equidistant from a point (the focus) and a line (the

Sec 6.1 – Conic Sections

Parabolas Name:

What is a parabola? It is geometrically defined by a set of points or locus of points that are

equidistant from a point (the focus) and a line (the directrix).

Parabolas can be found in many places in everyday life. A few examples are shown below. Can you guess where

the focus point should be in the flash light or the satellite dish?

Algebraically, parabolas are usually defined in two different forms: Standard Form and Vertex Form

Let’s start with the most basic graph of a parabola,

Fountains and Projectiles Flash Light Reflectors

Satellite Dishes

To see this idea visually try

drawing a straight line at the

bottom of a piece of paper with a ruler. Then, place a

point in the middle just

above the line as shown. Next, fold the paper multiple

times in various locations so

that the line folds on top of the point and make a crease.

The creases will outline a parabola. Why does this happen?

M. Winking (Section 6-1) ©

p.99

So, where would the focus point and the directrix be in the basic

equation of y = x2. This is not a trivial task. We know that the

directrix should be somewhere below the vertex and the distance

from the vertex to the line should be the same as the distance from

the vertex to the focus point.

Consider putting an arbitrary point on the y – axis above the vertex

and call it the focus point at (0, p). We will need to figure out what

p is to find the location of the focus. Then, we know that the

directrix is the same distance away from the vertex but on the other

side of the parabola and it is a line. So, it the directrix would need

to be the line at y = – p.

The parabola is geometrically defined as the set of point equidistant

from a point and a line. So, we already know algebraically the

parabola is given by the equation y = x2 which would suggest every

point on the parabola is basically of the form (x, x2) (eg. (1,1), (2, 4),

(3,9), etc.). We described the point of the focus point as (0,p) and then, if you think about it carefully the point

directly below the general point of the parabola shown in the diagram on the directrix would have to be the point

(x,–p). Now, we can just use the distance formula to say that AF = AB.

( ) ( ) ( ) ( )2 2 2 2

A F A F A B A Bx x y y x x y y− + − = − + −

( ) ( ) ( ) ( )222 22 20x x p x x x p−− + − = − + −

We can square both sides:

( ) ( ) ( ) ( )222 22 20x x p x x x p−− + − = − + −

Clean things up a bit:

( ) ( )2 2

2 2 2 20x x p x p+ − = + +

Expand (use F.O.I.L. if necessary): 2 4 2 2 4 2 22 2x x x p p x x p p+ − + = + +

Cross out items on both sides (subtract from both sides) 2 4x x+ 2 22x p p− + 4x= 2 22x p p+ +

Move the 2x2p to the right: 2 2 2

2 2

2 2

2 2

2 2

4

x x p x p

x p x p

x x p

− =

+ +

=

Solve for p 2 2

2 2

4

4 4

x x p

x x=

Which reduces

1

4p=

So the focus is located at 1

0,4

and the directrix is located at 1

4y = −


p.100

y p= −

(x,x2)

(x,-p)

F

(0,p)

A

B

It turns out doing nearly the same thing for the vertex form of the parabola,

( ) ( )2

y k a x h− = − OR ( ) ( )2

x h a y k− = −

we can show in a similar manner that the focus is located ‘inside the mouth’ of the parabola a vertical distance

of 1

4a from the vertex whereas the directrix is located the same distance away from the vertex on the other

side of the parabola from the focus.

1. Sketch a graph the following Parabolas (Label the Vertex, Focus, and Directrix)

a. ( ) ( )2

2 1 3y x− = + b. ( ) ( )21

1 48

y x+ = −

c. ( ) ( )2

4 2 2y x− = − − d. ( ) ( )21

3 24

x y+ = −


p.101

e. ( ) ( )21

4 212

x y+ = − − f. 2 8y x=

2. Find each of the requested components of a parabola.

a. Given the vertex of a parabola is located at (2,2)

and has a directrix of y = 4, find the location of

the focus.

b. Given the vertex of a parabola is located at (–

2,1) and has a focus at (–1,1), find the equation

of the directrix.

c. Given the directrix of a parabola is y = – 3 and a

focus located at (4,1), find the location of the

vertex. Which way does the parabola open?

d. Find the vertex of the parabola defined by 2 6 2y x x= − +


p.102

3. Find each of the requested components of a parabola.

a. Given the vertex of a parabola is located at

(-2, -3) and has a directrix of y = -6, find the

location of the focus and the equation of

the parabola in standard form.

b. Given the focus of a parabola is located at

(1.5,1) and has a directrix at x = 2.5, find the

coordinates of the vertex and the equation of

the parabola in standard form..

c. Find the vertex, directrix, and focus of the

following parabola defined by:

xxy 41 2 +−=+

d. Find the vertex, directrix, and focus of the

following parabola defined by:

142 2 +−= yyx


p.103

1. Sec 6.2 – Conic Sections Circles Name:

What is a circle? It is geometrically defined

by a set of points or locus of points that are equidistant

from a point (the center). Consider the circle at the

right. What is the length of every segment drawn from

center O to a point on the edge of the circle?

