SOLUTION: It is said that i must convert the following equation to the standard form of a conic section: x^2-y^2-6x+13=0 I grouped the x's together and completed the square to get (x-3

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: It is said that i must convert the following equation to the standard form of a conic section: x^2-y^2-6x+13=0 I grouped the x's together and completed the square to get (x-3      Log On


   

Question 726334: It is said that i must convert the following equation to the standard form of a conic section:
x^2-y^2-6x+13=0
I grouped the x's together and completed the square to get (x-3)^2, but am not sure how to deal with the y^2 and 13 parts. Do I move those parts to the other side and add 9 to it to get
(x-3)^2=y^2-4 ?
What must I do after this point? The lonely y^2 is throwing me off a bit.
I appreciate any help you can me! Happy π Day.


Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
convert the following equation to the standard form of a conic section:
x^2-y^2-6x+13=0
complete the square:
x^2-6x-y^2+13=0
(x^2-6x+9)-y^2=-13+9
(x-3)^2-y^2=-4
divide by -4
y%5E2%2F4-%28x-3%29%5E2%2F4=1
This is an equation of a hyperbola with vertical transverse axis.
Its standard form: %28y-k%29%5E2%2Fa%5E2-%28x-h%29%5E2%2Fb%5E2=1, (h,k)=(x,y) coordinates of the center.
For given hyperbola:
center: (3,0)
a^2=b^2=4