Question
My equation for an ellipse is:
(y+6)^2 /104 + (x-46)^2/155 = 1
Would this be the co-vertices?
(46, -6 +/- squareroot 104)
(y+6)^2 /104 + (x-46)^2/155 = 1
Would this be the co-vertices?
(46, -6 +/- squareroot 104)
Answers
Damon
Those are indeed the top and bottom of your ellipse.
of course you have them along the horizontal axis as well
x = 46 +/- sqrt 155 and y = -6
of course you have them along the horizontal axis as well
x = 46 +/- sqrt 155 and y = -6