find and sketch important points ( i.e, x, y intercept, vertex, etc.)3x^2 - 2y^2-6x-12y-27=0

To sketch  3x^2 -2y^2-6x-12y-27=
0.

We bring this to the standard form (x-h)a^2+(y-k)^2/b^2 =
1.

3x^2-6x  - (2y^2 +12y) =
27.

3(x^2-2y )  -  2(y^2 +6y) =
27.

3(x^2-2y+1)-3 - 2(y^2+6y+3^2) + 18 =
27

3(x-1)^2 -2(y+1)^2 = 27 -18+3 =
12.

3(x-1)^2/(12) - 2(y+1)/12 =
12.

(x-1)/(12/3) - 2(y+1)/(12/2) =
1.

(x-1)^2/2^2 - (y+1)^2 / 6 = 1.

This
is a hyperbola of the form (x-h)^2/a^2 -(y-k)^2/k^2 = 1.

Its centre
is (h,k) = (1, -1).

The length of axes are 2a and 2b  and in our
case 2*2 = 4 and 2sqrt6.

The vertices are at (1+4 ,-1 ) and (1-4 ,
-1) or at (5,-1) and (-3 , -1).

Accentricity is e given by e^2 =
1+b^2/a^2 = 1+6/4 = 5/2. So e = sqrt(10/4) = +or- sqrt(5/2)

Focii S
ans S' are  are at  (ae+h , k) = (4sqrt(5/2) + 1 , -1) = (10 , -1)  and S' = (-ae+1 , -1) = (-
4sqrt(5/2) , -1).

The equations of the directrices : x = a/e Or x=
4/sqrt(5/2) . And x = -a/e . Or x = -4/sqrt(5/2).

The x intercepts
are given by putting y = 0 and solving for x in the given equation of the
hyperbola:

3x^2-6x-27 = 0.

x^2 -2x-9 =
0

x1 = {2+sqrt(2^2+36}/2 = 1+2sqrt10
and

x2 = 1- 2sqrt10.

y intercepts are
got by putting y = 0 in the given parabola and solving for
y:

-2y^2-12y-27 = 0.

2y^2+12y+27 =
0.

 y1 = {-12 +sqrt(12^2-4*27)}/2*2 = - 3 + 3sqrt
3.

 y2 = -(3+3sqrt3).