ellipse with vertices of (6,3) and (6,-3) and a focus at (6,-2). this is what i have so far...
(x-6)^2+y^2/9=1. How do I get the value under (x-6)^2
To find the value under (x-6)^2 in the equation of the ellipse, we need to determine the length of the major or minor axis. The formula for the equation of an ellipse in standard form is:
(x-h)^2/a^2 + (y-k)^2/b^2 = 1
Where:
- (h, k) represents the center of the ellipse.
- "a" is the length of the major axis, which determines the horizontal radius.
- "b" is the length of the minor axis, which determines the vertical radius.
From the given information, we know that the vertices of the ellipse are at (6, 3) and (6, -3), which means the center (h, k) is (6, 0). The distance between the center and each vertex represents the length of the major axis, so the major axis is 3 + 3 = 6 units long.
Therefore, in your equation (x-6)^2 + y^2/9 = 1, the value under (x-6)^2 is 9.