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>> Hi, this is Julie Harland
and I'm YourMathGal.
Please visit my website
at yourmathgal.com
where you could search for any
of my videos organized
by topic.
This is part 4 of hyperbolas.
And this one we are discussing
hyperbola centered at HK,
and we state the standard form
of a hyperbola centered at HK
and we do the
following problem.
Recall that the standard form
of hyperbola centered
at the origin is one
of these two things you see
right here,
X squared over A squared minus
Y squared over B squared
equals 1.
Or where the Y--
the sign for another Y square
is positive so it's Y squared
over B squared minus X squared
over A squared equals 1.
It's easy to tell what the X
intercepts are
for the first hyperbola
and the Y intercepts
for the second hyperbola
and the asymptotes are.
In the general form,
you're just going
to multiply all of these out.
There's one thing I did not
make clear
about this general form is
that A and B must be
different signs.
So one is positive
and one is negative, okay?
Some people assume
that means they're both
positive but remember these
are just constants, A,
B and C. So, we're going
to take this and I'll think,
what would happen
if it was not centered
at the origin?
And this is what happens.
It looks very similar except
if we have a center HK.
Then if it's just an X square
or a Y squared,
we'll have an X minus H
squared or a Y minus
K squared.
And again,
either the X squared is
positive, the coefficient
of the squared is positive
or the coefficient
of the Y squared is positive.
The slopes
of the asymptotes will be
the same.
It will be plus or minus,
B over A. Remember,
B always goes with the Y
and A always goes with the X.
But the actual equation,
since they go
through the center HK,
it will be this Y minus K
equals plus or minus BA,
that's the slope,
X minus H. That's sort
of the form of the line.
So the general form
of hyperbola now,
we'll have AX squared plus--
and then you could have
capital letters.
I just happen to put
in lower case letter.
Here's an extra one I put
here, CXY.
We aren't going
to be doing any
where we have an XY term
but there are parabolas
that actually have XY terms.
All the ones we're doing are
going to be vertical
or horizontal hyperbolas
so this-- you won't see this
term, in other words
so it will be zero.
All right,
so let's try a problem.
All right, so we're going
to graph this parabola.
Y minus 2 squared over 9,
minus X plus 1 squared
over 16 equals 1.
And so notice
that 9 is 3 squared
and 16 is 4 squared.
So if you want,
you could rewrite those
in that form
so you could see this is 3
squared and that's 4 squared.
Well let's go ahead and start
with the center.
Where's the center?
We're asked
to state the center anyway.
So what will the center be?
Well, if it's an X plus 1
that will give us a negative 1
for the X coordinate.
And Y minus 2
that will give us a 2
for the Y coordinates.
So there's our center
so we can go over here,
and negative 1, 2,
we know that's the center
of the parabola right--
I'm sorry of the hyperbola
right here.
And we want
to make our little box.
This is going to be done
in the same way
as if it was centered
at the origin.
So, in the Y direction,
think of this as 3 squared,
we're going to go up 3--
1, 2 and 3, okay and down 3,
and those are going to be--
I went from the wrong place.
Down 3 from the center here,
which is right here.
So this-- those are
the vertices.
I should have asked
for vertices.
So the vertices
of this hyperbola are,
let's see what that is, well,
I got negative 1,
5 and I've got negative 1,
negative 1.
So those are our vertices.
All right, now the--
in the X direction,
16 is on the denominators,
that's 4 squared so I'm going
to go over 4--
1, 2, 3, 4
and then the other direction,
1, 2, 3, 4.
Remember, we could just sort
of make this little box
to help us draw
the asymptotes.
So if we do this,
our box should go
right through.
Those are our asymptotes
that line.
Again, I'm just sketching it.
I know it's not exact.
Okay, so there're
our asymptotes.
And since the coefficient
of the Y squared is positive,
this is going to go up.
So we just use the asymptotes
as guides.
Okay, the same thing here.
So that is how we graph it.
Now we're just sketching it,
right?
Okay, now I found the center,
I found vertices.
I'm asking for the equations
of the asymptotes.
There's two ways you can
get it.
You could actually just look
at your graph and figure
out what it is by trying
to see where it goes
through the Y axis here.
I'm not sure
that I drew it very
accurately,
so it isn't obvious
to me what the Y
intercepts are.
So instead, let's do this way.
We know the asymptotes are
going to be form.
Instead Y equals plus
or minus A over B,
it's Y minus 2 equals,
we're going to have plus
or minus, so we'll do first
the plus, and what will it be?
It's always be Y
over the X value.
But remember, that's the 3
over 4, so that's going
to 1 asymptote.
And of course
that can be simplified
or written
in the different form.
You could put
in slope intercept form.
And the other one is plus
or minus 3/4--
I'm sorry,
minus 3/4, X plus 1.
So these are the equations o f
the asymptotes.
[ Pause ]
And if you wanted to put this
in slope intercept form,
we could do that as well.
