### Conic Sections

- Know how to put an equation in standard form by completing the square. Be able to identify which conic section is the graph of the equation.
- Know how to analyze a general equation of degree 2 to determine which conic section is the graph of the equation.

### Conic Sections

x

^{2}− 2x + y + 6 = 0

- Since only one of the variables is raised to the second power, this is a Parabola
- Isolate the x terms to one side of the equation.
- y + 6 = − x
^{2}+ 2x - Factor a negative in order to complete the square
- y + 6 = − (x
^{2}− 2x) - y + 6 − [(b
^{2})/4] = − (x^{2}− 2x + [(b^{2})/4]) - y + 6 − [(( − 2)
^{2})/4] = − (x^{2}− 2x + [(( − 2)^{2})/4]) - y + 6 − [4/4] = − (x
^{2}− 2x + [4/4]) - y + 6 − 1 = − (x
^{2}− 2x + 1) - y + 5 = − (x − 1)
^{2}

^{2}− 5

x + 2y

^{2}+ 20y + 48 = 0

- Since only one of the variables is raised to the second power, this is a Parabola
- Isolate the y terms to one side of the equation.
- x + 48 = − 2y
^{2}− 20y - Factor a − 2 in order to complete the square
- x + 48 = − 2(y
^{2}+ 10y) - x + 48 + − 2([(b
^{2})/4]) = − 2(y^{2}+ 10y + [(b^{2})/4]) - x + 48 + − 2([(10
^{2})/4]) = − 2(y^{2}+ 10y + [(10^{2})/4]) - x + 48 + − 2([100/4]) = − 2(y
^{2}+ 10y + [100/4]) - x + 48 + − 2(25) = − 2(y
^{2}+ 10y + 25) - x − 2 = − 2(y + 5)
^{2}

^{2}+ 2

x

^{2}+ y

^{2}− 6x − 2y + 6 = 0

- Since both x and y are raised to the second power, complete the square twice to see which conic
- section this belongs to
- Group the x and y terms together, move the constant to the other side of the equation
- (x
^{2}− 6x) + (y^{2}− 2y) = − 6 - Complete the square for both the x and y variables
- (x
^{2}− 6x + [(b^{2})/4]) + (y^{2}− 2y + [(b^{2})/4]) = − 6 + [(b^{2})/4] + [(b^{2})/4] - (x
^{2}− 6x + [(( − 6)^{2})/4]) + (y^{2}− 2y + [(( − 2)^{2})/4]) = − 6 + [(( − 6)^{2})/4] + [(( − 2)^{2})/4] - (x
^{2}− 6x + [36/4]) + (y^{2}− 2y + [4/4]) = − 6 + [36/4] + [4/4] - (x
^{2}− 6x + 9) + (y^{2}− 2y + 1) = 4 - (x − 3)
^{2}+ (y − 1)^{2}= 4

x

^{2}+ y

^{2}+ 6x − 8y + 21 = 0

- Since both x and y are raised to the second power, complete the square twice to see which conic
- section this belongs to
- Group the x and y terms together, move the constant to the other side of the equation
- (x
^{2}+ 6x) + (y^{2}− 8y) = − 21 - Complete the square for both the x and y variables
- (x
^{2}+ 6x + [(b^{2})/4]) + (y^{2}− 8y + [(b^{2})/4]) = − 21 + [(b^{2})/4] + [(b^{2})/4] - (x
^{2}+ 6x + [((6)^{2})/4]) + (y^{2}− 8y + [(( − 8)^{2})/4]) = − 21 + [((6)^{2})/4] + [(( − 8)^{2})/4] - (x
^{2}+ 6x + [36/4]) + (y^{2}− 8y + [64/4]) = − 21 + [36/4] + [64/4] - (x
^{2}+ 6x + 9) + (y^{2}− 8y + 16) = − 21 + 9 + 16 - (x + 3)
^{2}+ (y − 4)^{2}= 4

2x

^{2}+ 2y

^{2}+ 14x − 14y + 31 = 0

- Since both x and y are raised to the second power, complete the square twice to see which conic
- section this belongs to
- Group the x and y terms together, move the constant to the other side of the equation
- (2x
^{2}+ 14x) + (2y^{2}− 14y) = − 31 - Complete the square for both the x and y variables, factor out a 2 from both x an y first
- 2(x
^{2}+ 7x + [(b^{2})/4]) + 2(y^{2}− 7y + [(b^{2})/4]) = − 31 + [(b^{2})/4] + [(b^{2})/4] - 2(x
^{2}+ 7x + [(7^{2})/4]) + 2(y^{2}− 7y + [( − 7^{2})/4]) = − 31 + 2( [(b^{2})/4] ) + 2( [(b^{2})/4] ) - 2(x
^{2}+ 7x + [49/4]) + 2(y^{2}− 7y + [49/4]) = − 31 + 2( [49/4] ) + 2( [49/4] ) - 2(x
^{2}+ 7x + [49/4]) + 2(y^{2}− 7y + [49/4]) = 18 - 2(x + [7/2])
^{2}+ 2(y − [7/2])^{2}= 18 - [(2(x + [7/2])
^{2})/2] + [(2(y − [7/2])^{2})/2] = [18/2] - (x + [7/2])
^{2}+ (y − [7/2])^{2}= 9

