SOLUTION: find the vertices and foci of the hyperbola 4x²-y²-48x-4y+40=0

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: find the vertices and foci of the hyperbola 4x²-y²-48x-4y+40=0       Log On


   

Question 724435: find the vertices and foci of the hyperbola 4x²-y²-48x-4y+40=0
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
find the vertices and foci of the hyperbola
4x²-y²-48x-4y+40=0
4x^2-48x-y^2-4y=-40
complete the square:
4(x^2-12x+36)-(y^2+4y+4)=-40+144+4
4(x-6)^2-(y+2)^2=108
%28x-6%29%5E2%2F27-%28y%2B2%29%5E2%2F108=1
This is an equation of a hyperbola with horizontal transverse axis
Its standard form: %28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1, (h,k)=(x,y)coordinates of the center.
For given equation:
center: (6,-2)
a^2=27
a=√27≈5.2
b^2=108
c^2=a^2+b^2=27+108=135
c=√135≈11.6
vertices: (6±a,-2)=(6±5.2,-2)=(.8,-2) and (11.2,-2)
foci: (6±c,-2)=(6±11.6,-2)=(-5.6,-2) and (17.6,-2)