Question

Convert the equation to the standard form for a hyperbola by completing the square on x and y.

x^2 - y^2 + 6x - 4y + 4 = 0

A. (x + 3)^2 + (y + 2)^2 = 1

B. (y + 3)^2/16 - (x + 2)^2/36 = 1

C. (x + 3)^2 - (y + 2)^2 = 1

D. (y + 3)^2- (x + 2)^2 = 1

Answer 1

i choose A am i correct

Answer 2

You can always just multiply out the answer choices and see which one matches. The problem I see right away with A is that the equation we're trying to reach has some negatives. There are no negatives in A.

Answer 3

oh okay

Answer 4

B, C, and D all have negative signs in them.

Answer 5

o yeah they do i was'nt looking correctly

Answer 6

wasn't

Answer 7

oh so its not A isn't C...am i right or wrong

Answer 8

For C, (x+3)^2 - (y+2)^2=1
(x+3)^2=x^2+6x+9
-(y+2)^2=-y^2-4y-4
x^2+6x+9-y^2-4y-4
x^2-y^2+6x-4y+5=0

It doesn't match either, so yes you are right.

Answer 9

Oh but I didn't get you to the right answer yet, it isn't C! xD

Answer 10

ooooooooooooo lol i thought you were saying is c lmao haha...whats the right answer then

Answer 11

Well I gave you the values of (x+3)^2 and (y+2)^2, I'm sure you can handle this now =P I can't do all of the work ^^

Answer 12

yeah your right let me figure it out brb

Answer 13

Ohhh before you solve for what D is, they mixed up the numbers inside! Make sure you reevaluate those!

Answer 14

And B as well.

Answer 15

ok

Answer 16

Oh my god, I am so sorry. I made a fatal mistake when solving for C -.- I set the answer=0, not =1 like it should have been:

x^2-y^2+6x-4y+5=1
x^2-y^2+6x-4y+4=0

Answer 17

And with that one small change, it is the answer. I am infamous for my stupid mistakes, so if I happen to help you with another problem in the future, make sure you double-check my work >.

Answer 18

oh okay its fine everyone makes mistakes....so it is C again thank you so much for all your help your a great person @smokeydabear ;-)

Answer 19

Yeah no problem! Glad you were willing to try some on your own too, most people just want answers -.-

Answer 20

yeah I know ;-)