x

^{2}− 2y

^{2}− 20 = 0

- Move the constant to the other side of the equation
- x
^{2}− 2y^{2}= 20 - Divide both sides by 20

^{2})/20] − [(2y

^{2})/20] = [20/20]

^{2})/20] − [(y

^{2})/10] = 1

− x

^{2}+ y

^{2}+ 4y + 3 = 0

- Move the constant to the other side of the equation, group the variables
- (y
^{2}+ 4y) − x^{2}= − 3 - Complete the square for y
- (y
^{2}+ 4y + [(b^{2})/4]) − x^{2}= − 3 + [(b^{2})/4] - (y
^{2}+ 4y + [(4^{2})/4]) − x^{2}= − 3 + [(4^{2})/4] - (y
^{2}+ 4y + [16/4]) − x^{2}= − 3 + [16/4] - (y
^{2}+ 4y + 4) − x^{2}= 1 - (y + 2)
^{2}− x^{2}= 1

16x

^{2}+ 9y

^{2}+ 128x − 36y + 148 = 0

- Move the constant to the other side of the equation, group the variables
- (16x
^{2}+ 128x) + (9y^{2}− 36y) = − 148 - Complete the square for x and y, factor out the coefficient first
- 16(x
^{2}+ 8x) + 9(y^{2}− 4y) = − 148 - 16(x
^{2}+ 8x + [(b^{2})/4]) + 9(y^{2}− 4y + [(b^{2})/4]) = − 148 + 16( [(b^{2})/4] ) + 9( [(b^{2})/4] ) - 16(x
^{2}+ 8x + [(8^{2})/4]) + 9(y^{2}− 4y + [(( − 4)^{2})/4]) = − 148 + 16( [(8^{2})/4] ) + 9( [(( − 4)^{2})/4] ) - 16(x
^{2}+ 8x + [64/4]) + 9(y^{2}− 4y + [16/4]) = − 148 + 16( [64/4] ) + 9( [16/4] ) - 16(x
^{2}+ 8x + 16) + 9(y^{2}− 4y + 4) = − 148 + 16( 16 ) + 9( 4 ) - 16(x + 4)
^{2}+ 9(y − 2)^{2}= 144 - Divide both sides by 144
- [(16(x + 4)
^{2})/144] + [(9(y − 2)^{2})/144] = [144/144] - This conic section is an ellipse

^{2})/9] + [((y − 2)

^{2})/16] = 1

9x

^{2}+ 25y

^{2}+ 36x + 50y − 164 = 0

- Move the constant to the other side of the equation, group the variables
- (9x
^{2}+ 36x) + (25y^{2}+ 50y) = 164 - Complete the square for x and y, factor out the coefficient first
- 9(x
^{2}+ 4x) + 25(y^{2}+ 2y) = 164 - 9(x
^{2}+ 4x + ( [(b^{2})/4] )) + 25(y^{2}+ 2y + ( [(b^{2})/4] )) = 164 + 9( [(b^{2})/4] ) + 25( [(b^{2})/4] ) - 9(x
^{2}+ 4x + ( [(4^{2})/4] )) + 25(y^{2}+ 2y + ( [(2^{2})/4] )) = 164 + 9( [(4^{2})/4] ) + 25( [(2^{2})/4] ) - 9(x
^{2}+ 4x + ( [16/4] )) + 25(y^{2}+ 2y + ( [4/4] )) = 164 + 9( [16/4] ) + 25( [4/4] ) - 9(x
^{2}+ 4x + 4) + 25(y^{2}+ 2y + 1) = 164 + 9( 4 ) + 25( 1 ) - 9(x + 2)
^{2}+ 25(y + 1)^{2}= 225 - Divide both sides by 225
- [(9(x + 2)
^{2})/225] + [(25(y + 1)^{2})/225] = [225/225] - This conic section is an ellipse

^{2})/25] + [((y + 1)

