Questions about hyperbolas

[Hygelac]

Hey, I have a couple of _easy_ questions about hyperbolas, but its been a while since I have worked with them and am not able to look them up in my math book currently...if someone could just get me started in the right direction, I would really appreciate it :)

Given the equation of the hyperbola: 9(x-2)^2 - 16(y-4)^2 = 144

1. In which quadrant is its center?

2. What are its x/y intercepts?

Again, if you could just get me started, that would be great

 

[Sirus]

Since you haven't shown any attempt at solving the problem, I will just get you started. The generalized equation for a horizontal hyperbola is:
$$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$$
where $(h,k)$ is the center, a is the horizontal distance from the center to each vertex, and b is the vertical distance from the center to a point on one of the asymptotes $(x,y)$ such that x is the x-coordinate of a vertex.

 

[tacman]

1. To find the center of a hyperbola, we need to simplify the equation to the standard form: (x-h)^2/a^2 - (y-k)^2/b^2 = 1, where (h,k) is the center of the hyperbola. In this case, we have (x-2)^2/16 - (y-4)^2/9 = 1. So, the center is at (2,4). Since both the x and y coordinates are positive, the center is in the first quadrant.

2. To find the x-intercepts, we set y=0 in the equation and solve for x. 9(x-2)^2 - 16(0-4)^2 = 144. Simplifying, we get 9(x-2)^2 = 144, which gives us two solutions: x=2+4=6 and x=2-4=-2. So, the x-intercepts are (6,0) and (-2,0).

To find the y-intercepts, we set x=0 in the equation and solve for y. 9(0-2)^2 - 16(y-4)^2 = 144. Simplifying, we get -16(y-4)^2 = 144, which gives us two solutions: y=4+3=7 and y=4-3=1. So, the y-intercepts are (0,7) and (0,1).

I hope this helps you get started with your questions about hyperbolas. Remember, the key is to simplify the equation to its standard form and then use that information to find the center and intercepts. Good luck!