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

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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: $y=\pma/b x$

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

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

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