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December 19, 2014

Homework Help: math30

Posted by alejandro on Wednesday, October 10, 2012 at 12:32pm.

The equation of the hyperbola is (x-3)^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:
(x-h)^2/a^2 - (y-k)^2/b^2 =1.

So the (y-k)^2/b^2 term is subtracted the (x-h)^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.
(x-3)^2/4 - (y+1)^2/16 =-1.
16(x-3)^2-4(y+1)^2=-1(16*4)
16(x-3)^2-4(y+1)^2=-63
4(y+1)^2=16(x-3)^2+63
(y+1)^2=[63+16(x-3)^2]/4
y=square[63+16(x-3)^2]/4 -1

Till here I don't know what to do anymore!! help

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