of the coordinates of the vertex (h,k) and the focus.
We saw that:
y = ɑ(x  h)^{2} + k
Using Pythagoras's Theorem we can prove that the coefficient ɑ = 1/4p, where p is the distance from the focus to the vertex. Substituting this into the equation above gives us an equation in terms of the focus:
When the axis of symmetry is parallel to y axis:
y = ɑ(x  h)^{2} + k = 1/(4p)(x  h)^{2} + k
Multiply both sides of the equation by 4p:
4py = (x  h)^{2} + 4pk
Rearrange:
4p(y  k) = (x  h)^{2}
or
(x  h)^{2} = 4p(y  k)
Similarly:
When the axis of symmetry is parallel to x axis:
(y  k)^{2} = 4p(x  h)
Example:
Find the focus for the simplest parabola y = x^{2}
Answer:
Since the parabola is parallel to the y axis, we use the equation we learned about above
(x  h)^{2} = 4p(y  k)
First find the vertex, the point where the parabola intersects the y axis (for this simple parabola, we know the vertex occurs at x = 0)
So set x = 0, giving y = x^{2} = 0^{2 }= 0
and the therefore the vertex occurs at (0,0)
But the vertex is (h,k), therefore h = 0 and k = 0
Substituting for the values of h and k, the equation (x  h)^{2} = 4p(y  k) simplifies to
(x  0)^{2} = 4p(y  0)
giving us
x^{2} = 4py
Now compare this to our original equation for the parabola y = x^{2}
We can rewrite this as x^{2} = y, but the coefficient of y is 1, so 4p must equal 1 and p = 1/4.
From the graph above, we know the coordinates of the focus are (h, k + p), so substituting the values we worked out for h, k and p gives us the coordinates of the vertex as
(0, 0 + 1/4) or (0, 1/4)
A Quadratic Function is a Parabola
Consider the function y = ɑx^{2} + bx + c
This is called a quadratic function because of the square on the x variable.
This is another way we can express the equation of a parabola.
How to Determine Which Direction a Parabola Opens
Irrespective of which form of equation that is used to describe a parabola, the coefficient of x^{2} determines whether a parabola will "open up" or "open down". Open up means that the parabola will have a minimum and the value of y will increase on both sides of the minimum. Open down means it will have a maximum and the value of y decreases on both sides of the max.
 If ɑ is positive, the parabola will open up
 If ɑ is negative the parabola will open down
Parabola Opens Up or Opens Down
How to Find the Vertex of a Parabola
 Expand out the equation into the form ɑx^{2} + bx + c
 Identify the coefficients a and b
 The vertex occurs at x = b/2ɑ (we can deduce this from simple calculus)
 Plug the value b/2ɑ into the equation to get the value of y
Example:
Find the vertex of the equation y = 5x^{2}  10x + 7
 The coefficient a is positive, so the parabola opens up and the vertex is a minimum
 ɑ = 5 and b = 10 so the minimum occurs at b/2ɑ =  (10)/(2(5)) = 1
 Substitute for x giving:
y = 5x^{2}  10x + 7
= 5(1)^{2}  10(1) + 7
= 5  10 + 7
= 2
So the vertex occurs at (1,2)
How to Find the XIntercepts of a Parabola
A quadratic function y = ɑx^{2} + bx + c is the equation of a parabola.
If we set the quadratic function to zero, we get a quadratic equation
i.e. ɑx^{2} + bx + c = 0 .
Graphically, equating the function to zero means setting a condition of the function such that the y value is 0, in other words, where the parabola intercepts the x axis.
The solutions of the quadratic equation allow us to find these two points. If there are no real number solutions, i.e. the solutions are imaginary numbers, the parabola doesn't intersect the x axis.
The solutions of a quadratic equation are given by the equation:
x = b ± √(b^{2} 4ac) / 2ɑ
Finding the Roots of a Quadratic Equation
Example 1: Find the xaxis intercepts of the parabola y = 3x^{2} + 7x + 2
Solution
 y = ɑx^{2} + bx + c
 In our example y = 3x^{2} + 7x + 2
 Identify the coefficients and constant c
 So ɑ = 3, b = 7 and c = 2
 The roots of the quadratic equation 3x^{2} + 7x + 2 = 0 are at x = b ± √(b^{2}  4ɑc) / 2ɑ
 Substitute for ɑ, b and c
 The first root is at x = 7 + √(7^{2} 4(3)(2)) / (2(3) = 1/3
 The second root is at 7  √(7^{2} 4(3)(2)) / (2(3) = 2
 So the x axis intercepts occur at (2, 0) and (1/3, 0)
Example 2: Find the xaxis intercepts of the parabola with vertex located at (4, 6) and focus at (4, 3)
Solution
 The equation of the parabola in focus vertex form is (x  h)^{2} = 4p(y  k)
 The vertex is at (h,k) giving us h = 4, k = 6
 The focus is located at (h, k + p). In this example the focus is at (4, 3) so k + p = 3. But k = 6 so p = 3  6 = 3
 Plug the values into the equation (x  h)^{2} = 4p(y  k) so (x  4)^{2} = 4(3)(y  6)
 Simplify giving (x  4)^{2} = 12(y  6)
 Expand out the equation gives us x^{2}  8x + 16 = 12y + 72
 Rearrange 12y = x^{2} + 8x + 56
 Giving y = 1/12x^{2} + 2/3x + 14/3
 The coefficients are a = 1/12, b = 2/3, c = 14/3
 The roots are at 2/3 ± √((2/3)^{2}  4(1/12)(14/3))/(2(1/12)
 This gives us x = 4.49 approx and x = 12.49 approx
 So the x axis intercepts occur at (4.49, 0) and (12.49, 0)
How to Find the YIntercepts of a Parabola
To find the yaxis intercept (yintercept) of a parabola, we set x to 0 and calculate the value of y.
Example 3: Find the yintercept of the parabola y = 6x^{2} + 4x + 7
Solution:
y = 6x^{2} + 4x + 7
Set x to 0 giving
6(0)^{2} + 4(0) + 7 = 7
The intercept occurs at (0, 7)
Summary of Parabola Equations
Equation Type
 Axis Parallel to YAxis
 Axis Parallel to XAxis


Quadratic Function
 y = ɑx² + bx + c
 x = ɑy² + by + c

Vertex Form
 y = ɑ(x  h)² + k
 x = ɑ(y  h)² + k

Focus Form
 (x  h)² = 4p(y  k)
 (y  k)² = 4p(x  h)

Parabola with Vertex at the Origin
 x² = 4py
 y² = 4px

How the Parabola is Used in the Real World
The parabola isn't just confined to math. The parabola shape appears in nature and we use it in science and technology because of its properties.
 When you kick a ball into the air or a projectile is fired, the trajectory is a parabola
 The reflectors of vehicle headlights or flashlights are parabolic shaped
 The mirror in a reflecting telescope is parabolic
 Satellite dishes are in the shape of a parabola as are radar dishes
Related reading: Deriving Projectile Motion Equations
Acknowledgements
All graphics were created using GeoGebra Classic.
© 2019 Eugene Brennan