How would you find the length of segment OB?

1. Basic Circles Graph the following:

A. 422 =+ yx B. 3622 =+ yx C. 2022 =+ yx

2. Translated Circles Graph the following:

A. ( ) ( ) 92422=++− yx B. ( ) ( ) 2513

22=+++ yx C. ( ) ( ) 1832

22=−++ yx

M. Winking © Unit 6-2 page 104

3. Equations of Circles Find the equation of each of the following:

a. ( ) ( ) __________________22=+ yx b. ( ) ( ) __________________

22=+ yx

4. Equations of Circles The following design is composed of 3 full circles and 2 semi-circles. Can you find the

equations of each and put them in your calculator?

a. ( ) ( ) __________________22=+ yx

b. ( ) ( ) __________________22=+ yx

c. ( ) ( ) __________________22=+ yx

d. ( ) ( ) __________________22=+ yx

e. ( ) ( ) __________________22=+ yx

** When you put these in your TI-83/84 calculator you will have to solve for y using the square root method

you may have to use two equations to describe a complete circle. For example, if you wanted to graph the

complete circle ( ) ( ) 92422=++− yx . It would require that you use two equations ( ) 249

2

1 −−−= xy and

( ) 2492

2 −−−−= xy

M. Winking © Unit 6 - 2 page 105

5. Finding Standard form of circles. Put the following circles in standard form and graph them.

A. 091022 =+−+ yyx B. 88222 −=+−+ yxyx

C. 152622 =−−+ xyyx D. 612822 22 =+−+ yxyx


6. Find the Equations of the Following Circles in Standard form.

A. If the endpoints of a diameter of a

circle are A:(– 4, – 2) and B:(– 2, – 2),

what is the equation of the circle.

B. If the endpoints of a diameter of a

circle are A:(– 3, – 2) and B:(7, 4),

what is the equation of the circle.

C. If the endpoints of a radius of a circle are

A:(2,3) and B:(6,1) with the center at point

A, what is the equation of the circle.

( ) ( )2 2

______ _____ ______x y+ = ( ) ( )2 2

______ _____ ______x y+ = ( ) ( )2 2

______ _____ ______x y+ =

7. Applications of circles.

A. A new company E-Mobile is setting up Ad based free Wi-Fi Access

Points. Each mobile tower that E-Mobile creates covers the area of a

circle with a radius of 4 miles with the tower at the center. Sketch the

following circles that show on the map the coverage areas.

• ( ) ( )2 2

7 4 16x y+ + − = • ( ) ( )2 2

1 12 16x y− + − =

• ( ) ( )2 2

6 8 16x y+ + + = • ( )2 22 16x y− + =

B. A Bus travels the entire length of U.S. 29 and stops at the bust stops

located at A:(– 8, – 6), B:(– 4, – 3), C:(4, 1), D:(8, 3). Which bus

stops have access to the free Wi-Fi Access by E-mobile?

C. A company will be building an Urgent Care facility in

Pickens County. Ideally, the company would like to be

within 6 miles reach of the most citizens. Create an

equation of a circle that encapsulates as many cities as

possible using a 6-mile radius and the center being

where the Urgent Care facility should be created.


A B A

B A

B

4 miles

Gwinnett County, GA

4 miles

( ) ( )2 2

______ _____ ______x y+ =

1. Sec 6.3 – Conic Sections Ellipses Name:

An ELLIPSE could be accurately described as circle that has been

stretched or compressed by a constant ratio towards a diameter of a

circle. A circle is actually a special type of ellipse. More explicitly, an

ELLIPSE is the locus of points whose distance from two focal points is

constant. To better understand what this means consider the following:

Paste or tape a coordinate grid to a piece of cardboard. Put two push pins in

the grid at (– 4, 0) and (4, 0). These pins will represent the focal points. Next

tie a string to each push pin such that the length of the string between the two

pins is approximately 10 units long (using the unit length of the coordinate

grid). Finally, take a pencil and stretch out the string until the string is

straightened and trace out all of the places the string allows the pencil to

move with the string remaining taut. This will ensure that the combined

distance from each focal point will always be a total of 10 units long.

