hyperbola with vertices at (0,+/-2), and asymptotes y=+- 1/2x
since the vertices are on the y-axis, we will have
y^2/a^2 - x^2/b^2 = 1
the slope of the asymptotes is b/a, so
y^2 - x^2/4 = 1
but that has vertices at (0,+/-1) so
y^2/4 - x^2/16 = 1
To verify, see
http://www.wolframalpha.com/input/?i=hyperbola+y%5E2%2F4+-+x%5E2%2F16+%3D+1
To sketch the hyperbola with vertices at (0, +/-2) and asymptotes y = +/- 1/2x, we can follow these steps:
Step 1: Plot the vertices
The vertices of the hyperbola are located at (0, +/-2). Mark these points on the coordinate plane as (0, 2) and (0, -2).
Step 2: Plot the asymptotes
The asymptotes of the hyperbola are given by the equations y = +/- 1/2x. These are straight lines that intersect at the origin (0, 0). Draw these lines on the coordinate plane.
Step 3: Determine the distance between the vertices
The distance between the vertices is called the "transverse axis" of the hyperbola. In this case, the distance between the vertices is 2 + 2 = 4.
Step 4: Determine the distance between the center and the vertices
The distance between the center of the hyperbola and the vertices is called the "semi-major axis". In this case, since the center is at (0, 0) and the vertices are at (0, +/-2), the semi-major axis is 2.
Step 5: Sketch the hyperbola
Now, let's sketch the hyperbola using the information we have gathered. The hyperbola will be symmetric with respect to the x-axis and the y-axis. We know that the center is at (0, 0), and the vertices are at (0, +/-2). Connect the vertices with smooth curves that pass through the asymptotes. These curves should be slightly steeper and closer to the asymptotes near the center, and gradually become more horizontal as they move away from the center.
The final sketch will show a hyperbola with the center at the origin, vertices at (0, +/-2), and asymptotes y = +/- 1/2x.
To understand the properties of a hyperbola with given vertices and asymptotes, we can start by understanding the standard form equation of a hyperbola.
The standard form equation of a hyperbola with vertical transverse axis is:
((y - k)^2 / a^2) - ((x - h)^2 / b^2) = 1
Where:
- The center of the hyperbola is given by the coordinates (h, k).
- The length of the transverse axis is given by 2a, and it extends vertically.
- The length of the conjugate axis is given by 2b, and it extends horizontally.
In this case, the vertices are located at (0, +/−2), which means the center of the hyperbola is at (0, 0), as it lies at the midpoint between the two vertices.
Given that the asymptotes are y = ±1/2x, we can determine the value of b.
The slope of the asymptotes is given by ±(a/b), so we can write:
±1/2 = ±(a/b)
Simplifying the equation, we get:
b = ±2a
Since the conjugate axis has a length of 2b, we have:
2b = ±4a
Since we already know that the vertices lie at (0, +/−2), and the center is at (0, 0), we can deduce that the value of a is 2. This is because the distance between the center and the vertices represents a.
Now, let's calculate the equation of the hyperbola using the standard form.
((y - k)^2 / a^2) - ((x - h)^2 / b^2) = 1
Plugging in the values:
k = 0
h = 0
a = 2
b = ±4a = ±8
The equation of the hyperbola becomes:
(y^2 / 4) - (x^2 / 64) = 1
Therefore, the equation of the hyperbola with given vertices at (0, +/−2) and asymptotes y = ±1/2x is:
(y^2 / 4) - (x^2 / 64) = 1