What are the asymptotes of the hyperbola (x+1)^2/4 - (y-2)^2/1 =1
I'm homeschooled, so the concept is hard to grasp.
Show steps please.
as you know, the hyperbola
x^2/a^2 - y^2/b^2 = 1
has asymptotes of
y = ± b/a x
You have a=2, b=1, so the asymptotes have slope ±1/2.
Also, your hyperbola is shifted, so its center is not at (0,0). Thus, your asymptotes are
y-2 = 1/2 (x+1)
y-2 = -1/2 (x+1)
or,
y = x/2 + 5/2
y = -x/2 + 3/2
google and wikipedia are your friends. A simple search will turn up many useful articles.
Also, wolframalpha is very useful for solving equations and plotting graphs. For this one, see
http://www.wolframalpha.com/input/?i=plot+%28x%2B1%29^2%2F4+-+%28y-2%29^2%2F1+%3D1+%2C+y+%3D+x%2F2+%2B+5%2F2%2C+y+%3D+-x%2F2+%2B+3%2F2
Thank you very much, I understand it a little bit better now...
Sure! I can help you with that. To find the asymptotes of the hyperbola, we need to analyze the equation and determine its standard form.
The standard form of a hyperbola equation is given by:
[(x - h)^2 / a^2] - [(y - k)^2 / b^2] = 1
In the given equation, (x + 1)^2 / 4 - (y - 2)^2 / 1 = 1, we have:
h = -1 (x-coordinate of the center)
k = 2 (y-coordinate of the center)
a^2 = 4 (square of the semi-major axis length)
b^2 = 1 (square of the semi-minor axis length)
Now, let's determine the slope of the asymptotes. The slope of the asymptotes is given by the formula ±b/a.
Substituting the values of a and b, we have:
Slope of the asymptotes = ±√(b^2 / a^2)
= ±√(1 / 4)
= ±1/2
Therefore, the slopes of the asymptotes are ±1/2.
Next, we need to find the equations of the asymptotes. The general equation of a straight line with slope m and passing through a point (x1, y1) is given by:
(y - y1) = m(x - x1)
Since the asymptotes pass through the center of the hyperbola (h, k), our point is (-1, 2).
For the first asymptote (with positive slope), m = 1/2:
(y - 2) = (1/2)(x + 1)
Simplifying this equation:
2y - 4 = x + 1
2y = x + 5
y = (1/2)x + (5/2)
For the second asymptote (with negative slope), m = -1/2:
(y - 2) = (-1/2)(x + 1)
Simplifying this equation:
2y - 4 = -x - 1
2y = -x + 3
y = (-1/2)x + (3/2)
Therefore, the equations of the asymptotes are:
y = (1/2)x + (5/2)
y = (-1/2)x + (3/2)
These are the equations of the asymptotes for the given hyperbola.