What are the vertices, foci, and asymptotes of the hyperbola with the equation 16x2–4y2=64 ?
To find the vertices, foci, and asymptotes of the hyperbola defined by the equation 16x^2 - 4y^2 = 64, we need to first rewrite the equation in standard form:
16x^2/64 - 4y^2/64 = 1
x^2/4 - y^2/16 = 1
This equation represents a hyperbola with a transverse axis parallel to the x-axis.
Now we can determine the center which is at (0,0) and calculate the vertices, foci, and asymptotes.
- Vertices:
The distance from the center to the vertices along the x-axis is a = 2, so the vertices are located at (2,0) and (-2,0).
- Foci:
To find the foci, we can use the formula c^2 = a^2 + b^2, where c is the distance from the center to the foci. In this case, b^2 = 16, so b = 4. Then, c^2 = 2^2 + 4^2 = 4 + 16 = 20. Taking the square root of 20 gives us √20, so the foci are located at (√20,0) and (-√20,0).
- Asymptotes:
The equation of the asymptotes for a hyperbola centered at (h,k) and with semi-major axis a and semi-minor axis b is given by the formula (y-k) = ±(b/a)(x-h).
In this case, the equation becomes y = ±(4/2)x or y = ±2x. Thus, the asymptotes are y = 2x and y = -2x.
In summary:
- Vertices: (2,0) and (-2,0)
- Foci: (√20,0) and (-√20,0)
- Asymptotes: y = 2x and y = -2x