Consider the hyperbola 16x^2 - 4y^2 + 64 = 0

Consider the hyperbola 16x^2 - 4y^2 + 64 = 0
Write this equation in standard form and give the coordinates of the vertices and the equations of the asymptotes. Then graph.

16x^2 - 4y^2 = 64

(16x^2) / 64 - (4y^2) / 64 = 1

x^2 / 0.25 - y^2 / 0.0625 = 1 < simplified

Let x = 0:

y^2 / 0.25 - (0)^2 / 0.0625 = 1

y = +- 0.5

Let y = 0:

(0)^2 / 0.25 - x^2 / 0.0625 = 1

x = +- 0.25

Now the vertex coordinates are:
(0, 0.5), (0, -0.5), (0.25, 0), (-0.25, 0)

Asymptotes:
y = (0.5/0.25) x

and

y = (-0.5/0.25) x

now how do you graph the coordinates?

Re: Help me with graphing this problem which i already did

NEHA said:

16x^2 - 4y^2 = 64
16x^2 / 64 - 4y^2 / 64 = 1
x^2 / 0.25 - y^2 / 0.0625 = 1 <== No good.

Click to expand...

You wandered off right away on this one.

16/64 = 1/4 NOT 1/(¼).
4/64 = 1/16 NOT 1/(1/16).

NEHA said:

Consider the hyperbola 16x^2 - 4y^2 + 64 = 0...

16x^2 - 4y^2 = 64

Click to expand...

Which equation is it? The first one would be "16x² - 4y² = -64", not "+64", so it is different from the second one.

NEHA said:

now how do you graph the coordinates?

Click to expand...

How do you graph which coordinates? Aren't you supposed to graph the entire hyperbola...?

Please clarify. Thank you.

Eliz.