Answer:
To find the vertices and name two points on the minor axis of the ellipse represented by the equation 9x^2+y^2-18x-6y+9=0, we need to first put it in standard form by completing the square for both x and y terms.
Starting with the x terms:
9x^2 - 18x = 0
9(x^2 - 2x) = 0
We need to add and subtract (2/2)^2 = 1 to complete the square inside the parentheses:
9(x^2 - 2x + 1 - 1) = 0
9((x-1)^2 - 1) = 0
9(x-1)^2 - 9 = 0
9(x-1)^2 = 9
(x-1)^2 = 1
x-1 = ±1
x = 2 or 0
Now we can do the same for the y terms:
y^2 - 6y = 0
y^2 - 6y + 9 - 9 = 0
(y-3)^2 - 9 = 0
(y-3)^2 = 9
y-3 = ±3
y = 6 or 0
So the center of the ellipse is (1, 3), the major axis is along the x-axis with a length of 2a = 2√(9/1) = 6, and the minor axis is along the y-axis with a length of 2b = 2√(1/9) = 2/3.
The vertices are the points on the major axis that are farthest from the center. Since the major axis is along the x-axis, the vertices will be (1±3, 3), or (4, 3) and (-2, 3).
To find two points on the minor axis, we can use the center and the length of the minor axis. Since the minor axis is along the y-axis, we can add or subtract the length of the minor axis from the y-coordinate of the center to find the two points. Therefore, the two points on the minor axis are (1, 3±1/3), or approximately (1, 10/3) and (1, 8/3).
Explanation: