You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
Help!!!
- Thread starter Mj12
- Start date
D
Deleted member 4993
Guest
Mj12 said:Find the center, foci, and vertices of the hyperbola 9x^2- y^2+54x+2y+71=0. I have completed the square but after that I am lost! Help please...- Megan Breanne
Show your work of completing the square.
D
Deleted member 4993
Guest
Mj12 said:After I complete the square I get. (x^2+3)/ -81 + (y^2+1) / -81 =1..... which I think I maybe wrong, but after that I am lost.
I do not see completed squares in your answer. Please show complete work.
How did you get the answer you have posted?
Sorry lol it's a lot a type on an iPhone sorry haha.9x^2-y^2+54x+2y+71...... make it 9(x^2+6x+_) + -y^2+2y+_=-71..... and I do half of 6 which is 3 and square it which makes 9. And half of 2y^2 which is one. Square it and get 1... So I plug those in and subtract from -71. Which makes... 9(x^2+6x+9) - y^2+ 2y+1= -81.. and I divide all by -81 and get... (x^2+3)/-81 + (y^2+1)/-81= 1... and that is what I did...
Sorry lol it's a lot a type on an iPhone sorry haha.9x^2-y^2+54x+2y+71...... make it 9(x^2+6x+_) + -y^2+2y+_=-71..... and I do half of 6 which is 3 and square it which makes 9. And half of 2y^2 which is one. Square it and get 1... So I plug those in and subtract from -71. Which makes... 9(x^2+6x+9) - y^2+ 2y+1= -81.. and I divide all by -81 and get... (x^2+3)/-81 + (y^2+1)/-81= 1... and that is what I did...