There are three conic sections that result when a cone is sliced by a plane a certain way.

1.

If the plane slices both sides of a cone, the conic section that is created is an ellipse. (The light blue area inside the cone is the ellipse.) If the slice is made perpendicular to the cone's axis, then a circle is formed. A circle is a special case of an ellipse.

2.

If the plane slices one side of the cone and is parallel to the other, the resulting conic section is a parabola. (The light green plane shows the outline of the parabola.)

3.

If the plane slices the cone parallel to the cone's axis, then the resulting conic section is a hyperbola. (The tan colored plane gives the outline of the hyperbola.)

There are also three degenerate conic sections.

1.

A point

It is the degenerate ellipse, and it occurs when the plane slices the cone through its vertex.

2.

A line

It is a degenerate parabola, and it occurs when the plane that slices the cone is tangent to the cone.

3.

A pair of intersecting lines

It is a degenerate hyperbola, and it occurs when the plane that slices the cone cuts along the cone's axis.

FACT:

A circle is the set of points in a plane whose distance from a given fixed point in the plane is constant. The fixed point is the center of the circle, and the constant distance is the radius.

FACT:

The equation of a circle with radius a, centered at the origin is x ² + y ² = a ².

FACT:

The equation of a circle with radius a, centered at the point (h, k) is (x - h) ² + (y - k) ² = a ².

EXAMPLE 1:

Find the center and radius of this circle: x ² + y ² - 4x + 6y = 10.

SOLUTION:

To find the center and the radius of this circle, I will have to complete the square.

FACT:

A set that consists of all points in a plane equidistant from a given fixed point and a given fixed line is a parabola. The fixed point is the focus of the parabola. The fixed line is the directrix. (see the figure below)

The blue dot above and inside the parabola is the focus, and the blue line below the parabola is the directrix.

FACT:

The vertex of a parabola lies halfway between the directrix and the focus.

STANDARD FORM EQUATIONS FOR PARABOLAS WITH VERTICES AT THE ORIGIN (p > 0)

EQUATION

FOCUS

DIRECTRIX

AXIS

DIRECTION IT OPENS

x ² = 4py

(0, p)

y = - p

y-axis

Up

x ² = - 4py

(0, - p)

y = p

y-axis

Down

y ² = 4px

(p, 0)

x = - p

x-axis

To the right

y ² = -4px

(- p, 0)

x = p

x-axis

To the left

EXAMPLE 2:

Given the equation of the parabola, find the focus and the directrix.

x ² = - 8y

SOLUTION:

This equation, x ² = - 8y, matches the equation x ² = - 4py, so let - 4p = - 8 and solve for p.

-4 p = - 8 ® p = 2

Therefore, the focus for this parabola is (0, -2) and y = 2.

I just used the related forms for the parabola of the form x ² = - 4py. Here is a picture of this parabola with the focus and directrix in blue.

Notice that the focus is in the inside of the parabola. If you ever have your focus outside of the parabola, then you have made a mistake.

EXAMPLE 3:

Given the equation of a parabola, find the focus and directrix.

x = 10y ²

SOLUTION:

First of all, I am going to solve for y ². Doing this will allow me to determine the form of this parabola.

Now, I will solve for p.

This equation fits the form y ² = 4px; therefore the focus and the directrix are the following.

Here is a picture of this parabola with the focus and directrix in blue.

Now, let us look at some shifted parabolas.

EXAMPLE 4:

The parabola y ² = -12x is shifted right 4 units and up 3 units to generate the parabola (y - 3) ² = - 12(x - 4). Find the new parabola's vertex, focus, and directrix.

SOLUTION:

Since the original parabola's vertex was at the origin, the new parabola's vertex is the translation (4, 3).

Since the original parabola was of the form y ² = -4px, -4p = -12, implies p = 3.

The focus for the original parabola would have had the form (-p, 0), so using the new vertex, I will subtract 3 from the x-coordinate to get the focus for the new parabola.

Focus: (4 - 3, 3) = (1, 3)

The directrix for the original parabola would have had the form x = p, so using the x-coordinate from the new vertex, I will add 3 to it and set it equal to x. This will be the directrix for the new parabola.

Directrix: x = 4 + 3 ® x = 7

Here is a picture of this parabola with the focus and directrix in blue.

EXAMPLE 5:

The parabola x ² = 6y is shifted right 3 units and down 2 units to generate the parabola (x - 3) ² = 6(y + 2). Find the new parabola's vertex, focus, and directrix.

SOLUTION:

Since the original parabola's vertex was at the origin, the new parabola's vertex is the translation (3, -2).

Since the original parabola was of the form x ² = 4py, then 4p = 6 and p = 3/2.

The focus for the original parabola has the form of (0, p), so using the vertex from the new parabola, I will add p to the y-coordinate.

The directrix for the original parabola has the form y = - p, so using the y-coordinate from the vertex, I will subtract p from it.

Here is a picture of this parabola with the focus and directrix in blue.

FACT:

An ellipse is the set of points in a plane whose distance from two fixed points in the plane has a constant sum. The two fixed points are the foci of the ellipse.

FACT:

The line through the foci is called the focal axis.

FACT:

The point halfway between the foci is the center of the ellipse.

FACT:

The points where the focal axis and the ellipse crosses are called the ellipse's vertices.

Here is a graph of an ellipse. The foci are indicated by blue dots.

