SOLUTION: For the hyperbola, find the equations of the asymptotes.
x^2 - 4y^2 - 6x - 8y - 11 = 0
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After working the equation into standard form and having found the cent
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Quadratic-relations-and-conic-sections
-> SOLUTION: For the hyperbola, find the equations of the asymptotes.
x^2 - 4y^2 - 6x - 8y - 11 = 0
--------
After working the equation into standard form and having found the cent
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Question 1204704: For the hyperbola, find the equations of the asymptotes.
x^2 - 4y^2 - 6x - 8y - 11 = 0
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After working the equation into standard form and having found the center and vertices, I wrote the equation of the asymptotes as
y = k +- b/a(x-h) => y = -1 +- 2/4(x-3)
But the answer I was provided was
x - 2y - 5 = 0
x + 2y - 1 = 0
What is this form? Found 2 solutions by mananth, MathLover1:Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! Simplify
y = -1 + 2/4(x-3)
y = -1 + 1/2(x-3)
multiply equation by 2
2y=-2+x-3
2y=x-5
2y-x+5=0
x-2y-5=0
y = -1 - 2/4(x-3)
2y=-2-x+3
2y= -x+1
x+2y-1=0
You can put this solution on YOUR website!
that is general form form
The general form is one of the many different forms you can write linear functions in.
so, in general form is
..multiply by switch sides
