SOLUTION: Please solve, show work and include all relevant points (i.e vertices, intercepts, foci,directrix,asymptotes, ect..) x^2-4y^2+2x=32y=67

Algebra ->  Algebra  -> Quadratic-relations-and-conic-sections -> SOLUTION: Please solve, show work and include all relevant points (i.e vertices, intercepts, foci,directrix,asymptotes, ect..) x^2-4y^2+2x=32y=67      Log On

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Question 147722: Please solve, show work and include all relevant points (i.e vertices, intercepts, foci,directrix,asymptotes, ect..)
x^2-4y^2+2x=32y=67
Answer by Fombitz(13828) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2-4y%5E2%2B2x-32y=67
Complete the square in both x and y.
x%5E2%2B2x-4y%5E2-32y=67
%28x%5E2%2B2x%2B1%29-1-4%28y%5E2%2B8y%2B16%29%2B64=67
%28x%5E2%2B2x%2B1%29-4%28y%5E2%2B8y%2B16%29=67%2B1-64
%28x%2B1%29%5E2-4%28y%2B4%29%5E2=4
%28x%2B1%29%5E2%2F4-%28y%2B4%29%5E2=1
This is an equation for a hyperbola which has a general form of
%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1
a=2
b=1
Center of the hyperbola is (h,k) or in this case (-1,-4)
Distance between the foci is 2c where
c%5E2=a%5E2%2Bb%5E2
c%5E2=4%2B1
c=sqrt%285%29
The eccentricity,e, is then
e=c%2Fa=sqrt%285%29%2F2
The foci of the hyperbola are located at
(h+%2B-+c,k)=(-1%2B-sqrt%285%29,-4)
The vertices of the hyperbola are located at
(h+%2B-+a,k)=(-1%2B-2,-4)=(-3,-4) and (1,-4)
The equation of the asymptote lines is given by
y=k+%2B-+%28b%2Fa%29%28x-h%29
y=-4+%2B-+%281%2F2%29%28x%2B1%29
The directrix of a hyperbola is given by
x=h+%2B-+a%5E2%2Fc
x=-1+%2B-+4%2F%28sqrt%285%29%2F2%29
x=-1+%2B-+8%2F%28sqrt%285%29%29