SOLUTION: Please sketch y^2-6y-4x^2+8x-1=0 Need help finding the aymptotes of the hyperbola

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Question 936865: Please sketch y^2-6y-4x^2+8x-1=0
Need help finding the aymptotes of the hyperbola
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
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:

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

http://www.purplemath.com/modules/hyperbola.htm