Hi i was given the equation (y-2)^2 - x^2/4 =1. I need to find the center, vertices and asymptotes of this hyperbola. I found the center (0,2) and the vertices were tricky but I think they are (0,1) (0,3) but I'm having trouble finding the asymptotes. Can you please help me find what they are thanks.
To find the asymptotes of a hyperbola, we can start by rewriting the equation in a standard form. The given equation is:
(y - 2)^2 - x^2/4 = 1
Let's rearrange the equation to isolate the y term:
(y - 2)^2 = 1 + x^2/4
Next, take the square root of both sides to solve for y:
y - 2 = ± √(1 + x^2/4)
Now, add 2 to both sides:
y = 2 ± √(1 + x^2/4)
The equation of the asymptotes for a hyperbola in standard form with center (h, k) is given by:
y = k ± a/b * (x - h)
where a and b are the lengths of the transverse and conjugate axes, respectively.
From the standard form of a hyperbola, we can see that the transverse axis is along the y-axis, and the conjugate axis is along the x-axis. Thus, a = √4 = 2 and b = √1 = 1.
The center of the hyperbola is (0, 2), as you correctly found.
Plugging in the values, we get:
y = 2 ± 2/1 * (x - 0)
y = 2 ± 2x
So, the equations of the asymptotes of the hyperbola are:
y = 2 + 2x
y = 2 - 2x
Hence, the asymptotes of the given hyperbola are y = 2 + 2x and y = 2 - 2x.