graphing a hyperbola

[softball1145]

if you have the following problem:

x^2/16 - y^2/4 = 1 solve as following:

(+-2 to the square root of 5)^2/16 - y^2/4 = 1

20/16 - y^2/4 = 1

5/4 - y^2/4 = 1

y^2/4 = 1/4

y = +-1


now would that be the same for the problem y^2/16 - x^2/4 = 1

 

[galactus]

What is it you want to do?. Solve for y so you can graph?.

\(\displaystyle \L\\\frac{x^{2}}{16}-\frac{y^{2}}{4}=1\)

LCD is 16:

\(\displaystyle \L\\x^{2}-16=4y^{2}\)

\(\displaystyle \L\\y=\pm\frac{\sqrt{x^{2}-16}}{2}\)

The lines \(\displaystyle y=\pm\frac{b}{a}x\) are asymptotes.

Foci can be found by \(\displaystyle a^{2}+b^{2}=c^{2}\)

 

[softball1145]

hyperbola

yes i need to graph the hyperbola, label the center, 2 vertices, 2 foci, and 2 asymptotes.