Question

Given the ellipse x^2 + 4y^2 - 2x - 8y + 1 = 0 find:
The Center C, Length of Major Axis, Length of Minor Axis, Distance from C to foci

Accepted Answer

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.

Math

Given the ellipse x^2 + 4y^2 - 2x - 8y + 1 = 0 find: The Center C, Length of Major Axis, Length of Minor Axis, Distance from C to foci

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