Question
Find the equation of the ellipse with vertices at (-1,3) and (5,3)and length of minor 4.
Answers
Reiny
first find the centre, which would be the midpoint of the vertices,
it would be (2,3)
the length of the minor axis is 4, so b = 2
sofar we would would have
(x-2)^2/a^2 + (y-3)^2/4 = 1
but (5,3) lies on it, so
(5-2)^2/a^2 + (3-3)^2/4 = 1
9/a^2 = 1
a^2 = 9
final equation
(x-2)^2/9 + (y-3)^2/4 = 1
it would be (2,3)
the length of the minor axis is 4, so b = 2
sofar we would would have
(x-2)^2/a^2 + (y-3)^2/4 = 1
but (5,3) lies on it, so
(5-2)^2/a^2 + (3-3)^2/4 = 1
9/a^2 = 1
a^2 = 9
final equation
(x-2)^2/9 + (y-3)^2/4 = 1
confused
thanks!
vero
8(16)