Posted by alejandro on Wednesday, October 10, 2012 at 12:32pm.
The equation of the hyperbola is (x3)^2/4  (y+1)^2/16 =1. What is the range?
All I know is this but I am not even sure what I am doing!!
My answer:
The general form of the equation of a horizontally aligned hyperbola is:
(xh)^2/a^2  (yk)^2/b^2 =1.
So the (yk)^2/b^2 term is subtracted the (xh)^2/a^2 term.
(h, k) is the center of the horizontally aligned hyperbola.
Center (3, 1)
a : is the distance from the center of the hyperbola to each vertex of the hyperbola.
Each vertex of the hyperbola lies on the transverse axis of the hyperbola.
Solve the equation for y in terms of x.
(x3)^2/4  (y+1)^2/16 =1.
16(x3)^24(y+1)^2=1(16*4)
16(x3)^24(y+1)^2=63
4(y+1)^2=16(x3)^2+63
(y+1)^2=[63+16(x3)^2]/4
y=square[63+16(x3)^2]/4 1
Till here I don't know what to do anymore!! help

math30  Steve, Wednesday, October 10, 2012 at 2:10pm
since it's (x3)^2/4  (y+1)^2/16 = 1
that's the same as
(y+1)^2/16  (x3)^2/4 = 1
so the hyperbola is vertical. The xvalues go from ∞ to ∞, and so do the yvalues, but the interval between the vertices (5,3) is excluded.
Look up any web page explaining the properties of hyperbolas (or maybe your textbook) to read about the values of h,k and a,b,c and the asymptotes.