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?
I answered this same question from you
http://www.jiskha.com/display.cgi?id=1212525764
ya but you make no sense to me...no offense
uioavxqm cxapwd sflxnjm kigezxj bgufn qxtvc yvkoeu
To find the value under (x-6)^2 in the equation of the ellipse, we need to determine the length of the major axis or the distance between the two vertices.
In this case, the vertices are given as (6, 3) and (6, -3). Since the x-coordinates of both vertices are the same (6), it means the major axis is vertical. Therefore, the length of the major axis is the difference between the y-coordinates of the two vertices.
Length of major axis = |3 - (-3)| = 6
The value under (x-6)^2 can be obtained by squaring half the length of the major axis. So, divide the length of the major axis by 2 and then square it.
Value under (x-6)^2 = (6/2)^2 = 3^2 = 9
Now, you can substitute this value into your equation:
(x-6)^2 + y^2/9 = 1
Thus, the equation of the given ellipse is correct:
(x-6)^2 + y^2/9 = 1