SOLUTION: Hi, I'm having trouble completing the square to find the standard form of the following equation.(I've only ever done this with one variable so not sure how to proceed) {{{x^2-9

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Hi, I'm having trouble completing the square to find the standard form of the following equation.(I've only ever done this with one variable so not sure how to proceed) {{{x^2-9      Log On


   

Question 372868: Hi, I'm having trouble completing the square to find the standard form of the following equation.(I've only ever done this with one variable so not sure how to proceed)
x%5E2-9y2%2B36y-72=0
Could you please let me know the steps I have to take to complete the square with two variables? Thanks for your time :)
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2+-+9%28y%5E2-4y%29+-+72+=+0,
x%5E2+-9%28y%5E2-4y+%2B+4%29+%2B+36+-+72+=+0,
x%5E2+-+9%28y-2%29%5E2+-+36+=+0,
x%5E2+-+9%28y-2%29%5E2+=+36,
x%5E2%2F36+-+%28y-2%29%5E2%2F4+=+1,
it's an hyperbola with vertex at (0,2), y-axis as the conjugate axis and y = 2 as the transverse axis.