Uploaded by TheIntegralCALC on 06.09.2011

Transcript:

Hi, everyone. Welcome back to integralcalc.com. Today, we’re going to be talking about the

equation of a parabola and using it to find the parabola’s vertex, focus, directix and

axis. So in this particular problem, we’ve been the given the equation for a parabola

x squared minus 4x minus 4y equals zero. and the first thing that we want to take notice

of is which of our variables is quadratic and which is linear. And that means which

one of them is squared and which one of them is not.

So notice that we have an x squared term. That means that x is our quadratic variable

and as far as y goes, we only have a y to the first power which means it’s a linear

variable. We don’t have y squared. So x is the quadratic variable, y is the linear

variable. So what we want to do is just keep that in mind and we want to separate the variables

so we have x’s on one side and y’s on the other. So we’ll go ahead and say x squared

minus 4x, we’ll add 4y to both sides so we’ll get 4y.And now, because x is the quadratic

factor, we want to complete the square with respect to x. Remember that to complete the

square, we take the coefficient on the first degree term so this is x to the first power.

We don’t want to deal with the x squared term. We’re taking the coefficient on the

first degree and we’re going to divide that by 2. We’ll get negative 2 and then we square

our answer and we get a positive 4. Remember too that when you’re completing the square,

if you have a coefficient on your x squared term right here, you need to divide through

both sides of the equation by that coefficient because the coefficient on the x squared term

has to be one before you can complete the square root with respect to x. in this case,

it already is so we could just go ahead and take this coefficient on the first degree

term. So 4 now is what we’re going to add to both sides in order to complete the square

with respect to x. Remember that we have to add it to both sides so that we don’t actually

fundamentally change the equation. So we add 4 to both sides. Now, we can factor the left-hand

side into x minus 2 times x minus 2 which is the same thing as x minus 2 squared. On

the right-hand side, let’s go ahead and factor out a 4 just to simplify things so

we get 4 times y plus 1 over here on the right. Now our whole goal here is to try to transform

our equation into something that’s as similar as possible to one of four other equations.

When we’re dealing with finding the vertex, focus, directix and axis of a parabola, I’ve

made a chart that I hope would help simplify things. So notice here before we go look at

the chart that x squared is our quadratic factor. We already knew that. We determined

that x was our quadratic factor and we can see over here on the right-hand side that

y is positive. We’ve got positive 4y plus 4. y is positive. If this were a negative

4y, y would be negative but in this case y is positive. So those are the two things we

need to look at. So now when we go and look at this chart, what we need to do is determine

which of these four instances applies to our particular equation. So what you want to do

is look at this equation section here. this second row for the equation and the first

two columns where the parabolas open right and open left, y is the quadratic factor because

you notice we’ve got squared over here with the variable y. These two are out because

x is our quadratic factor. These last two columns here, the x variable is squared. So

we know that our parabola either opens up or opens down because x is our quadratic variable.

Now, we just need to know whether y is positive or negative. And remember, we said that in

our case, y was positive. Here, notice that you have positive 4p times y and here you

have negative 4p times y. The first one is positive, the second one is negative so that

means that our parabola, because x is our quadratic variable and because y is positive,

that means that our parabola opens up and that this particular equation here is the

one that we’re going to use. So we need to get our equation into this format here.

X minus h squared, remember we already have that. We know it’s x minus 2 squared equals

4p times y minus k. So let’s talk about that. So 4p times y minus k. We already have y minus k. we have y plus

1 so we know that k is going to be equal to negative 1. It’s this p here that’\s the

problem. So all we need to do is since we have 4 here, and 4p here, we have to set those

equal to each other and solve for p. So if we say 4 equals 4p and then divide both sides

by 4, we’ll see that 4 equals 1. So now what we want to do is just go ahead and write

in p. So x minus 2 squared equals 4 and all we’ll do is 4 times 1 to account for the

p and then y plus 1. And the reason we do that is just so that we can easily identify

our variables. So from our formula thqat we looked at before, we know that we have the

format here. So this is going to be h, this is going to be k here and this is going to

be p. So now let’s go back and look over here. and since we’re in this third column

here, we know our vertex is at (h,k). So if our vertex is at (h,k). Because in our equation

it’s x minus h and y minus k, we have a negative here that matches the format of the

formula so h is going to be 2 and k, the format is normally y minus k. because we have y plus

1, we know that k has to be equal to negative 1 to get rid of that negative sign so that

means our vertex is at (2,-1). If we go look back at our chart here, the

focus is at (h,k+p). We already know that h is positive 2 so that means the x coordinate

of the focus is at positive 2 and then the y coordinate of the focus is at k plus p.

We know that k is negative and we know that p is 1 so negative 1 plus 1 is zero. So there’s

our focus. If we go back to our chart, we know that the directix is at y equals k minus p. We know k to be negative 1 and p is 1 so negative

1 minus 1 is going to be negative 2. So that’s the line for the directix. And then

the axis, x equals h, is x equals 2 because we know that h is 2. So now that we have those

four pieces of information, we can graph our parabola and kind of get a visual for what

this looks like. So we know our vertex is at (2,-1). Our focus is at (2,0) which means

our parabola is opening up because the focus is always inside the parabola so the parabola

is going to look something like this. We’ll call this the vertex, we’ll call this the

focus. And then the directix is the line equals negative 2 which is going to be right about

here. the focus is always the same distance away from the vertex as the vertex is from

the directix so those two distances are equal to each other. And then the axis is at the

line x equals 2. So the line x equals 2 is the axis of the parabola over which the parabola

is reflected. So that gives you a visual for how these four pieces of information fit together

to compose the parabola. So that’s it. I really hope this video helped you guys and

I will see you in the next one. Bye!

equation of a parabola and using it to find the parabola’s vertex, focus, directix and

axis. So in this particular problem, we’ve been the given the equation for a parabola

x squared minus 4x minus 4y equals zero. and the first thing that we want to take notice

of is which of our variables is quadratic and which is linear. And that means which

one of them is squared and which one of them is not.

