SOLUTION: What is the minor axis vertices written in (x1y1),(x2y2) form? (x+2)^2/4+(y+3)^2/16=1

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: What is the minor axis vertices written in (x1y1),(x2y2) form? (x+2)^2/4+(y+3)^2/16=1       Log On


   

Question 665233: What is the minor axis vertices written in (x1y1),(x2y2) form?
(x+2)^2/4+(y+3)^2/16=1

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
%28x%2B2%29%5E2%2F4%2B%28y%2B3%29%5E2%2F16=1

the major axis is vertical (because 4 is smaller than 16)
so, the sqrt%2816%29+=4
the center is at (-2,-3), so move over 4 vertically on both sides, and those are the vertices
so (-2,-3%2B4) = (-2,1)
and (-2, -3-4) = (-2,-7)
you can do the same for the minor axis, just moving horizontally up and down by 2