Graph ((y+3)^2)/9 - ((x-2)^2)/16 = 1?

Question

psymon · Answer

Well, the center would be (x-h) and (y-k). So your y-coordinate for the center would be (y-(-3)) and the x-coordinate would be (x-2). Now the negative sign is supposed to be there, so that means that the 2 there is actually positive. Thats part 1 xD

anonymous · Answer

Okay so I know the center is (2, -3). I kinda forgot the formula for finding vertices and foci of hyperbolas NOT on the origin. I just want to know the formulas and I can do the rest. I can graph it easily..

anonymous · Answer

Lolz, okay!

anonymous · Answer

"Part I: Identify the coordinates of the center of this hyperbola.
Part II: Use the values of a and b to locate the coordinates of the vertices.
Part III: What are the coordinates of the foci?
Part IV: Graph this hyperbola using the graph shown below. Clearly label the locations of the foci and the vertices. You do not need to label their coordinates.
**Part V. Write the equations of the two asymptotes"

psymon · Answer

Oh, lol. Well, when y^2 comes first (the one that is positive), your verticies are up and down a units from where your center is. So it looks like a^2 is going to be 9, meaning a is 3. As for the foci, those are just:
$$c = \sqrt{(a)^{2}+(b)^{2}}$$ Now the foci go inside of the bowl shapes that the hyperbola makes, so basically those will be up and down from the center, too. As for verticies, when y^2 is the first term (positive term), the asymptotes are:
$$y = \pm\frac{ ax }{ b }$$

anonymous · Answer

terrible drawing, but oh well |dw:1376195247643:dw|