+466 x2 - y2 - 2x - 4y - 4 = 0 is the general equation of a hyperbola. 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.
Please help, I also need this procedure broken down, my book does a poor job with explanation.
+130474 This is going to be similar to the ellipse one, Shades.....just a few minor differences
Standard form looks similar to an ellipse except that a minus separates the first two terms instead of a plus
(x - h)^2 - (y - k)^ 2 = 1
a^2 b^2
x^2 - y^2 - 2x - 4y - 4 = 0 we are gong to complete the square on x and y again.....this one is a little easier to do...
x^2 - 2x + 1 - [y^2 + 4y + 4] = 4 + 1 - 4 simplfy
(x^2 - 1)^2 - ( y + 2)^2 = 1
Note that we can write this as :
(x^2 - 1)^2 - ( y + 2)^2 = 1
1 1
And now, it's in S.F.
The center is ( 1, -2)
The vertices lie at (0, -2) and (2, -2)
The equations of the asymptotes are, in this case, are given by :
y = ±(b/a) (x- h) + k where a,b = 1 h = 1 and k = -2 ....so we have....
y = ±(1/1) (x - 1) - 2 which simplifies to :
y = ± (x -1) - 2
Here's the graph with the asymptotes added : https://www.desmos.com/calculator/i33hddox2i
Here's a pretty good resource for this one, as well
:http://www.purplemath.com/modules/hyperbola.htm