Rather than always having a fixed diameter an oblong ellipse can be described as having a major axis (the longest diameter)

and a minor axis (the shortest diameter). The endpoints of the major axis are considered the vertices of an ellipse and the

endpoints of the minor axis are consider the co-vertices.

1. Determine the length of the major & minor axis. List the coordinates of vertices and co-vertices of the following ellipses.

A. B.

Major Axis Length:

Minor Axis Length:

Vertices:

Co-Vertices:

Major Axis Length:

Minor Axis Length:

Vertices:

Co-Vertices:


• •

•

Rather than always having a fixed radius an oblong ellipse can be described as having a major radius/semi-major axis(the

longest radius) and a minor radius/semi-minor axis (the shortest radius). The major radius is commonly denoted by the

variable ‘a ’ and the minor radius is commonly denoted by the variable ‘b’. These different radii are used to write the

equation of the ellipse in standard form.

2. Determine the equations of the following ellipses in standard form (list the vertices and co-vertices).

A. B. C.

•

Standard Form of an ELLIPSE centered at the origin.

𝒙𝟐

𝒂𝟐+

𝒚𝟐

𝒃𝟐= 𝟏 OR

𝒙𝟐

𝒃𝟐+

𝒚𝟐

𝒂𝟐= 𝟏

** ‘a’ must always be the largest radius **

The Equation of the ELLIPSE shown at the left.

𝒙𝟐

𝟓𝟐+

𝒚𝟐

𝟑𝟐= 𝟏 OR

𝒙𝟐

𝟐𝟓+

𝒚𝟐

𝟗= 𝟏

Standard Form of an ELLIPSE

(𝒙−𝒉)𝟐

𝒂𝟐+

(𝒚−𝒌)𝟐

𝒃𝟐= 𝟏 OR

(𝒙−𝒉)𝟐

𝒃𝟐+

(𝒚−𝒌)𝟐

𝒂𝟐= 𝟏

** ‘a’ must always be the largest radius **

The Equation of the ELLIPSE shown at the left.

(𝒙−(−𝟑))𝟐

𝟐𝟐+

(𝒚−𝟏)𝟐

𝟑𝟐= 𝟏 OR

(𝒙+𝟑)𝟐

𝟒+

(𝒚−𝟏)𝟐

𝟗= 𝟏

•

3. Graph the following equations and list the center and radius of each ellipse.

A. 𝒙𝟐

𝟐𝟓+

𝒚𝟐

𝟒= 𝟏 B.

(𝒙−𝟐)𝟐

𝟗+

(𝒚+𝟑)𝟐

𝟏= 𝟏 C. 𝟒(𝒙 − 𝟏)𝟐 + 𝟗(𝒚 − 𝟐)𝟐 = 𝟑𝟔

4. Complete the square to put each of the following ellipses in standard form.

Then, list the center, vertices, and co-vertices. Finally, graph each ellipse.

𝟏𝟔𝒙𝟐 + 𝟒𝒚𝟐 − 𝟑𝟐𝒙 = 𝟒𝟖

5. Complete the square to put each of the following ellipses in standard form.

Then, list the center, vertices, and co-vertices. Finally, graph each ellipse.

𝟗𝒙𝟐 + 𝒚𝟐 + 𝟓𝟒𝒙 − 𝟒𝒚 = −𝟕𝟔


Finding the focal points algebraically, requires the use of the Pythagorean Theorem. First, consider the constant distance

found between any point on the ellipse and the two focal points is equal to the length of the major axis.

In the above example, you can see that m + n = 10 which is also the length of the major axis. Now, if we position the

‘string’ at one of the co-vertices then a right triangle is formed which is shown below. The hypotenuse is half the length

of the major axis which would equal to the length of the major radius. One of the legs of the right triangle is the minor

radius and the other leg is represented by the distance from the center of the ellipse to a focal point. It may seem a little

out of order but you can see based on the defined variables we could state the following.

6. Find the focal points of each ellipse shown below.

A. B.

The focal points are always located on the major axis and

are a distance of ‘c ’ units away from the center of the

ellipse where ‘c ’ is defined by the equation:

𝒃𝟐 + 𝒄𝟐 = 𝒂𝟐 OR 𝒂𝟐 − 𝒃𝟐 = 𝒄𝟐

** ‘a’ is the length of the major radius and ‘b’ the minor radius **

These equations correspond to the ellipse shown at the left.

𝟑𝟐 + 𝟒𝟐 = 𝟓𝟐 OR 𝟓𝟐 − 𝟑𝟐 = 𝟒𝟐


Eccentricity of an Ellipse is a measure of how close the ellipse is to being a circle and is given by the formula:

𝑬𝒄𝒄𝒆𝒏𝒕𝒓𝒊𝒄𝒊𝒕𝒚 = 𝒄

𝒂 , where ‘c’ is the distance from the center to a focal point and ‘a’ is the length of the major radius.

