How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola #y^2/36-x^2/4=1#?

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Answer 1 · 2018-04-01T05:38:26

#"Vertices : "(0,+-6)," Asymptotes : "x+-3y=0, and,#

#Focii : (0,+-2sqrt10)#.

We know that, for the #"Hyperbola S : "y^2/b^2-x^2/a^2=1#,

# (1) :" Vertices are "(0,+-b), (2)" Focii are "(0,+-be)#

# (3) :" the Asymptotes are "y=+-b/a*x#.

The Eccentricity #e# is given by, #a^2=b^2(e^2-1)#.

We have, #b=6, a=2#.

#:.# Vertices are #(0,+-6)#, and the eqns. of the Asymptotes are

# x+-3y=0#.

For #e#, we have, #4=36(e^2-1), or, e=sqrt10/3#.

#:.# Focii are #(0,+-2sqrt10)#.