Question 596617
{{{x^2-4y^2=16}}}
First let's put the equation in standard form:
{{{(x-h)^2/a^2-(y-k)^2/b^2 = 1}}}
Dividing by 16 (to get the 1 on the right):
{{{x^2/16-y^2/4=1}}}
Since x = x - 0 and y = y - 0, we can make these substritutions (so we can see the full form):
{{{(x-0)^2/16-(y-0)^2/4=1}}}<br>
We can now see that the center is (0, 0) and that {{{a^2 = 16}}} and {{{b^2 = 4}}} making a = 4 and b = 2. And since the x is before the minus, this is a horizontally-oriented hyperbola.<br>
With horizontally-oriented hyperbolas, the slopes of the asymptotes are b/a and -b/a, making the slopes for your asymptotes 2/4 and -2/4 or, in reduced form, 1/2 and -1/2.<br>
The asymptotes pass through the center of the hyperbola. So your asymptotes will be:
a line passing through (0, 0) with a slope of 1/2; and
a line passing through (0, 0) with a slope of -1/2<br>
Writing the equations of these lines will be very easy because (0, 0) is not just any point on the asymptotes, but it is the y-intercept for both asymptotes (since the x-coordinate is 0). So we can directly write the slope-intercept forms of these equations:
{{{y = (1/2)x + 0}}} or simply {{{y = (1/2)x}}}
and
{{{y = (-1/2)x + 0}}} or simply {{{y = (-1/2)x}}}