How would I find the equation for the asymptotes of the hyperbola y[squared]/4 - x[squared]/16=1?
Umm, it's the season for hyperbolas now. The parabolas went silent.
[y^2 / 4] -[x^2 / 16] = 1 -----------(1)
The x^2 term is negative, so the hyperbola does not open left and right ---not along the x-axis, not "horizontal".
So the parabola is "vertical", or its focal axis is vertical, or it opens up and down.
[You can key your clue on the positive variable-squared ---the y^2 here---to mean the focal axis is always there. But I am used to key mine on the negative variable-squared as the parabola not opening along there.]
The semi-transverse axis, a, is "under" the positive variable-squared always---because the transverse axis is always along the focal axis.
The asymptotes are straight lines passing through the center of the hyperbola and the corners of the "rectangle" between the vertices. Here, the slopes of the two asymptotes are +,-a/b, since "a" here is vertical and b is horizontal.
The standard form of the equation of a vertical hyperbola is
[(y-k)^2 / a^2] -[(x-h)^2 / b^2] = 1 ------------------------(i)
where
(h,k) is the center
a is semi-transverse axis
b is semi-cojugate axis
In the given hyperbola:
-----the x^2 and the y^2 have no "companions", so the center (h,k) is (0,0)
-----the a^2 is 4, so a = +,-2.
-----the b^2 is 16, so b = +,-4
So the slopes of the asymptotes are
m = +,-a/b = +,-2/4 = +,-1/2.
We have a point, (0,0), and a slope for any of the two asymptotes, so, using the point-slope form of the equation of a line,
(y -y1) = m(x -x1),
(y -0) = +,-(1/2)(x-0)
y = +,-x/2 ----------------the asymptotes, answer.
Or,
y = x/2 and y = -x/2.
Hello, Gretchen!
How would I find the equation for the asymptotes of the hyperbola: .y²/4 - x²/16 .= .1 ?
There is a formula for this problem.
Given the hyperbola: .x²/a² - y²/b² .= .± 1
. . the equations of the asymptotes are: .y .= .±(b/a)x
You problem has: .a = 4, b = 2 . . . Hence: b/a = ½
The asymptotes are: .y .= .± ½x