Conic sections are shapes that come from the intersection of a plane and 2 cones. All of the Ellipses and Hyperbolas that will be explained are centered at (0,0). There are 4 shapes that fall into this category.

 

1)      Parabolas

2)      Circles

3)      Ellipses

4)      Hyperbolas

 

Parabolas are the first of the conic sections. Here is a picture of a parabola with the important parts labeled. .

 

 

 

Focus

Vertex

Directrix

 

 

Axis of Symmetry

 

 

 

 

 

 

The vertex is the point that would be considered the center of the parabola. It will always lie on the axis of symmetry.

The focus is on the inside of the parabola and will also always lie on the axis of symmetry.

The axis of symmetry is a line that if the graph were folded along it would look exactly the same on both sides.

The directrix is a line behind the parabola that is the same distance to the vertex as the focus.

 

All parabolas can be written in the form y = a(x-h)² + k. This is true for all parabolas whether they open up or down. For parabolas that open left and right the general form is x = a(y-h)² + k. Notice that the difference that determines if a parabola opens up and down, or left and right is if the x term is squared (up and down) or the y term is squared (left and right). By first writing an equation of a parabola in this form it is easy to calculate the location of the 4 main parts; vertex, focus, axis of symmetry, and directrix.

Depending upon which term is squared, here is a chart that states how to find all the important parts of the parabola once it is in general form.

 

Form	y = a (x-h)2 + k	x = a (y-k)2 + h
Vertex	(h , k)	(h, k)
Focus	(h, k + )	(h + , k)
Axis of Symmetry	x = h	y = k
Directrix	y = k -	x = h -
Direction it opens	If a > 0 it opens up If a < 0 it open down	If a > 0 it opens right If a < 0 it opens left

 

 

All that needs to be done to solve a problem like this is to first put it in general form, identify if x or y is squared, and identify a, h, and k. From there, plug in to the chart to find each point on the graph.

 

Example

Find the vertex, focus, axis of symmetry, and directrix of y = 2 (x-3)² + 4

 

To find the vertex, first notice that this parabola is already in standard form with the x term being squared. That means to use the left column on the chart. Next identify the a, h, and k.

 

a=2, h=3, and k= 4

 

Use the fact that the vertex is at (h, k) to find the vertex.

Vertex (3,4)

 

Use (h, k + ) to find the focus.

Focus ( 3, 4 + )

 

Simplify.

Focus (3, )

 

Use x = h to find the axis of symmetry.

 

Axis of symmetry x = 3

 

y = k - is the equation to find the directrix. Plug in k and a.

 

Directrix y = 3 -

 

Simplify.

 

Directrix y =

 

Answer!

 

 

 

 

 

Example

 

Find the vertex, focus, axis of symmetry, and directrix of y = (x-1)² – 2 and graph.

 

The x term is squared so use the left side again. Find a, h, and k.

 

a = , h = 1, k = -2

 

Find the vertex using (h, k).

Vertex (1, -2)

 

Find the focus using (h, k + ).

Focus (1, -2 + )

 

Simplify.

Focus (1,-2 +1)

 

Keep simplifying.

Focus (1,-1)

 

Find the axis of symmetry with x = h.

Axis of symmetry x = 1

 

The equation for the directrix is y = k - .

Directrix y = -2 -

 

Simplify.

Directrix y = -2-1

 

Keep simplifying.

 

Directrix y = -3

 

To graph this parabola, first plot the vertex (1, -2).

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The next step would be to plot the focus.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

The axis of symmetry is x = 1, so graph it.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Graph y = -3 as the directrix.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

All the parts are now graphed except for the parabola itself. a > 0 since it is . That means the parabola opens up from the vertex. Now graph it. Here is the graph.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Answer!

 

 

 

 

Example

 

Find the vertex, focus, axis of symmetry, and directrix of x = -y² –6y- 2 and graph.