So let's go ahead
and just write
that in slope intercept form--
intercept form
and you'll see why
that was hard to figure
by looking
at the graph 'cause you're
going to have some
fractions here.
So let's say we wanted
to put this
in slope intercept form.
Here are our two equations.
The way I like to do it is
to multiply both sides
by the least common
denominator, getting rid
of the fraction.
So on both of this,
the least common denominator
happens to be 4.
So for the first 1,
I've got to remember
to multiply 4
by this whole quantity
on the left so--
or by each term,
however you want to think
about that,
that will give me 4Y minus 8.
It will cancel with this 4,
so I only have
to do 3 times the X plus 1
so that gives 3X plus 3.
And so then I would just add 8
to both side, just 11
and then I just divide it
by 4.
Now you don't have
to do it the same way 'cause I
did it.
But if you did write it
in slope intercept form,
that's one of the equations
of the asymptotes
in the slope intercept form.
Let's do the same thing
over here.
We'll have 4Y minus 8,
and now the 4 cancels
so we have a negative 3 times
X plus 1.
Make sure you distribute this
negative 3 all the way
through the parenthesis.
So it's a negative 3X minus 3.
And again, I'm going to add 8
to both sides.
So negative 3 plus 8 is 5
and then divide both sides
by 4, so that will give you
the other asymptote
which is Y equals negative
3/4X plus 5/4.
So like I was saying,
it was hard to see
where it went
through the Y axis
from the picture.
But these are the equations
of the asymptotes
in slope intercept form.
So let's see
if that looks right.
So one of them is Y equals
3/4X plus 11/4.
So slope at 3 /4, okay.
But we've got a Y intercept
of 11/4 and the other one has
the Y intercept to 5/4.
So let's see.
11/4 and 5/4,
let's check it out.
11/4. So, nope--
it doesn't really look
like it to me.
Oh yes, it does,
there it is, 11 /4.
So it's around, you know,
almost 3, right?
So that's this first one
in slope of 3/4.
The other one is 5/4
and that's this either one
right here.
I was looking at the wrong.
I was looking up here.
I want to look
at where the asymptotes
go through.
So it does look
like that would make sense,
those two equations
for the asymptotes.
All right,
let's do the last part
which is to write the equation
in general form.
All right,
to write the equation
in general form, what we want
to do is multiply everything
at the least common
denominator
to eliminate fractions.
So let's begin with that
and let's see,
that would be 9 time 16.
No, you could just write
and we'll play everything
by 9 times 16,
that's actually 144
but it will be easier
when I distribute here.
I'm going to multiply this
first term,
9 times 16 times Y minus 2
squared over 9 and it's easy
to see the 9 is
cancelling minus.
Then we're going
to have the 9 times 16 times
the next term.
And it's easy to see,
the 16 is cancelling.
And then I have
to multiply 9 times 16 times 1
which is just 144.
Okay, so now what do I do?
Well, I've got 16 times the
quantity Y minus 2 squared.
I'm going to have to square
that binomial
and that's Y minus 2 times Y
minus 2, and I'm going
to use the formula--
because of space,
I'm not showing all the steps
using the FOIL method
but we'll get 16 times Y
squared minus 4Y plus 4.
And then, for the next one,
I'm going to have
to square X plus 1 before I
multiply it by 9
and that's X squared plus 2X
plus 1.
Everybody remembers
that middle term, right?
Don't just write X squared
plus 1 or Y squared plus 4.
If you need help on that,
go back on multiplying
polynomials,
squaring polynomials.
Okay, now we can distribute
the 16 and distribute the
negative 9.
We're getting really
close here.
So we can either write the Y
squared or the X squared
term first.
I'm going to go ahead
and just write the X squared
term first.
So it's negative 9X squared
plus 16Y squared.
All right, next we're looking
for the X term
which is the negative 18X.
And then I'm looking
for the Y term
which is negative 64Y.
And let's see.
I've got 64 minus 9.
All right, so that's going
to be, I think 55.
And I'm going to subtract 44
from both sides.
It's almost done.
Negative 9X squared plus 16Y
squared minus 18X minus 64.
And then I've 55 minus 144.
Let's see, so what's happens.
Let's do 144 minus 55,
maybe 9 so I have minus 89
equals 0.
Now notice I have a minus sign
in front of the X squared.
If I multiplied everything
by negative 1,
all the signs would change.
[ Pause ]
Both of these are exactly the
same hyperbola so it's hard
to tell when you only look
at the equation
in general form,
if it's a vertical
or a horizontal hyperbola
by the sign.
You really can't tell
because in one equation,
it looks like the minus sign
in front of the X squared.
And the next equation,
it looks like the minus sign
in front of the Y squared.
You can't tell
until you put it in,
in the standard form,
which one it is.
So that's important,
so both of these are correct.
The main thing
that let us know
that is hyperbola is if--
is that the coefficients
of X squared
and the Y squared term have
to be different signs,
one must be positive
and one must be negative.
[ Pause ]
Please visit my website
at YourMathGal.com
where you can view all
of my videos
which are organized by topic.