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complex numbers- geometry of ellipse (1 Viewer)
- Thread starter zinc
- Start date
To find them algebraically:
The major axis is easy. It is just 10. NOTE: the semi major axis is 5
To find the minor axis:
First find the centre of the ellipse.
The two foci are (1 + 3i) and (9 + 3i) so the centre is the midpoint of the foci, so is (5 + 3i)
Now let z = x + iy
|(x-1) + i(y-3)| + |(x-9) + i(y-3)| = 10
Therefore rt[(x-1)2 + (y-3)2] + rt[(x-9)2 + (y-3)2] = 10
The top and bottom of the ellipse must lie on the line x=5 as the centre lies on this line so let x = 5
Therefore rt[16 + (y-3)2] + rt[16 + (y-3)2] = 10
rt[16 + (y-3)2] = 5
16 + (y-3)2 = 25
(y-3)2 = 9
Therefore y -3 = +-3
y = 6, 0
Therefore the minor axis is the distance between these points which is 6
The semi minor axis is 3
The major axis is easy. It is just 10. NOTE: the semi major axis is 5
To find the minor axis:
First find the centre of the ellipse.
The two foci are (1 + 3i) and (9 + 3i) so the centre is the midpoint of the foci, so is (5 + 3i)
Now let z = x + iy
|(x-1) + i(y-3)| + |(x-9) + i(y-3)| = 10
Therefore rt[(x-1)2 + (y-3)2] + rt[(x-9)2 + (y-3)2] = 10
The top and bottom of the ellipse must lie on the line x=5 as the centre lies on this line so let x = 5
Therefore rt[16 + (y-3)2] + rt[16 + (y-3)2] = 10
rt[16 + (y-3)2] = 5
16 + (y-3)2 = 25
(y-3)2 = 9
Therefore y -3 = +-3
y = 6, 0
Therefore the minor axis is the distance between these points which is 6
The semi minor axis is 3
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