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February 28, 2015

February 28, 2015

Posted by **confused** on Monday, October 5, 2009 at 9:01pm.

- Algebra -
**Reiny**, Monday, October 5, 2009 at 9:16pmfirst 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

- Algebra -
**confused**, Monday, October 5, 2009 at 9:19pmthanks!

- Algebra -
**vero**, Tuesday, October 6, 2009 at 9:50pm8(16)

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