 

This parabola is obviously not in the general form needed to find out the necessary information. So the first step will be to get it in general form. Move the –2 to the right side by adding 2 to both sides.

 

x+2 = -y² –6y

 

Use the completing the square technique to this problem. In order to do that, make the y² term positive. To do this divide every term by –1.

 

-x-2 = y² +6y

 

Use the completing the square technique on the right side. Take half of 6, which is 3 and square it, which results in 9. Add this to both sides.

 

-x-2+9 = y² +6y +9

 

Now factor the right side and simplify the left.

 

-x + 7 = (y + 3)²

 

Move the 7 to the right side.

 

-x = (y + 3)² – 7

 

Divide both sides by –1 to make sure and get the x a positive so that it is in general form.

 

x = -(y + 3)² + 7

 

Now it is in general form. The next step is to find a, h, and k. The – in front of (y + 3)² means that a = -1. Also, keep in mind that in the parentheses there is a + sign, not a negative, this makes the h negative.

 

a = -1, h = 7, and k = -3

 

Before finding any of the important parts, it is important to realize that the y term is squared in this case. That means to use the right column of the chart to find all the necessary parts. Find the vertex with (h, k).

 

Vertex ( 7, -3)

Find the focus with (h + ,k).

Focus (7 + ,-3)

 

Simplify.

 

Focus (,7)

 

Find the axis of symmetry using y = k.

 

Axis of Symmetry y = -3

 

Find the directrix using x = h - .

Directrix x =

 

Now all the important parts have been found. One important thing to note is with the y term squared and a < 0 the parabola will open left. Here is a graph of this equation.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Example

Write the equation of the parabola with vertex (1,2) and focus (1,6).

Draw a quick sketch of the information already known.

 

 

Focus

 

Vertex

 

 

 

 

 

 

 

 

 

 

 

From looking at this graph, notice that the vertex is on bottom and therefore the parabola opens up. This means that a is positive and the parabola is of the general form y = a (x-h)² + k. With the vertex being (1,2) it is easy to know what h and k are.

 

h = 1, k= 2

 

The last piece of information needed is the a value. By knowing that the focus is (1,6) and k = 2, set up the following equation.

2 + =6

 

Solve for a.

 

a =

 

Substitute a = , h = 1, and k = 2 into the general form equation.

y=(x-1)² +2

 

Answer!

 

 

 

 

Example

Write the equation of the parabola with focus (0,2) and directrix x = 4, and graph.

 

Note that the directrix is x =, referring to the chart that means that the parabola is of the general form x = a (y-k)² + h. With it being in that form, the focus is (0,2) which makes k easy to find and set up an equation.

k=2

 

h + = 0

 

With the directrix being x = 4, write an equation with that information.

 

h - = 4

 

Use substitution on the last 2 equations. This will find h and a.

 

h= 2

 

a =

 

Put all the information into the general form equation.

x = (y-2)² + 2

 

Now graph the equation.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Answer!

 

 

 

 

Completing the square is an important concept in putting certain conic sections in the proper form. Some may teach it in the section of solving equations of degree two but it is really useful in the area of conics. Following is a quick lesson on how to complete the square in an equation.

 

The first thing to do with any equation is to look at it to see what can break it down. All equations with one variable and of degree two can be written in the form

ax² + bx = c

To complete the square first make the a, 1. This can be done by dividing everything by a. After that, the equation should be in the following form.

x² + bx = c

Take b and divide it by 2. Take the result of that and square it. Add this amount to both sides of the equation.

 

This may seem confusing without an example, but following is an example with numbers and a step-by-step explanation of the previous process.

 

 

 

 

Example

Complete the square and rewrite x² + 6x = 16

 

Notice that the a value is 1 so don’t divide by anything to get it in the proper form. 6 is the value for b so divide it by 2. That results in 3. Now square that value 3, which gets 9, and add it to both sides.

x² + 6x + 9 = 16 + 9

 

Rewrite the left side and simplify the right.

(x + 3)² = 25

 

Answer!