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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.
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.

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cutiecomittee123
 one year ago
Best ResponseYou've already chosen the best response.0Vertical (0,0) How do i find a and b?

cutiecomittee123
 one year ago
Best ResponseYou've already chosen the best response.0and the equation?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0asymptotes \(y =\pm \dfrac{a}{b}\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0hiihiii, y = + (a/b) x

cutiecomittee123
 one year ago
Best ResponseYou've already chosen the best response.0okay and the equation?

Loser66
 one year ago
Best ResponseYou've already chosen the best response.0You have formula, right? just plug the a and the b in , DAT SIT

cutiecomittee123
 one year ago
Best ResponseYou've already chosen the best response.0x^2/16+y^2/9=1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So, because the vertices are at different ycoordinates, that is how you know your hyperbola is oriented up and down. The xcoordinates 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 xcoordinate doesn't change, the center is at (0,0). Now if the ycoordinates 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 avalue comes from the fraction that is positive and the bvalue comes from the fraction that is negative. Since our hyperbola opens up, this means that the fraction containing the yvarable 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: \[\frac{ (yk)^{2} }{ a^{2} }  \frac{ (xh)^{2} }{ b^{2} } = 1\] 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.

cutiecomittee123
 one year ago
Best ResponseYou've already chosen the best response.0awesome thanks for that! that will definitely help me with my big test coming up!:)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You'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.
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