Asked by confused
Find the equation of the ellipse with vertices at (-1,3) and (5,3)and length of minor 4.
Answers
Answered by
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
Answered by
confused
thanks!
Answered by
vero
8(16)
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