Sunday
April 20, 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

Answer this Question

First Name:
School Subject:
Answer:

Related Questions

Algebra 2 Mannn - Can someone please explain to me how to get the 16 in this ...
algebra 2 - Write an equation in standard form for the hyperbola with center (0,...
algebra - a hyperbola has vertices (+-5,0) and one focus (6,0) what is the ...
Algebra 2 - a hyperbola is centered at (3,7). The vertices are (9,7) and (-3,7...
algebra - Can someone please help... a hyperbola has vertices (+-5,0) and one ...
maths - a hyperbola of eccentricity 3/2 has one focus at (1,-3). The ...
Maths Important - How would you draw the graph of this x^2/16 + y^2 = 1 ( this ...
math - the vertices are at (2,1) and (2,7) and focus is at (2,*) write the ...
algebra - 1)What is the 4x^2=y^2+8y+32 answer= either hyperbola or parabola 2)...
Math Word Problem - A cross section of a nuclear cooling tower is a hyperbola ...

Search
Members