^{2})/9] = 1

x

^{2}+ 10x + y + 29 = 0

- Move the constant to the other side of the equation, group the variables
- y + 29 = − x
^{2}− 10x - Complete the square for x , factor out the coefficient first
- y + 29 = − (x
^{2}+ 10x) - y + 29 − ( [(b
^{2})/4] ) = − (x^{2}+ 10x + ( [(b^{2})/4] )) - y + 29 − ( [(10
^{2})/4] ) = − (x^{2}+ 10x + ( [(10^{2})/4] )) - y + 29 − ( [100/4] ) = − (x
^{2}+ 10x + ( [100/4] )) - y + 29 − ( 25 ) = − (x
^{2}+ 10x + ( 25 )) - y + 4 = − (x + 5)
^{2} - Subtract 4 from both sides
- y = − (x + 5)
^{2}− 4 - This conic section is a parabola

^{2}− 4

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Conic Sections

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Conic Sections 0:16
- Double Cone Sections
- Standard Form 1:27
- General Form
- Identify Conic Sections 2:16
- B = 0
- X and Y
- Identify Conic Sections, Cont. 4:46
- Parabola
- Circle
- Ellipse
- Hyperbola
- Example 1: Identify Conic Section 8:01
- Example 2: Identify Conic Section 11:03
- Example 3: Identify Conic Section 11:38
- Example 4: Identify Conic Section 14:50

### Algebra 2

### Transcription: Conic Sections

*Welcome to Educator.com.*0000

*In the past four lectures, we have discussed various conic sections: parabolas, circles, ellipses, and hyperbolas.*0003

*And this lecture is designed to bring that information together and to give you some context about this.*0010

*First of all, what are conic sections? We know we can name them; we know what they are; but where do they come from?*0016

*Well, they are literally sections of a cone: when you take a double cone (it is actually a double cone, as follows, with the points together),*0023

*and you section them (sectioning is slicing)--when you take slices of them, using a plane, you come up with these four types of curves.*0034

*So, as you can see, when you take a plane and section, or slice, the cone across, you are going to end up with a circle.*0043

*If you tip that plane at an angle, the result is an ellipse.*0054

*If you encompass the edge of the come, you end up with a parabola.*0065

*And if you slice through in such a way that you capture the edges of both cones, then you end up with a hyperbola; and there you can see the two branches.*0071

*So, this is where conic sections come from; and they have many applications in science.*0079

*We have talked about the standard form of each conic section (for example, the standard form of a circle, or the standard form of an ellipse).*0087

*This standard form that I am talking about now is a very general form.*0095

*It gives you a general equation, ax ^{2} + bxy + cy^{2} + dx + ey + f = 0.*0098

*So, what we are going to do in a minute is talk about how you can look at this general form and determine*0109

*which type of conic section you are working with, so that you can put the equation*0115

*in the standard form particular to that type of conic section.*0119

*And as we have been going through, I have mentioned some ways that you can tell, if you just have an equation in the general form,*0124

*what type of conic section you are working with; and now I am going to bring that all together.*0131

*OK, if b = 0, we can analyze that standard form of the conic section to determine what type of conic section the equation represents.*0137

*Looking back at that general standard form again, ax ^{2} + bxy + cy^{2} + dx + ey + f = 0.*0146

*Here, we are having the limitation that b = 0.*0168

*And throughout this course, when we work with conic sections, we have only worked with ones where b is 0.*0171

*When b is 0, you end up with this.*0178

*Once you have this standard form, then you can go ahead and analyze it in ways we are going to discuss in a minute*0192

*to determine which type of conic section you have (what the equation describes).*0197

*But let's talk for a minute about what the xy tells you.*0201

*So far, we have worked with shapes such as parabolas; and some were oriented vertically; some, horizontally.*0204

*We also worked with ellipses (some had a horizontal major axis; some had a vertical major axis); and the same with hyperbolas.*0215

*So, even though the center may have been shifted, these were all either strictly vertical or strictly horizontal.*0230

*What this bxy term does is rotates it so that instead of, say, having an ellipse that has*0237

*a completely vertical major axis or horizontal major axis, you could end up with an ellipse like this--*0243

*the major axis is at an angle--or a parabola that is like this.*0250

*And that is definitely more complicated to work with; and it doesn't allow us to complete the square, then,*0259

*to shift an equation from the general form to a specific standard form.*0265

*So, later on, if you continue on in math, you may end up working with these shapes.*0270

*But for this course, we are limiting it to conic sections that are either vertical or horizontal; but they are not tipped at any other type of angle.*0274

*In order to identify conic sections, you need to look at the coefficients of the x ^{2} and y^{2} terms.*0286

