Question

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`?

Answer 1

Answer is below

Explanation:

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)`