Question 936865
start with y^2 - 6y - 4x^2 + 8x - 1 = 0


isolate the x's and y's with parentheses and add 1 to both sides of the equation to get:


(y^2 - 6y) - (4x^2 - 8x) = 1


factor the 4 out of (4x^2 - 8x) to get:


(y^2 - 6y) - 4 * (x^2 - 2x) = 1


complete the squares on (y^2 - 6y) and (x^2 - 2x) to get:


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


simplify to get:


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


simplify further to get:


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


add 9 and subtract 4 from both sides of the equation to get:


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


simplify to get:


(y-3)^2 - 4(x-1)^2 = 6


divide both sides of the equation by 6 to get:


(y-3)^2 / 6 - 4(x-1)^2 / 6 = 1


multiply 4(x-1)^2 / 6 by (1/4) / (1/4) to get (x-1)^2 / (6/4) which makes your equation look like:


(y-3)^2 / 6 - (x-1)^2 / (6/4) = 1


the equation is now in the standard form of:


(y-k)^2 / a^2 - (x-h)^2 / b^2 = 1


you have a^2 = 6 which makes a = sqrt(6)


you have b^2 = 6/4 which makes b = sqrt(6) / 2


the equation for the asymptote for a vertically oriented hyperbola is:


y = plus or minus a/b (x-h)^2 + k


a/b = sqrt(6) / (sqrt(6) / 2) which becomes a/b = 2


h is equal to 1 and k is equal to 3


the equation for the asymptote becomes:


y = plus or minus 2(x-1) + 3


the graph of your hyperbola is shown below:


<img src = "http://theo.x10hosting.com/2015/010702.jpg" alt="$$$" </>


here's an excellent reference to help you understand how to work with hyperbolas.


<a href = "http://www.purplemath.com/modules/hyperbola.htm" target = "_blank">http://www.purplemath.com/modules/hyperbola.htm</a>