*So, let's rewrite this; and again, the assumption is that b = 0.*0293

*So, I am just going to have ax ^{2} + cy^{2} + dx + ey + f = 0.*0299

*Parabola: Recall that, with a parabola, you have an x ^{2} term or a y^{2} term, but not both.*0310

*Therefore, either a is 0 (so this drops out) or c is 0 (so this drops out).*0325

*An example would be something like x = 3y ^{2} + 2y + 6.*0331

*Or you might have y = 2x ^{2} - 4x + 8.*0338

*So, neither of these has both an x ^{2} and a y^{2} term in the same equation.*0344

*For a circle, recall that what you are going to end up with is an x ^{2} and a y^{2} term*0352

*on the same side of the equation, with the same sign; and they are going to have the same coefficients.*0361

*Therefore, a is going to equal c.*0366

*An example would be x ^{2} + y^{2} + 3x - 5y - 10 = 0.*0369

*Here, a equals 1, and c equals 1; those are the same coefficients; x ^{2} and y^{2}*0378

*have the same sign and the same side of the equation; so it is a circle.*0386

*If we are working with an ellipse, this time the x ^{2} term and y^{2} terms are going to be*0392

*on the same side of the equation, with the same sign (like with the circle), but a and c are different.*0399

*They are unequal; that tells me that I am working with an ellipse.*0405

*For example, 12x ^{2} + 9y^{2} + 25x + 28y + 40 = 0.*0409

*Here, I have a = 12 and c = 9; so this is the equation describing an ellipse.*0423

*Finally, with a hyperbola, these are pretty straightforward to recognize, because you are going to have*0431

*an x ^{2} term and a y^{2} term, but they are going to have opposite signs.*0436

*Their coefficients will have opposite signs.*0440

*For example, 4x ^{2} - 8y^{2} + 10x + 6y - 34 = 0.*0443

*So, I have a = 4 and c = -8; since a and c have opposite signs, this is an equation describing a hyperbola.*0458

*You can use these rules to allow you to identify conic sections when you are given an equation in what we are going to call "general form."*0467

*It is standard form, but it is a very general standard form for any type of conic section.*0476

*OK, now we are going to work on identifying the various conic sections by looking at their equations.*0481

*First, write in standard form, and identify the conic section.*0488

*OK, so general standard form is what I am talking about right now: it is x ^{2} + 2y^{2}.*0495

*I need to subtract 4x from both sides, subtract 12, and set everything equal to 0.*0503

*What this tells me is that I have a = 1 and c = 2.*0510

*Since a = 1 and c = 2, these have the same sign (the x ^{2} and the y^{2} terms); but they have different coefficients.*0515

*And that means that what I am working with is an ellipse.*0525

*You could go on, then, and write this in the specific standard form for an ellipse.*0531

*Let's do that by completing the square: start out by grouping...let's rewrite it here.*0537

*And then, let's group the x and the y terms; so x ^{2} terms group together; y terms group together.*0545

*And now, add 12 to both sides to move that over, to make completing the square a little bit easier.*0555

*To complete the square for x ^{2} - 4x, I need to add b^{2}/4.*0563

*b ^{2}/4 is equal to 4^{2}/4, is 16/4; it is 4.*0571

*So, I add x ^{2} - 4x + 4; and it is very important to remember to add the 4 to the right side, as well.*0582

*There is no factor out here; I don't need to multiply--it is just 1; so 4 times 1 is 4; that gives me 12 + 4.*0592

*All right, that is x ^{2} - 4x + 4 + 2y^{2} = 16.*0601

*This can be rewritten as (x - 2) ^{2} + 2y^{2} = 16.*0609

*But recall, in standard form for an ellipse, you need to have a 1 on the right.*0616

*So, rewrite this up here, and then divide both sides by 16.*0622

*This is just (x - 2) ^{2}/16; this will cancel; this will become 8; and then 16/16 is 1.*0633

*So, we started out with this equation, put it into the general standard form to identify that this is an ellipse,*0645

*and then went on to complete the square; and now I have it written in standard form specifically for an ellipse,*0652

*which is much more useful when you are working with that and trying to graph.*0657

*This time, without completing the square, all we are going to do is identify the conic section.*0664

*And this is already in standard form; therefore, a = 2; c = -3.*0668

*Since a and c have opposite signs, this is the equation for a hyperbola.*0678

*I have an x ^{2} term and a y^{2} term, both, so it is not a parabola.*0687

*They have opposite signs; therefore, it must describe a hyperbola.*0691

*OK, write in standard form and identify the conic section.*0698

*Right now, this is not in any type of standard form; so I am going to work with the general standard form.*0702