So notice that we have an x squared term. That means that x is our quadratic variable

and as far as y goes, we only have a y to the first power which means it’s a linear

variable. We don’t have y squared. So x is the quadratic variable, y is the linear

variable. So what we want to do is just keep that in mind and we want to separate the variables

so we have x’s on one side and y’s on the other. So we’ll go ahead and say x squared

minus 4x, we’ll add 4y to both sides so we’ll get 4y.And now, because x is the quadratic

factor, we want to complete the square with respect to x. Remember that to complete the

square, we take the coefficient on the first degree term so this is x to the first power.

We don’t want to deal with the x squared term. We’re taking the coefficient on the

first degree and we’re going to divide that by 2. We’ll get negative 2 and then we square

our answer and we get a positive 4. Remember too that when you’re completing the square,

if you have a coefficient on your x squared term right here, you need to divide through

both sides of the equation by that coefficient because the coefficient on the x squared term

has to be one before you can complete the square root with respect to x. in this case,

it already is so we could just go ahead and take this coefficient on the first degree

term. So 4 now is what we’re going to add to both sides in order to complete the square

with respect to x. Remember that we have to add it to both sides so that we don’t actually

fundamentally change the equation. So we add 4 to both sides. Now, we can factor the left-hand

side into x minus 2 times x minus 2 which is the same thing as x minus 2 squared. On

the right-hand side, let’s go ahead and factor out a 4 just to simplify things so

we get 4 times y plus 1 over here on the right. Now our whole goal here is to try to transform

our equation into something that’s as similar as possible to one of four other equations.

When we’re dealing with finding the vertex, focus, directix and axis of a parabola, I’ve

made a chart that I hope would help simplify things. So notice here before we go look at

the chart that x squared is our quadratic factor. We already knew that. We determined

that x was our quadratic factor and we can see over here on the right-hand side that

y is positive. We’ve got positive 4y plus 4. y is positive. If this were a negative

4y, y would be negative but in this case y is positive. So those are the two things we

need to look at. So now when we go and look at this chart, what we need to do is determine

which of these four instances applies to our particular equation. So what you want to do

is look at this equation section here. this second row for the equation and the first

two columns where the parabolas open right and open left, y is the quadratic factor because

you notice we’ve got squared over here with the variable y. These two are out because

x is our quadratic factor. These last two columns here, the x variable is squared. So

we know that our parabola either opens up or opens down because x is our quadratic variable.

Now, we just need to know whether y is positive or negative. And remember, we said that in

our case, y was positive. Here, notice that you have positive 4p times y and here you

have negative 4p times y. The first one is positive, the second one is negative so that

means that our parabola, because x is our quadratic variable and because y is positive,

that means that our parabola opens up and that this particular equation here is the

one that we’re going to use. So we need to get our equation into this format here.

X minus h squared, remember we already have that. We know it’s x minus 2 squared equals

4p times y minus k. So let’s talk about that. So 4p times y minus k. We already have y minus k. we have y plus

1 so we know that k is going to be equal to negative 1. It’s this p here that’\s the

problem. So all we need to do is since we have 4 here, and 4p here, we have to set those

equal to each other and solve for p. So if we say 4 equals 4p and then divide both sides

by 4, we’ll see that 4 equals 1. So now what we want to do is just go ahead and write

in p. So x minus 2 squared equals 4 and all we’ll do is 4 times 1 to account for the

p and then y plus 1. And the reason we do that is just so that we can easily identify

our variables. So from our formula thqat we looked at before, we know that we have the

format here. So this is going to be h, this is going to be k here and this is going to

be p. So now let’s go back and look over here. and since we’re in this third column

here, we know our vertex is at (h,k). So if our vertex is at (h,k). Because in our equation

it’s x minus h and y minus k, we have a negative here that matches the format of the

formula so h is going to be 2 and k, the format is normally y minus k. because we have y plus

1, we know that k has to be equal to negative 1 to get rid of that negative sign so that

means our vertex is at (2,-1). If we go look back at our chart here, the

focus is at (h,k+p). We already know that h is positive 2 so that means the x coordinate

of the focus is at positive 2 and then the y coordinate of the focus is at k plus p.

We know that k is negative and we know that p is 1 so negative 1 plus 1 is zero. So there’s

our focus. If we go back to our chart, we know that the directix is at y equals k minus p. We know k to be negative 1 and p is 1 so negative

1 minus 1 is going to be negative 2. So that’s the line for the directix. And then

the axis, x equals h, is x equals 2 because we know that h is 2. So now that we have those

four pieces of information, we can graph our parabola and kind of get a visual for what

this looks like. So we know our vertex is at (2,-1). Our focus is at (2,0) which means

our parabola is opening up because the focus is always inside the parabola so the parabola

is going to look something like this. We’ll call this the vertex, we’ll call this the

focus. And then the directix is the line equals negative 2 which is going to be right about

here. the focus is always the same distance away from the vertex as the vertex is from

the directix so those two distances are equal to each other. And then the axis is at the

line x equals 2. So the line x equals 2 is the axis of the parabola over which the parabola

is reflected. So that gives you a visual for how these four pieces of information fit together

to compose the parabola. So that’s it. I really hope this video helped you guys and

I will see you in the next one. Bye!