# Find and sketch through the important points, i.e, x, y intercept, vertex, etc.3x^2 -2y^2-6x-12y-27=0

*print*Print*list*Cite

### 1 Answer

To sketch 3x^2 -2y^2-6x-12y-27= 0.

We bring this to the standard form.

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

whose 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) = sqrt(5/2) Or e = -sqrt(5/2)

Focii S and S' 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 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 hyperbola 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).