How to Understand the Equation of a Parabola, Directrix and Focus

"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 is the same distance from the focus as it is from the directrix as you can see in the animation below.

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.

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 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.}

The simplest parabola is y = x² 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 = ɑx²

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 = ɑx² 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.

If we turn the parabola y = x² on its side, we get a new function y² = x or x = y². 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 = y² = 4

when y = 3, x = y² = 9

when y = 4, x = y² = 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 = ɑy²

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 = x²

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.

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:

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)

Consider the function y = ɑx² + 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.

Irrespective of which form of equation that is used to describe a parabola, the coefficient of x² 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

1. Expand out the equation into the form ɑx² + bx + c
2. Identify the coefficients a and b
3. The vertex occurs at x = -b/2ɑ
4. Plug the value -b/2ɑ into the equation to get the value of y

Example: Find the vertex of the equation y = 5x² - 10x + 7

1. The coefficient a is positive, so the parabola opens up and the vertex is a minimum
2. ɑ = 5 and b = -10 so the minimum occurs at -b/2ɑ = - (-10)/(2(5)) = 1
3. Substitute for x giving:

y = 5x² - 10x + 7

= 5(1)² - 10(1) + 7

= 5 - 10 + 7

= 2

So the vertex occurs at (1,2)

A quadratic function y = ɑx² + bx + c is the equation of a parabola.

If we set the quadratic function to zero, we get a quadratic equation

i.e. ɑx² + 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 ± √(b² -4ac) / 2ɑ

Example 1: Find the x-axis intercepts of the parabola y = 3x² + 7x + 2

Solution

• y = ɑx² + bx + c

• In our example y = 3x² + 7x + 2

• Identify the coefficients and constant c

• So ɑ = 3, b = 7 and c = 2

• The roots of the quadratic equation 3x² + 7x + 2 = 0 are at x = -b ± √(b² - 4ɑc) / 2ɑ

• Substitute for ɑ, b and c

• The first root is at x = -7 + √(7² -4(3)(2)) / (2(3) = -1/3

• The second root is at -7 - √(7² -4(3)(2)) / (2(3) = -2

• So the x axis intercepts occur at (-2, 0) and (-1/3, 0)

Example 2: Find the x-axis 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 x² - 8x + 16 = -12y + 72
• Rearrange 12y = -x² + 8x + 56
• Giving y = -1/12x² + 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)

To find the y-axis intercept (y-intercept) of a parabola, we set x to 0 and calculate the value of y.

Example 3: Find the y-intercept of the parabola y = 6x² + 4x + 7

Solution:

y = 6x² + 4x + 7

Set x to 0 giving

6(0)² + 4(0) + 7 = 7

The intercept occurs at (0, 7)

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

All graphics were created using GeoGebra Classic.