How do you graph the hyperbola #y^2/16-x^2/9=1#?

1 Answer
Aug 8, 2017

Answer:

Crosses at #y=+-4#

Explanation:

As this equation is already in the correct format, put #x and y# equal to zero and you will see that when #x=0#, #y^2=16# which gives the y intercepts.

If you put #y=0#, you get #x^2=-9# which doesn't give any real roots. The graph will therefore be a hyperbola in the shape of a V and an upside down V (as opposed to being on their sides like ><).

You can use the formula: #x/a = +-y/b# to determine the equations of the asymptotes where #a=sqrt(9), b=sqrt(16)# and you will get #y=+-(4/3)x#