Final answer:
The equation represents an ellipse centered at (-3/16, 0) with vertices at (111/16, 0) and (-31/16, 0) and foci at (-3/16 ± √(261)/16, 0).
Step-by-step explanation:
The equation (x+3/16)^2+y^2/49=1 represents an ellipse. To find the vertices and foci, we must first rewrite the equation in standard form, which is (x-h)^2/a^2 + (y-k)^2/b^2 = 1 where (h,k) are the coordinates of the center of the ellipse, a is the semi-major axis, and b is the semi-minor axis.
For the given ellipse, we can see that it is centered at (-3/16, 0) because the x-term is (x + 3/16), and it has a semi-major axis of 'a' = 7 (since 7^2 = 49) and a semi-minor axis of 'b' = 4/16 = 1/4 (since (4/16)^2 = 1/16). The vertices are at (-3/16 ± a, 0) and (-3/16, ± b).
The distance of the foci from the center is given by the formula c = √(a^2 - b^2), for an ellipse centered at the origin. For this ellipse, we calculate c = √(7^2 - (1/4)^2) = √(49 - 1/16) = √(784/16 - 1/16) = √(783/16) = √(261/16).
Therefore, the foci of the ellipse are at (-3/16 ± √(261)/16, 0), which simplifies to (-3/16 ± √(261)/16, 0). Finally, we can write the coordinates of the vertices and foci:
- Vertices: (-3/16 ± 7, 0) = (111/16, 0) and (-31/16, 0)
- Foci: (-3/16 ± √(261)/16, 0)