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Question 590198: I have to name the center, vertices, foci and slopes of the asymptotes
x^2 - y^2 -4x +2y = -2 add 4
(x-2)^2 add 4 to the right side
now I am stuck because I am not sure what the negative in front of the y^2
I am not sure how to do the (y- or +2)^2 because of the - in front of the y^2 and do I add a neg or positive 2 on the left side?
What would my equation be?
Thank you!
Connie
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! I have to name the center, vertices, foci and slopes of the asymptotes
x^2 - y^2 -4x +2y = -2 add 4
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I will assume this is the original equation:
x^2 - y^2 -4x +2y = -2
To get all the answers, you must first put this equation in standard form by completing the squares:
(x^2-4x+4) -(y^2-2y+1) = -2+4-1
(x-2)^2-(y-1)^2=1
This is an equation of a hyperbola with horizontal transverse axis of the standard form:
(x-h)^2/a^2-(y-k)^2/b^2=1, (h,k)=(x,y) coordinates of the center.
..
For given equation: (x-2)^2-(y-1)^2=1
center: (2,1)
a^2=1
a=1
vertices:(2±a,1)=(2±1,1)=(1,1) and (3,1)
..
b^2=1
b=1
..
c^2=a^2+b^2=1+1=2
c=√2
Foci: (2±c,1)=(2±√2,1)=(2-√2,1) and (2+√2,1)
..
slopes of asymptotes=±b/a=±1/1=±1
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