How do you find the center, vertices, foci and asymptotes of #(x^2/16)-(y^2/9)=1#?

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Answer 1 · 2018-05-01T13:07:39

Center #(0,0)#, Vertices are #-4,0) and (4,0)#
Foci are # (-5,0) and (5,0)# and equation of
asymptotes are # y= +-3/4#

# (x^2/4^2-y^2/3^2=1#

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 : # Center #(0,0),a=4 , b=3#

The vertices are #a=4# units from the center, and the foci are

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

#:. c^2=16+9 =25 :. c = +-5 # Vertices are #-4,0) and (4,0)#

Foci are # (-5,0) and (5,0)# The asymptotes pass through the

vertices of a rectangle of dimensions #2 a=8 and 2 b= 6 # with its

centre at #(0,0) :.# slope #=+-b/a= +-3/4)# Equation of

asymptotes are is #y-0= +-3/4(x-0) or y= +-3/4#

graph{x^2/16-y^2/9=1 [-10, 10, -5, 5]} [Ans]