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

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justaguide | College Teacher | (Level 2) Distinguished Educator

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Here we are given the equation: 3x^2 - 2y^2 - 6x - 12y - 27= 0.

First let's rewrite this as follows:

3x^2 - 6x - 2y^2 - 12y = 27

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

Let's complete the squares

=> 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

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

divide all terms by 12

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

=> (x-1)^2 / 4 - (y+1)^2/ 6 = 1

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

Now this is the standard equation of a hyperbola of the form  (x - h)^2/a^2 - (y - k)^2/k^2 = 1, where the center is ( h , k)

Here it is ( 1, -1)

The length of axes for the given hyperbola are 2*2 = 4 and 2*sqrt 6

The vertices are: (5,-1) and (-3 , -1)

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neela | High School Teacher | (Level 3) Valedictorian

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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).

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