How do you find the coordinates of the vertices and foci and the equations of the asymptotes for the hyperbola #(y+6)^2/20-(x-1)^2/25=1#?

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Answer 1 · 2017-01-11T03:24:35

Given: #(y-k)^2/a^2-(x-h)^2/b^2=1" [1]"#
vertices: #(h,k+-a)#
foci:#(h,k+-sqrt(a^2+b^2))#
asymptotes: #y=+-a/b(x-h)+k#

This is a good reference for Hyperbolas

Write the given equation in the same form as equation [1]:

#(y - -6)^2/(2sqrt5)^2 - (x - 1)^2/5^2 = 1" [2]"#

The vertices are:

#(h,k-a) and (h,k+a)#

#(1,-6-2sqrt5) and (1,-6+2sqrt5)#

The foci are:

#(h,k-sqrt(a^2+b^2)) and (h,k+sqrt(a^2+b^2))#

#(1,-6-sqrt(20+25)) and (1,-6+sqrt(20+25))#

#(1,-6-sqrt(45)) and (1,-6+sqrt(45))#

The equations of the asymptotes are:

#y=-a/b(x-h)+k# and #y=+a/b(x-h)+k#

#y=-2sqrt5/5(x-1)-6# and #y=2sqrt5/5(x-1)-6#