Sample Ellipses Eccentricity = 0 Eccentricity ≈ 0.75 Eccentricity ≈ 0.94 Eccentricity ≈ 0.99

7. Find the eccentricity of each of the following ellipses also used in problem number 6.

A. B.

8. Find the equations of ellipse given the following parameters and sketch a graph.

A. Find the equation and graph of an ellipse that

has vertices at (-2, 5) and (-2, -3) and co-vertices

at (-4,1) and (0,1).

B. Find the equation and graph of an ellipse that has

vertices at (-6, 1) and (4, 1) and focal points at (-5,1)

and (3,1).


9. Consider the following describes the orbit of Halley’s Comet. It orbits the sun in a highly elliptical orbit with the

sun being one of the focal points of the ellipse. Determine the following given that a = 17.834 astronomical units

and b = 4.534 astronomical units.

A. Determine an equation that describes the orbit of Halley’s comet.

B. Find the position of the Sun on the coordinate grid.

C. What is the Eccentricity of the orbit?

10. In a Introduction to Theater Course, a student

noticed that the spotlight actually cast a

spotlight on the stage floor in the shape of an

ellipse. The student laid out a coordinate grid

on the floor. The units are in feet.

A. What is the equation of the ellipse created

by the spotlight on the coordinate grid?

B. What are the coordinates of the focal points?

C. What is the eccentricity of the ellipse?

Halley’s Comet


1. Sec 6.4 – Conic Sections Hyperbolas Name:

A HYPERBOLA is the locus of points whose difference in distance from two focal points is constant. To better understand

what this means consider the following example of the graph of 𝒙𝟐

𝟏𝟔−

𝒚𝟐

𝟗= 𝟏 with focal points at (−𝟓, 𝟎) and (𝟓, 𝟎).

Example 1 Example 2

|d1 – d2| = difference in distance from the focal points

|10.25 – 2.25| = 8

|d1 – d2| = difference in distance from the focal points

|11.57 – 3.57| = 8

Notice, that the distance remains a constant 8 units which is

also equal to the distance between the two vertices or 2a. As

shown the graph at the right . The distance from the left focal

point to the vertex on the right is 9 units. The distance from

the right focal point to the vertex on the right is 1 unit. So, the

difference is again 8 units.

|9.00 – 1.00| = 8

Next let’s look at the graph shown to the left of

with some descriptions of points and axes.

𝒙𝟐

𝟏𝟔−

𝒚𝟐

𝟗= 𝟏

P:(6, 3.4)

P:(5, 2.25)

P:(4, 0)


Determine the equations of the following Hyperbolas in standard form (list the vertices, co-vertices, and the equations of the asymptotes).

1. 2. 3.

Graph the following equations and list the center and radius of each hyperbola (label the vertices and co-vertices)

4. 𝒚𝟐

𝟏−

𝒙𝟐

𝟗= 𝟏 5.

(𝒙+𝟐)𝟐

𝟒−

(𝒚−𝟏)𝟐

𝟒= 𝟏 6.

(𝒚−𝟐)𝟐

𝟏−

𝒙𝟐

𝟗= 𝟏


7. Complete the square to put the following hyperbola in standard form.

Then, list the center, vertices, and co-vertices. Finally, graph the hyperbola.

𝟒𝒙𝟐 − 𝟗𝒚𝟐 − 𝟏𝟔𝒙 = 𝟐𝟎

8. Complete the square to put the following hyperbola in standard form.

Then, list the center, vertices, and co-vertices. Finally, graph the hyperbola.

𝒚𝟐 − 𝟗𝒙𝟐 + 𝟏𝟖𝒙 − 𝟒𝒚 = 𝟏𝟒


Given a HYPERBOLA of the form 𝒙𝟐

𝒂𝟐 −𝒚𝟐

𝒃𝟐 = 𝟏 the focal points are located a distance of ‘c’ units away from the center on

the transverse or major axis (i.e. in this example at (−𝒄, 𝟎) and (𝒄, 𝟎) given the relationship 𝒂𝟐 + 𝒃𝟐 = 𝒄𝟐. The proof of

this claim is an optional exploration for this course.

Consider the following diagram. Use the distance formula

to describe d1 and d2.

d1 =

d2 =

Next, substitute those descriptions in to the statement below and simplify a lot. We discussed earlier that the constant difference in distance is always equal to ‘2a’.

d1 – d2 = 2a


Given a HYPERBOLA of the form 𝒙𝟐

𝒂𝟐 −𝒚𝟐

𝒃𝟐 = 𝟏 the focal points are located a distance of ‘c’ units away from the center on

the transverse or major axis (i.e. in this example at (−𝒄, 𝟎) and (𝒄, 𝟎) given the relationship 𝒂𝟐 + 𝒃𝟐 = 𝒄𝟐.

