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Mathematics 18 Online
OpenStudy (anonymous):

Help Please! Will fan & medal!!

OpenStudy (anonymous):

Write the general form for the equation of the following hyperbola: (x-1)^2/9 - (y-3)^2/4 = 1

OpenStudy (anonymous):

x^2 - y^2 - 32x - 96y - 113 = 0 4x^2 - 9y^2 -4x + 108y - 55 = 0 4x^2 - 9y^2 -8x + 54y - 113 = 0 x^2 - y^2 - 8x + 54y - 113 = 0

OpenStudy (anonymous):

Those are answers !

OpenStudy (anonymous):

First step, multiply both sides by 9 and 4, and simplify it a bit.

OpenStudy (anonymous):

After that, expand the squares. Then distribute the 4 and 9 through their groups. At this point, you'll be able to look at the x^2 term and the x term to get your answer. Only one choice has the same number in front of those two (the coefficient).

OpenStudy (anonymous):

However, to do it completely, you'd just simplify, and then subtract 36 from both sides.

OpenStudy (anonymous):

Can you show me please

OpenStudy (anonymous):

Any particular part?

OpenStudy (anonymous):

9 *(x-1)^2/9 * - (y-3)^2/4 = 9

OpenStudy (anonymous):

(x-1)^2 * - (y-3)^2/4 = 9

OpenStudy (anonymous):

(x-1)^2 * - 4(y-3)^2/4 = 9* 4 (x-1)^2 * - (y-3)^2 = 36 (x-1)^2 * - (y-3)^2 -36 = 1

OpenStudy (anonymous):

like that^?

OpenStudy (anonymous):

Close, very close! When you multiply by 9, the y term is also multiplied by 9. \[(x-1)^2 - 9*\frac{(y-3)^2}{4} = 9\]

OpenStudy (anonymous):

Apply the same principle when you multiply by 4.

OpenStudy (anonymous):

okay thats what i thought but it doesnt divide

OpenStudy (anonymous):

That's ok. It'll work out well in just another step or so.

OpenStudy (anonymous):

hmm so 4(x-1)^2 * - 36(y-3)^2 =36 ?

OpenStudy (anonymous):

What happened to the 4 underneath the y?

OpenStudy (anonymous):

4(x-1)^2 * - 9(y-3)^2 =36

OpenStudy (anonymous):

canceled out ?

OpenStudy (anonymous):

There you go :)

OpenStudy (anonymous):

Now, expand the (x-1)^2 and the (y-3)^2

OpenStudy (anonymous):

4(x-2x+1) * - 9(y-3y+ =36

OpenStudy (anonymous):

idk how to complete the square for that second part

OpenStudy (anonymous):

Very close! \[(x-1)^2 = (x-1)(x-1) = x^2 - 2x + 1\] \[(y-3)^2 = (y-3)(y-3) = y^2 - 6y + 9\] If I may ask, why were you completing the square?

OpenStudy (anonymous):

Oops ugh its cuz im not use to solving these ones im use to solving something in general form to standard form haha

OpenStudy (anonymous):

Now, distribute the 9 and 4 to their groups. Then you'll be able to see the answer :)

OpenStudy (anonymous):

But okay so 4(x-2x+1) * - 9(y-6y+9) = 36 4x-8x2 - 9y+54y+81= 36 4x^2 - 9y^2 -8x + 54y - 113 = 0

OpenStudy (anonymous):

??

OpenStudy (anonymous):

yep! Good job!

OpenStudy (anonymous):

Thanks!! :)) Hey i have another question do you mind me opening a new page and tagging you in it to help me or are you busy?

OpenStudy (anonymous):

Small note, you messed up the sign on the 81 in the middle step, but it may have just been a typo. I can help with another question, sure.

OpenStudy (anonymous):

yeah nice catch and okay! :)

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