STANDARD FORM EQUATIONS FOR ELLIPSES CENTERED AT THE ORIGIN

FOCI ON THE X-AXIS

CENTER-TO-FOCUS DISTANCE

FOCI

(± c, 0)

VERTICES

(± a, 0)

FOCI ON THE Y-AXIS

CENTER-TO-FOCUS DISTANCE

FOCI

(0, ± c)

VERTICES

(0, ± a)

EXAMPLE 6:

Find the center, foci, and vertices for 2x ² + y ² = 2.

SOLUTION:

I will first divide both sides by 2 to get the equation of the ellipse into standard form.

The center of this ellipse is the origin.

Now I determine the value for c.

Since the y term has the value of a ², the foci will be on the y-axis.

Foci: (0, ± 1)

Here is the graph of this ellipse. The foci are indicated in blue.

EXAMPLE 7:

Find the center, foci, and vertices for 4x ² + y ² + 8x - 2y = -1.

SOLUTION:

I will first complete the square to get the ellipse into standard form.

Since a circle is a special case of an ellipse, then the center of an ellipse can be determined just like the way we determine the center of a circle. Therefore, the center of this ellipse is (-1, 1).

For this ellipse a ² = 4 and b ² = 1, therefore the foci will lie on the line x = -1. Now I will determine c.

To find the vertices, I will solve the following equation for y.

The vertices for this ellipse are (-1, -1) and (-1, 3).

Here is the graph of this ellipse. The foci are indicated in blue.

Notice that this ellipse has its center at (-1, 1), so this is a translated ellipse. The ellipse that would have its center at the origin would have had the following form.

EXAMPLE 8:

The ellipse

is shifted to the left 4 units and down 5 units to generate the ellipse

Find the new ellipse's center, foci, and vertices.

SOLUTION:

To determine the center (h, k), read the coordinates off as if you are determining the center of a circle.

Center: (-4, -1)

Now, let us find the center-to-focus distance, c. Let a ² = 25. This implies that the focal axis will be y = -5.

Foci: (-4 ± 4, -1) or (-8, -1) and (0, -1)

To determine the vertices of this ellipse, I will solve the following equation for x.

Therefore, the vertices are (-9, -1) and (1, -1).

Remember that the vertices and foci must be on the focal axis.

Here is the graph of this ellipse. The foci are indicated in blue.

The green ellipse is the ellipse in standard position. The translated ellipse is in purple.

EXAMPLE 9:

Find the equation of the ellipse centered at the origin with foci at

(± 4, 0) and vertices at (± 5, 0).

SOLUTION:

I am given the value for c and the value for a, but I need to determine the value for b. To determine this value, I will use the center-to-focus formula.

a ² = 25 and b ² = 9. Using the fact that the foci and vertices are on the x-axis, I will now write the equation for this ellipse.

FACT:

A hyperbola is the set of points in a plane whose distances from two fixed points in the plane have constant difference. The two fixed points are the foci of the hyperbola.

FACT:

The line through the foci is called the focal axis.

Here is a graph of the hyperbola. Foci are indicated in blue, and asymptotes are indicated in green.

STANDARD FORM EQUATION FOR HYPERBOLAS CENTERED AT THE ORIGIN

FOCI ON THE X-AXIS

CENTER-TO-FOCUS DISTANCE

FOCI

(± c, 0)

VERTICES

(± a, 0)

ASYMPTOTES:

FOCI ON THE Y-AXIS

CENTER-TO-FOCUS DISTANCE

FOCI

(0, ± c)

VERTICES

(0, ± a)

ASYMPTOTES:

EXAMPLE 10:

Find the foci and asymptotes for the hyperbola 9x ² - 16y ² = 144.

SOLUTION:

First of all, I will put the hyperbola into standard position by dividing everything by 144.

With the x term being before the negative sign, that implies that the foci will be on the x-axis. Now I will determine the value of c.

The foci are (± 5, 0).

To determine the asymptotes for this hyperbola, I will solve the following equation for y in terms of x.

Here is the graph of this hyperbola. Foci are indicated in blue and asymptotes are indicated in green.

EXAMPLE 11:

Find the foci and asymptotes for the hyperbola 8y ² - 2x ² = 16.

SOLUTION:

First, I will put the hyperbola into standard position by dividing everything by 16.

With the y term being before the negative sign, that implies that the foci will by on the y-axis. Now to determine the value of c.

Now to determine the asymptotes for this hyperbola.

Here is the graph of this hyperbola. Foci are indicated in blue and the asymptotes in green.

EXAMPLE 12:

The hyperbola

is shifted 5 units to the left and 1 unit down to produce the hyperbola

Find the center, foci, and asymptotes for the new hyperbola.

SOLUTION:

Treating the equation like a circle, we can determine that the center of the hyperbola is (-5, -1). We can also determine that the focal axis is y = -1. Now to determine the foci.

Now to determine the asymptotes.

Here is the graph of this hyperbola. Foci are indicated in blue and asymptotes in green.

EXAMPLE 13:

Find the center, foci, and asymptotes for y ² - 4x ² + 16x = 24.

SOLUTION:

First, I will complete the square so I can put this hyperbola into standard form.

The center for this hyperbola is (2, 0) and the focal axis is x = 2. Now to determine the foci.

Now to determine the asymptotes.

Here is the graph of this hyperbola. Foci are indicated in blue and asymptotes in green.