## cutiecomittee123 one year ago A hyperbola has vertices (0,-6) and (0,6) and asymptotes y=3/4x and y=-3/4x In which direction is the hyperbola oriented? What are the coordinates pf the center of this hyperbola? What are the values of a and b for this hyperbola? Write the equation of this hyperbola.

1. Loser66

|dw:1433643256376:dw|

2. cutiecomittee123

Vertical (0,0) How do i find a and b?

3. cutiecomittee123

and the equation?

4. cutiecomittee123

@Loser66

5. Loser66

asymptotes $$y =\pm \dfrac{a}{b}$$

6. Loser66

hiihiii, y = +- (a/b) x

7. Loser66

hence a = 3, b =4

8. cutiecomittee123

okay and the equation?

9. cutiecomittee123

@Loser66

10. Loser66

You have formula, right? just plug the a and the b in , DAT SIT

11. cutiecomittee123

x^2/16+y^2/9=1

12. anonymous

So, because the vertices are at different y-coordinates, that is how you know your hyperbola is oriented up and down. The x-coordinates would be different should the hyperbola be oriented left and right. The two vertices are equidistant from the center. So since one vertex is at y = 6 and the other is at y = -6, you can see that the center would be right in between, at y = 0. Since the x-coordinate doesn't change, the center is at (0,0). Now if the y-coordinates were a bit more complex, what we would do is find the difference between the two. As in we would do 6-(-6) = 12. Then we would divide by two. In this case, dividing by 2 gets you 6. This 6 value is how far away the center is from each vertex. So you could do that as well to find your center. In a hyperbola, the a-value comes from the fraction that is positive and the b-value comes from the fraction that is negative. Since our hyperbola opens up, this means that the fraction containing the y-varable is positive and thus a will come from this fraction. Since a comes from the fraction with y, a will represent a vertical distance from the center while b will represent a horizontal distance from the center. Now if x were positive, then it would be reversed and a would be the horizontal distance and b would be the vertical distance. So since a represents vertical distance here, a must be equal to 6, since is this how far we go up and down to reach a vertex. Since our hyperbola does not open left and right, we have no landmark to tell us what the value of b might be. This is where the asymptotes help us. As you know, slope is rise over run and the asymptotes given have a slope of 3/4 or -3/4. Well, since a is a distance up and down (a rise) and b is a distance left and right (a run), we can figure out what a is via this relationship $$\frac{3}{4} = \frac{rise}{run} = \frac{6}{b}$$ Solving this equation for b, we would get b = 8. Now the general equation of a hyperbola that opens up (y is positive) is this: $\frac{ (y-k)^{2} }{ a^{2} } - \frac{ (x-h)^{2} }{ b^{2} } = 1$ So (h,k) is the center, which is (0,0). a is 6 and b is 8, thus a^2 is 36 and b^2 is 64. Therefore your equation is this: $\frac{ y^{2} }{ 36 }-\frac{ x^{2} }{ 64 } = 1$ Thats a lot of info, so hopefully it makes sense.

13. cutiecomittee123

awesome thanks for that! that will definitely help me with my big test coming up!:)

14. anonymous

You're welcome. I want to make a correction to something loser said, though, which brings up a common mistake. Even though the asymptotes in this equation are defined as $$y = \pm \frac{a}{b}$$, that doesnt mean a and b are equal to the values you see. the asymptotes having a slope of $$\pm\frac{3}{4}$$ just mean that a could be any multiple of 3 and b could be any multiple of 4, it does not mean a is 3 and b is 4. Notice how I found a to be 6 and b to be 8, which just happens to reduce to 3/4. Just wanted to point that out.