How do you find the coordinates of the vertices, foci, and the equation of the asymptotes for the hyperbola #y^2-3x^2+6y+6x-18=0#?

Question

1962 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-19T14:24:18

Answer is below

Given equation of hyperbola:

#y^2-3x^2+6y+6x-18=0#

#(y^2+6y+9)-3(x^2-2x+1)-9+3-18=0#

#(y+3)^2-3(x-1)^2=24#

#\frac{(y+3)^2}{24}-\frac{3(x-1)^2}{24}=1#

#\frac{(y+3)^2}{(2\sqrt6)^2}-\frac{(x-1)^2}{(2\sqrt2)^2}=1#

The above vertical hyperbola: #y^2/a^2-x^2/b^2=1# has

Center: #(x-1=0, y+3=0)\equiv(1, -3)#

Eccentricity: #e=\sqrt{1+b^2/a^2}=\sqrt{1+8/24}=2/\sqrt3#

The vertices: #(x-1=0, y+3=\pma)\equiv(1, -3\pm2\sqrt6)# &

#(x-1=\pm b, y+3=0)\equiv(1\pm 2\sqrt2, -3)#

Focii: #(x-1=0, y+3=\pm ae)\equiv(1, -3\pm4\sqrt2)#

Asymptotes: #y=\pma/b x#

#y+3=\pm({2\sqrt6}/{2\sqrt2})(x-1)#

#y=-3\pm\sqrt3(x-1)#