Final answer:
The vertices of the ellipse (x²)/(49) + (y²)/(9) = 1 are at the points (7, 0) and (-7, 0), and the co-vertices are at the points (0, 3) and (0, -3).
Step-by-step explanation:
To identify the vertices and co-vertices of the ellipse given by the equation (x²)/(49) + (y²)/(9) = 1, we need to recognize the standard form of an ellipse's equation, which is (x²)/(a²) + (y²)/(b²) = 1, where 'a' and 'b' are the lengths of the semi-major and semi-minor axes, respectively.
For our ellipse, 'a' is the square root of the larger denominator, so a = √49 = 7, and 'b' is the square root of the smaller denominator, so b = √9 = 3.
The vertices are the points where the ellipse intersects the major axis. In this case, the major axis is horizontal and the vertices are at (a, 0) and (-a, 0), giving us the points (7, 0) and (-7, 0).
The co-vertices are the points where the ellipse intersects the minor axis. In this case, the minor axis is vertical, and the co-vertices are at (0, b) and (0, -b), giving us the points (0, 3) and (0, -3).