Using this relationship find the focus points for the following hyperbolas.

9. 10. 11. 𝒙𝟐

𝟗−

𝒚𝟐

𝟏= 𝟏

12. Find the equation of a hyperbola with a focal

points at (4, 0) and (-4,0) and vertices at (3,0)

and (-3, 0).

13. Some lamps cast light on the wall in the shape

of a hyperbola. Determine the equation of

that describes the outline of the light.

● ● ● ● ●


1. Sec 6.5 – Conic Sections Systems Name:

Find all of the solutions to the following conic systems of equations by graphing and verify the solutions.

1. 𝒙𝟐 + 𝒚𝟐 = 𝟐𝟓 2. 𝒙𝟐 + 𝒚𝟐 = 𝟐𝟓 3. 𝒙𝟐

𝟔𝟒+

𝒚𝟐

𝟏𝟐= 𝟏

−𝒙𝟐 + 𝟓 = 𝒚 (𝒙−𝟑)𝟐

𝟗+

𝒚𝟐

𝟏𝟔= 𝟏

𝒙𝟐

𝟒−

𝒚𝟐

𝟑= 𝟏

Find the solutions to the nearest hundredth of the systems by graphing with your calculator and sketch a graph

4. 𝒚 = 𝒙𝟐 − 𝟐 and 𝒙𝟐 + 𝒚𝟐 = 𝟗

M. Winking © Unit 6-5 pg. 119

Find the solutions to the nearest hundredth of the systems by graphing with your calculator and sketch a graph

5. 𝒙𝟐

𝟒+

(𝒚−𝟏)𝟐

𝟐𝟓= 𝟏 and

𝒚𝟐

𝟏𝟔−

𝒙𝟐

𝟒= 𝟏

Find the exact solutions to the systems of conic equations by substitution.

6. 𝒚 = 𝟐𝒙 + 𝟏 and 𝒙𝟐 + 𝒚𝟐 = 𝟏𝟎

7. 𝒙𝟐 + 𝒚𝟐 = 𝟐𝟓 and 𝟒𝒙𝟐 + 𝟑𝟔𝒚𝟐 = 𝟏𝟒𝟒


Find the exact solutions to the systems of conic equations by substitution.

8. 𝒙 + 𝟏 = 𝒚𝟐 and 𝒙𝟐 + 𝒚𝟐 = 𝟕

Find the exact solutions to the systems of conic equations by elimination.

9. 𝟒𝒙𝟐 + 𝟗𝒚𝟐 = 𝟑𝟔

𝟒𝒙𝟐 − 𝟒𝒚𝟐 = 𝟏𝟔

10. 𝒙𝟐 + 𝒚𝟐 = 𝟐𝟓

−𝟐𝒙𝟐 + 𝒚 = 𝟓


Find the exact solutions to the systems of conic equations by elimination.

11. 𝟔𝒙𝟐 − 𝟖𝒚𝟐 = 𝟒𝟖

6𝒚𝟐 − 𝟐𝒙𝟐 = 𝟏𝟐

12. The comet 41P and Mars’ Orbits come in very close proximity at times. Luckily, their orbits are not in the same plane as shown at the right. The two planes in which each other orbit just miss one another. However, if we look at a top down or orthogonal view of the two orbits we can find where their orthogonal projections intersect and when they could potentially be at one of their closest points to each other if Mars and 41P’s timing is just right.

Consider laying out a coordinate grid that goes through the plane in which the orbit of Mars exists. Let’s use Astronomical Units (AU) for measurements which represents the mean distance from the Earth to the Sun or roughly 150 million Km

Let the Sun be located at roughly (2, 0) which would be a focal point of the comet’s orbital projection

41P’s orbit could be roughly described by 𝟓. 𝟒𝒙𝟐 + 𝟗. 𝟓𝒚𝟐 = 𝟓𝟏. 𝟑

Mars’ orbit could be roughly described by (𝒙 − 𝟐)𝟐 + 𝒚𝟐 = 𝟐. 𝟑

At what coordinate points would the two orbits potentially almost intersect?

13. Draw a sketch of graph where an ellipse and a hyperbola intersect the requested number of times or put not possible.

a. 0 intersections b. 1 intersection c. 2 intersections d. 3 intersections e. 4 intersections


equidistant from a point (the focus) and a line (the

Documents