*First, I am going to subtract 36x ^{2} from both sides.*0708

*Then, I am going to subtract 128 from both sides.*0717

*This means that I have a = -36, and c = 16; since these two are opposite signs, this is an equation describing a hyperbola.*0727

*OK, now, let's go ahead and put this in standard form specific to a hyperbola.*0741

*And let's start out by moving this 128 back over to the right; this is actually 32.*0747

*Next, I do have a common factor of 4, so I am going to divide both sides by that, so that I am working with smaller numbers.*0763

*That is -9x ^{2} + 4y^{2} + 8y =...128/4 would give 32.*0770

*All right, now to make this already move it more towards looking like a hyperbola, I am going to put the positive terms here in front:*0786

*4y ^{2} + 8y - 9x^{2}, because I am going to have a difference.*0793

*To complete the square, I first need to factor out that 4; then I need to add b ^{2}/4 to this expression.*0801

*This is going to equal 2 ^{2}/4; that is 4/4, which is 1.*0816

*Here is where I need to be careful, because I need to make sure I add 4 times 1 to the right, which is 4, to keep the equation balanced.*0825

*At this point, I am going to rewrite this as (y + 1) ^{2} - 9x^{2} = 36.*0837

*The last step is: I want the right side to be 1, so I am going to divide both sides by 36.*0844

*4 goes into 36 nine times; 9 goes into 36 four times; and then this cancels out to 1.*0860

*OK, so I started out with an equation that wasn't in any kind of standard form.*0872

*I put it in general standard form, and then determined it was a hyperbola, completed the square, and ended up*0876

*with an equation in the standard form for a hyperbola, so that I can use that to graph the hyperbola, if needed.*0882

*Write in standard form and identify the conic section.*0891

*So, this is almost in the general standard form, but not quite.*0894

*I have 4x ^{2}; I need to move this -3y^{2} next, then -16x - 18y - 12 = 0.*0897

*Now, I can easily see that a = 4 and c = -3; since these have opposite signs, that means that this is an equation describing a hyperbola.*0907

*OK, the next task is to complete the square.*0921

*I am going to first add 12 to both sides to remove the constant from the left side.*0926

*Then, I am going to group the x terms, which is 4x ^{2} - 16x.*0937

*And then, I have a -3y ^{2} - 18y, and that all equals 12.*0946

*I have a leading coefficient that is something other than 1, so I am going to factor out the 4, leaving behind x ^{2} - 4x.*0955

*Here, I need to factor out a -3; that is going to leave behind y ^{2} + 6y.*0965

*You need to be careful with the signs here; just double-checking: -3 times y ^{2} is -3y^{2}.*0969

*-3 times positive 6y is -18y, when you factor out with that negative sign; equals 12.*0976

*Completing the square: b ^{2}/4, in this case, is 4^{2}/4, is 16/4; that is 4.*0987

*So, I am going to add 4 here; I am also going to add 4 times 4, or 16, to the right, to keep the equation balanced.*0998

*For the y expression, I have y ^{2} + 6y; therefore, b^{2}/4 = 6^{2}/4, which is 36/4, which is 9.*1012

*-3 times 9 is -27; so I am going to subtract 27 from the right side, again keeping the equation balanced.*1029

*I am rewriting this as (x - 2) ^{2} - 3(y + 3)^{2} = 16 + 12, is 28, minus 27; conveniently, I end up with a 1 on the right.*1041

*Now, this is almost in standard form; generally, with standard form for a hyperbola, this term will be in the denominator.*1059

*So, it is possible to rewrite it like this; and it might be easier to look at it that way,*1071

*so that you can immediately know that this is a ^{2}, instead of having to think it out.*1077

*Putting it in truly standard form is also a good idea, because recall that, if I have the numerator divided by 1/4, that is the same as this times 4.*1082

*And that tells me that I have a hyperbola with a center at (2,-3); you have to watch out for this positive sign.*1096

*And it has a horizontal transverse axis.*1103

*So, today we learned exactly what conic sections are, where they come from,*1109

*and how to look at an equation and determine what type of conic section it describes.*1114

*Thanks for visiting Educator.com; see you again soon!*1120

1 answer

Last reply by: Rafael Mojica

Sun Jul 26, 2015 2:46 PM

Post by Andy Choi on July 26, 2015

i don't understand what 'b' is in the first example. Could you specify?

1 answer

Last reply by: Dr Carleen Eaton

Wed Dec 28, 2011 9:17 PM

Post by Jonathan Taylor on December 27, 2011

Dr Carleen ex.1 are u sure' it seem like it should be a circle rather then ellipse