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Given the ellipse x^2 + 4y^2 - 2x - 8y + 1 = 0 find: The Center C, Length of Major...
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You need to convert the given equation of ellipse into its standard form `(x-h)^2/a^2 + (y-k)^2/b^2 = 1` , hence, you need to complete the squares such that:
`(x^2 - 2x + 1) + (4y^2 - 8y + 4) - 1 - 4 + 1 = 0`
Reducing like members yields:
`(x - 1 )^2 + (2y - 2)^2 = 4`
Dividing by 4 yields:
`(x - 1 )^2/4 + 4(y - 1)^2/4 = 1`
`(x - 1 )^2/4 + (y - 1)^2/1 = 1`
Since the length of semi-major axis is `a = sqrt4 = 2` , yields that the length of major axis is `2a = 4` . Evaluating the length of minor axis yields `2b = 2` .
Since the center of ellipse is given by the point (h,k), you need to identify h and k, such that:
`h = 1, k = 1`
You need to identify `a^2 = 4` and `b^2 = 1` to evaluate the foci `(h-c,k)` and `(h+c,k)` , such that:
`c = sqrt(a^2 - b^2) => c = sqrt(4 - 1) => c = sqrt3`
`(h-c,k) = (1 - sqrt3,1)`
`(h+c,k) = (1 + sqrt3,1)`
You need to evaluate the distance center and the two foci, such that:
`d_(1,2) = sqrt((h+-c - h)^2 + (k - k)^2)`
`d_(1,2) = c = sqrt3`
Hence, evaluating the lengths of major and minor axis yields `2a = 4 ` and `2b = 1` , the center of ellipse is at `(1,1)` and the distance from center to foci is of `c = sqrt3.`
Posted by sciencesolve on June 30, 2013 at 4:23 PM (Answer #1)
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