+466 The general equation of a hyperbola is 36x2 - 9y2 - 24x + 54y - 113 = 0. Write the equation of this hyperbola in standard form and give the coordinates of the center, the coordinates of the vertices, and the equations of the asymptotes.
I did learn how to solve these equations, but I can't get past putting these into S.F.
+128475 36x2 - 9y2 - 24x + 54y - 113 = 0 add 113 to each side
36x^2 - 9y^2 - 24x + 54y = 113 complete the square on x and y
36 [x^2 - (2/3)x + 1/9] - 9 [y^2 - 6y + 9] = 113 + 4 - 81 factor and simplify
36(x - 1/3)^2 - 9 (y - 3)^2 = 36 divide both sides by 36
( x - 1/3)^2 - (y - 3)^2 = 1
1 4
That's SF, Shades.....I think what may be giving you problems is completing the square, particulalrly when the lead coefficient isn't = 1.......here's a resource that might help.......http://regentsprep.org/Regents/math/algtrig/ATE3/quadcompletesquare.htm
Center = (1/3, 3)
Vertices = [x coordinate of the center ± √ [c^2 - b^2] , y coordinate of the center ]
In this case c = √5 and b = 2 so c^2 = 5 and b^2 = 4
[ 1/3 ± √[5 - 4] , 3 ] = [ 1/3 + 1 , 3] and [ 1/3 -1, 3] = [ 4/3, 3] and [-2/3, 3]
Asymptotes
y = ± 2/1 (x - 1/3) + 3
y = ± 2( x - 1/3) + 3
y = 2x -2/3 + 3 and y = -2x + 2/3 + 3
y = 2x + 7/3 and y = -2x + 11/3
Here's the graph with the pertinent info, Shades : https://www.desmos.com/calculator/l7ew2an7qj
+128475 36x2 - 9y2 - 24x + 54y - 113 = 0 add 113 to each side
36x^2 - 9y^2 - 24x + 54y = 113 complete the square on x and y
36 [x^2 - (2/3)x + 1/9] - 9 [y^2 - 6y + 9] = 113 + 4 - 81 factor and simplify
36(x - 1/3)^2 - 9 (y - 3)^2 = 36 divide both sides by 36
( x - 1/3)^2 - (y - 3)^2 = 1
1 4
That's SF, Shades.....I think what may be giving you problems is completing the square, particulalrly when the lead coefficient isn't = 1.......here's a resource that might help.......http://regentsprep.org/Regents/math/algtrig/ATE3/quadcompletesquare.htm
Center = (1/3, 3)
Vertices = [x coordinate of the center ± √ [c^2 - b^2] , y coordinate of the center ]
In this case c = √5 and b = 2 so c^2 = 5 and b^2 = 4
[ 1/3 ± √[5 - 4] , 3 ] = [ 1/3 + 1 , 3] and [ 1/3 -1, 3] = [ 4/3, 3] and [-2/3, 3]
Asymptotes
y = ± 2/1 (x - 1/3) + 3
y = ± 2( x - 1/3) + 3
y = 2x -2/3 + 3 and y = -2x + 2/3 + 3
y = 2x + 7/3 and y = -2x + 11/3
Here's the graph with the pertinent info, Shades : https://www.desmos.com/calculator/l7ew2an7qj
+128475 Sorry, Shades...I made this a little more complicated than I should have.....c is the focal distance = sqrt(5) = sqrt [ a^2 + b^2] = sqrt [1^2 + 2^2] = sqrt [1 + 4 ] = c
So, sqrt [c^2 - b^2] is just = a ......which, in this case = 1
So...I should have simplified the location of the vertices as
[ h + a , k ] and [ h - a, k] where h = 1/3 and k = 3 and a = 1
So....we get he same result as I gave originally
The vertices are located at
[1/3 + 1 , 3 ] and [ 1/3 - 1, 3] which simpliy to
[4/3 , 3 ] and [ -2/3, 3 ] as before
Sorry for the "overkill"
