Question 528281
how do i write an equation for a hyperbola with the give information, vertices (-4,0) (4,0) conjugate axis length of 8
<pre>
The transverse axis is the line segment between the vertices (-4,0) and
(4,0) which is 8 units long.  Its midpoint is the origin.  The transverse 
axis is horizontal so the equation is of the form:

{{{x^2/a^2}}} - {{{y^2/b^2}}} = 1

where a = one-half the transverse axis, which is half of 8 or 4, and
where b = one-half the conjugate axis, which is half of 8 or 4, so
a and b are both 4, so the equation is

{{{x^2/4^2}}} - {{{y^2/4^2}}} = 1

{{{x^2/16}}} - {{{y^2/16}}} = 1

That's the standard form, or you can clear of fractions and get

x² - y² = 16

Edwin</pre>