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.
Yo, boy/girl. Where have you went with this?
Math problems are like dates. You can't expect other people to do them for you and get the benefit associated as well.
i just need to know how to put it into standard form
then i can go from there
(x-h)^2/a^2 - (y-k)^2/b^2=1
or (y-k)^2/a^2 - (x-h)^2/b^2=1
Can you do complete square as its standard form suggest?
@Chlorophyll's process is correct. You complete the square. Do you know how to do that?
that's the part i don't understand
You have to complete the square twice.
Group x terms , then y terms together as preparing for complete the square TWICE :)
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?
yes
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 \(x^2-4x+k=(x-h)^2?\)
Let demonstrate the first one, then you can try with the second for y terms: ( x² - 4x + 2² ) - 2² = ( x -2 ) ² - 4
@kenneyfamily ? are you confused ?
yes....
@Chlorophyll, you kinda pulled that out of a hat?
Match what @Limitless 's step with what I show! Then try the second one, will you!
@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.
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.
ok thanks, lemme try and re read what youre saying
does k=4?
Yup! Now, add that to your original equation.
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?
yes!
sorry but where do we go from there?
Yay! (I'm back)
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?
3
?
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?
9
wait i thought k=4??
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.
oh ok
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?
sorry but can you keep going...
Hey, it's okay. Ok, so we have \((x-2)^2-(y-3)^2-6=4+9\). What can we do to simplify this?
add 4+9 to get 13
then add the 6 over?
Correct!
to get 19
So you have \((x-2)^2-(y-3)^2=19\). Now what?
this is where i'm lost
Big tip: on the right side of the equation = 1
subrtact 18 from both sides?
Always divide!
so divide eveything by 19??
Yep!
That's just because the standard form sets the entire left side equal to \(1.\)
@kenneyfamily Hold on, your calculation doesn't match with mine
(x-2)² - ( y +3)² + 9 - 4 - 6 = 0
- y² - 6y = - ( y² + 6y) = - ( y² + 6y + 3² - 9) = - ( y + 3)² + 9
Thus ( x-2)² - ( y +3)² = 1
so thats standard form?
which means the center is (2, -3)?
\[ \begin{align} x^2 - y^2 - 4x - 6y - 6&= 0\\ x^2-4x-y^2-6y&=6\\ (x^2-4x+4)-y^2-6y-9&=6+4-9\\ (x-2)^2-(y+3)^2&=1 \end{align} \] @Chlorophyll is correct here.
Yes, the center is \((2,-3).\)
and how do you find the vertices? sorry for so many ?s
@kenneyfamily, this is a rather special parabola. \(a=b=1.\) Just figured I'd point that out. As for the vertices, one moment
is it something like (h-a,k) (h+a,k)?
ok thanks
@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. :)
how do we know a=1?
and how do you find the foci?
that was my last question..promsie!!!
(x-h)²/ a² - (y-k)²/b² =1 Do you see why a = b = 1 now?
yes!
and finding the foci??
anybody?
If your post is correct, then the book is typo! Because this result is circle with center ( 2, -3) and radius = 1
weird but thanks!!
can you still find the foci of a circle?
NO, Circle is defined with center and radius! no foci, asymptote, ...
Let move on to other questions!
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