Question

Given: x^2 - y^2 - 4x - 6y - 6 = 0. Write the standard form of the hyperbola.

1) Find the center

2) Find the vertices

3) Find the foci

4) Find the slope of the asymptotes

5) Determine whether the transverse access is vertical or horizontal.

Answer 1

Yo, boy/girl. Where have you went with this?

Answer 2

Math problems are like dates. You can't expect other people to do them for you and get the benefit associated as well.

Answer 3

i just need to know how to put it into standard form

Answer 4

then i can go from there

Answer 5

(x-h)^2/a^2 - (y-k)^2/b^2=1

Answer 6

or (y-k)^2/a^2 - (x-h)^2/b^2=1

Answer 7

Can you do complete square as its standard form suggest?

Answer 8

@Chlorophyll's process is correct. You complete the square. Do you know how to do that?

Answer 9

that's the part i don't understand

Answer 10

You have to complete the square twice.

Answer 11

Group x terms , then y terms together as preparing for complete the square TWICE :)

Answer 12

Look at \(x^2-y^2-4x-6y-6=0\)
Rearrange it to look more beautiful and fabulous
\[x^2-4x-y^2-6y-6=0\]
Now, you need to add a constant to both sides so that you can factor the \(x^2-4x\) part into \((x-h)^2\) for some \(h\).
Follow me so far?

Answer 13

yes

Answer 14

Ok, so you have \(x^2-4x-y^2-6y-6=0\). What can you add to both sides to make \(x^2-4x\) factor? In other words, what is the \(k\) such that \(x^2-4x+k=(x-h)^2?\)

Answer 15

Let demonstrate the first one, then you can try with the second for y terms:
( x² - 4x + 2² ) - 2²
= ( x -2 ) ² - 4

Answer 16

@kenneyfamily ? are you confused ?

Answer 17

yes....

Answer 18

@Chlorophyll, you kinda pulled that out of a hat?

Answer 19

Match what @Limitless 's step with what I show!
Then try the second one, will you!

Answer 20

@kenneyfamily, what you're trying to do is find a number to add to \(x^2-4x\) so that it is a square binomial. that means \((x-h)^2=x^2-2hx+h^2\). If you let \(-4=-2h\), what must \(h\) be? Once you know \(h\), you have to add its square to the original equation.

Answer 21

Hey, Kenney, if you're confused--tell us where. We're here to help. Don't be embarrassed or anything. :) We all have been where you are.

Answer 22

ok thanks, lemme try and re read what youre saying

Answer 23

does k=4?

Answer 24

Yup! Now, add that to your original equation.

Answer 25

You should have \(x^2-4x-y^2-6y-6+4=4\). Now, you can say \((x-2)^2-y^2-6y-6=4.\) Does this make sense?

Answer 26

yes!

Answer 27

sorry but where do we go from there?

Answer 28

Yay! (I'm back)

Answer 29

Okay, so now you have \((x-2)^2-y^2-6y-6=4.\)
What should we add to this equation to factor the \(-y^2-6y\) part?
Recall from before \((y-k)^2=y^2-2ky+k^2\). We have \(-2k=-6\). So, what do we add to factor?

Answer 30

3

Answer 31

?

Answer 32

That's what \(k\) is. \(k=3\). So, we need to add the square (see how it's in \((y-k)^2=y^2-2ky+k^2\)?) So, add \(k^2\). What do you get?

Answer 33

9

Answer 34

wait i thought k=4??

Answer 35

Correct. You're doing well so far. Can we tie this together and factor?

Oh, and don't mind what we used above. We're just using letters to illustrate the point.

Answer 36

oh ok

Answer 37

So, after adding \(9\), we have
\((x-2)^2-y^2-6y-6+9=4+9\)
So, we can factor as follows
\((x-2)^2-(y-3)^2-6=4+9\)

Can you take it from here?

Answer 38

sorry but can you keep going...

Answer 39

Hey, it's okay.

Ok, so we have \((x-2)^2-(y-3)^2-6=4+9\). What can we do to simplify this?

Answer 40

add 4+9 to get 13

Answer 41

then add the 6 over?

Answer 42

Correct!

Answer 43

to get 19

Answer 44

So you have \((x-2)^2-(y-3)^2=19\).

Now what?

Answer 45

this is where i'm lost

Answer 46

Big tip: on the right side of the equation = 1

Answer 47

subrtact 18 from both sides?

Answer 48

Always divide!

Answer 49

so divide eveything by 19??

Answer 50

Yep!

Answer 51

That's just because the standard form sets the entire left side equal to \(1.\)

Answer 52

@kenneyfamily Hold on, your calculation doesn't match with mine

Answer 53

(x-2)² - ( y +3)² + 9 - 4 - 6 = 0

Answer 54

- y² - 6y = - ( y² + 6y) = - ( y² + 6y + 3² - 9)
= - ( y + 3)² + 9

Answer 55

Thus ( x-2)² - ( y +3)² = 1

Answer 56

so thats standard form?

Answer 57

which means the center is (2, -3)?

Answer 58

\[
\begin{align}
x^2 - y^2 - 4x - 6y - 6&= 0\\
x^2-4x-y^2-6y&=6\\
(x^2-4x+4)-y^2-6y-9&=6+4-9\\
(x-2)^2-(y+3)^2&=1
\end{align}
\]
@Chlorophyll is correct here.

Answer 59

Yes, the center is \((2,-3).\)

Answer 60

and how do you find the vertices? sorry for so many ?s

Answer 61

@kenneyfamily, this is a rather special parabola. \(a=b=1.\) Just figured I'd point that out.

As for the vertices, one moment

Answer 62

is it something like (h-a,k) (h+a,k)?

Answer 63

ok thanks

Answer 64

@kenneyfamily, I believe the vertices are indeed \((h+a, k)\) and \((h-a,k).\) :) We have that \(a=1\) (as said above). So, you know what your vertices are now. :)

Answer 65

how do we know a=1?

Answer 66

and how do you find the foci?

Answer 67

that was my last question..promsie!!!

Answer 68

(x-h)²/ a² - (y-k)²/b² =1
Do you see why a = b = 1 now?

Answer 69

yes!

Answer 70

and finding the foci??

Answer 71

anybody?

Answer 72

If your post is correct, then the book is typo!
Because this result is circle with center ( 2, -3) and radius = 1

Answer 73

weird but thanks!!

Answer 74

can you still find the foci of a circle?

Answer 75

NO, Circle is defined with center and radius!
no foci, asymptote, ...

Answer 76

Let move on to other questions!