How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola #(x+1)^2/4-(y+3)^2/9=1#?

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Answer 1 · 2018-03-25T13:03:00

The vertices are #(-3,-3) and (1,-3)# , focii are
#((-1-sqrt13),-3) and ((-1+sqrt13),-3)# asymptotes
are #y=3/2x-3/2 and y= -3/2x-9/2#

#(x+1)^2/4- (y+3)^2/9=1; h=-1,k=-3,a=2 , b= 3#

This is standard form of the equation of a hyperbola with center

#(h,k) ; (x-h)^2/a^2-(y-k)^2/b^2=1 :. # Centre #(-1,-3)#

The vertices are #a# units from the center, and the foci are

#c# units from the center. Moreover #c^2=a^2+b^2#

c #((-1-2),-3) and ((-1+2,3) # or

#(-3,-3) and (1,-3) ; c^2= 2^2+3^2=13#

#:. c= +- sqrt13 :. # Focii are at

#((-1-sqrt13),-3) and ((-1+sqrt13),-3)#

Each hyperbola has two asymptotes that intersect at the center of

the hyperbola. The asymptotes pass through the vertices of a

rectangle of dimensions #2a=4 and 2b=6 # with its

centre at #(-1,-3) :.# slope #+-b/a=+- 3/2# Equation of

asymptotes are #y+3= +-3/2(x+1) or y=-3+-3/2(x+1)#

or #y=3/2x-3/2 and y= -3/2x-9/2# [Ans]